id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-3900 | Asymptotic Growth of Euler Phi Function | Let $\phi$ be the Euler $\phi$ function.
For any $\epsilon > 0$ and sufficiently large $n$:
:$n^{1 - \epsilon} < \map \phi n < n$ | It is clear that $\map \phi n < n$ for all $n$, so it is sufficient to prove that:
:$\ds \lim_{n \mathop \to \infty} \frac {n^{1 - \epsilon} } {\map \phi n} = 0$
By Multiplicative Function that Converges to Zero on Prime Powers it is sufficient to prove that:
:$\ds \lim_{p^k \mathop \to \infty} \frac {p^{k \paren {1 - ... | Let $\phi$ be the [[Definition:Euler Phi Function|Euler $\phi$ function]].
For any $\epsilon > 0$ and [[Definition:Sufficiently Large|sufficiently large]] $n$:
:$n^{1 - \epsilon} < \map \phi n < n$ | It is clear that $\map \phi n < n$ for all $n$, so it is sufficient to prove that:
:$\ds \lim_{n \mathop \to \infty} \frac {n^{1 - \epsilon} } {\map \phi n} = 0$
By [[Multiplicative Function that Converges to Zero on Prime Powers]] it is sufficient to prove that:
:$\ds \lim_{p^k \mathop \to \infty} \frac {p^{k \pare... | Asymptotic Growth of Euler Phi Function | https://proofwiki.org/wiki/Asymptotic_Growth_of_Euler_Phi_Function | https://proofwiki.org/wiki/Asymptotic_Growth_of_Euler_Phi_Function | [
"Analytic Number Theory",
"Euler Phi Function"
] | [
"Definition:Euler Phi Function",
"Definition:Sufficiently Large"
] | [
"Multiplicative Function that Converges to Zero on Prime Powers",
"Euler Phi Function of Prime Power",
"Category:Analytic Number Theory",
"Category:Euler Phi Function"
] |
proofwiki-3901 | Finite Complement Space is T1 | Let $T = \struct {S, \tau}$ be a finite complement topology on any set $S$ which contains at least two points.
Then $T$ is a $T_1$ space. | Let $x, y \in S$ such that $x \ne y$.
This is possible because $S$ contains at least two points.
By definition, both $\set x$ and $\set y$ are finite.
Hence by definition of finite complement topology:
:$\relcomp S {\set x}$ is an open set of $T$ containing $y$ but not $x$
:$\relcomp S {\set y}$ is an open set of $T$ c... | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on any [[Definition:Set|set]] $S$ which contains at least two [[Definition:Point of Set|points]].
Then $T$ is a [[Definition:T1 Space|$T_1$ space]]. | Let $x, y \in S$ such that $x \ne y$.
This is possible because $S$ contains at least two [[Definition:Point of Set|points]].
By definition, both $\set x$ and $\set y$ are [[Definition:Finite Set|finite]].
Hence by definition of [[Definition:Finite Complement Topology|finite complement topology]]:
:$\relcomp S {\set ... | Finite Complement Space is T1 | https://proofwiki.org/wiki/Finite_Complement_Space_is_T1 | https://proofwiki.org/wiki/Finite_Complement_Space_is_T1 | [
"Finite Complement Topologies",
"Examples of T1 Spaces"
] | [
"Definition:Finite Complement Topology",
"Definition:Set",
"Definition:Element",
"Definition:T1 Space"
] | [
"Definition:Element",
"Definition:Finite Set",
"Definition:Finite Complement Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:T1 Space"
] |
proofwiki-3902 | Finite Complement Space is Irreducible | Let $T = \struct {S, \tau}$ be a finite complement space on an infinite set $S$.
Then $T$ is irreducible. | Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$.
By definition, $S$ is infinite.
By definition, $\relcomp S {U_1}$ and $\relcomp S {U_2}$ are finite.
From Complement of Finite Subset of Infinite Set is Infinite, it follows that $U_1$ and $U_2$ are both infinite.
From Infinite Subset of Finite Complement Space Int... | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Space|finite complement space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is [[Definition:Irreducible Space|irreducible]]. | Let $U_1, U_2 \in \tau$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open sets]] of $T$.
By definition, $S$ is [[Definition:Infinite Set|infinite]].
By definition, $\relcomp S {U_1}$ and $\relcomp S {U_2}$ are [[Definition:Finite Set|finite]].
From [[Complement of Finite Subset of Infi... | Finite Complement Space is Irreducible | https://proofwiki.org/wiki/Finite_Complement_Space_is_Irreducible | https://proofwiki.org/wiki/Finite_Complement_Space_is_Irreducible | [
"Finite Complement Topologies",
"Examples of Irreducible Spaces"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Irreducible Space"
] | [
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Infinite Set",
"Definition:Finite Set",
"Complement of Finite Subset of Infinite Set is Infinite",
"Definition:Infinite Set",
"Infinite Subset of Finite Complement Space Intersects Open Sets",
"Definition:Set Intersection",
"Def... |
proofwiki-3903 | Infinite Subset of Finite Complement Space Intersects Open Sets | Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $H \subseteq S$ be an infinite subset of $S$.
Then the intersection of $H$ with any non-empty open set of $T$ is infinite. | Let $U \in \tau$ be any non-empty open set of $T$.
Then by definition of finite complement topology:
:$\relcomp S U$ is finite.
We have that:
:$H = H \cap \paren {U \cup \relcomp S U} = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$
{{AimForCont}} $H \cap U$ is finite.
From Intersection is Subset:
:$H \cap \relco... | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Let $H \subseteq S$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$.
Then the [[Definition:Set Intersection|intersection]] of $H$ wi... | Let $U \in \tau$ be any [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] of $T$.
Then by definition of [[Definition:Finite Complement Topology|finite complement topology]]:
:$\relcomp S U$ is [[Definition:Finite Set|finite]].
We have that:
:$H = H \cap \paren {U \cup \relcomp S U} =... | Infinite Subset of Finite Complement Space Intersects Open Sets | https://proofwiki.org/wiki/Infinite_Subset_of_Finite_Complement_Space_Intersects_Open_Sets | https://proofwiki.org/wiki/Infinite_Subset_of_Finite_Complement_Space_Intersects_Open_Sets | [
"Finite Complement Topologies"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Infinite Set",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Infinite Set"
] | [
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Finite Complement Topology",
"Definition:Finite Set",
"Definition:Finite Set",
"Intersection is Subset",
"Subset of Finite Set is Finite",
"Definition:Finite Set",
"Definition:Set Union",
"Definition:Finite Set",
"Union of F... |
proofwiki-3904 | Irreducible Hausdorff Space is Singleton | Let $T = \struct {S, \tau}$ be a non-empty topological space which is irreducible and Hausdorff.
Then $S$ is a singleton. | Suppose $S$ has exactly one element.
Then by definition $T = \struct {S, \tau}$ is the trivial topological space.
Hence, from Trivial Topological Space is Irreducible, $S$ is irreducible.
{{qed|lemma}}
Suppose $S$ has at least two distinct elements:
:$x, y \in S, x \ne y$
By definition of irreducible space:
:there are ... | Let $T = \struct {S, \tau}$ be a non-[[Definition:Empty Topological Space|empty]] [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]] and [[Definition:Hausdorff Space|Hausdorff]].
Then $S$ is a [[Definition:Singleton|singleton]]. | Suppose $S$ has exactly one [[Definition:Element|element]].
Then by definition $T = \struct {S, \tau}$ is the [[Definition:Trivial Topological Space|trivial topological space]].
Hence, from [[Trivial Topological Space is Irreducible]], $S$ is [[Definition:Irreducible Space|irreducible]].
{{qed|lemma}}
Suppose $S$ h... | Irreducible Hausdorff Space is Singleton | https://proofwiki.org/wiki/Irreducible_Hausdorff_Space_is_Singleton | https://proofwiki.org/wiki/Irreducible_Hausdorff_Space_is_Singleton | [
"Irreducible Spaces",
"Hausdorff Spaces"
] | [
"Definition:Empty Topological Space",
"Definition:Topological Space",
"Definition:Irreducible Space",
"Definition:T2 Space",
"Definition:Singleton"
] | [
"Definition:Element",
"Definition:Trivial Topological Space",
"Trivial Topological Space is Irreducible",
"Definition:Irreducible Space",
"Definition:Distinct/Plural",
"Definition:Irreducible Space",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Definition:T2 Space",
"Definition:Ele... |
proofwiki-3905 | Finite Complement Space is not T2 | Let $T = \struct {S, \tau}$ be the finite complement topology on an infinite set $S$.
Then $T$ is not a $T_2$ (Hausdorff) space. | We have:
:Finite Complement Space is Irreducible
:Irreducible Hausdorff Space is Singleton
{{qed}} | Let $T = \struct {S, \tau}$ be the [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is not a [[Definition:T2 Space|$T_2$ (Hausdorff) space]]. | We have:
:[[Finite Complement Space is Irreducible]]
:[[Irreducible Hausdorff Space is Singleton]]
{{qed}} | Finite Complement Space is not T2 | https://proofwiki.org/wiki/Finite_Complement_Space_is_not_T2 | https://proofwiki.org/wiki/Finite_Complement_Space_is_not_T2 | [
"Finite Complement Topologies",
"Examples of Hausdorff Spaces"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:T2 Space"
] | [
"Finite Complement Space is Irreducible",
"Irreducible Hausdorff Space is Singleton"
] |
proofwiki-3906 | Double Pointed Finite Complement Topology is Compact | Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is a compact space. | The Finite Complement Topology is Compact.
Also, a Finite Topological Space is Compact.
So $D$, defined as the indiscrete space on a doubleton, is also compact.
The result follows directly from Tychonoff's Theorem.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
Then $T \times D$ is a [[Definition:Compact Topological Space|compac... | The [[Subspace of Finite Complement Topology is Compact|Finite Complement Topology is Compact]].
Also, a [[Finite Topological Space is Compact]].
So $D$, defined as the [[Definition:Indiscrete Space|indiscrete space]] on a [[Definition:Doubleton|doubleton]], is also [[Definition:Compact Topological Space|compact]].
... | Double Pointed Finite Complement Topology is Compact | https://proofwiki.org/wiki/Double_Pointed_Finite_Complement_Topology_is_Compact | https://proofwiki.org/wiki/Double_Pointed_Finite_Complement_Topology_is_Compact | [
"Double Pointed Topologies",
"Finite Complement Topologies",
"Examples of Compact Topological Spaces"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Double Pointed Topology",
"Definition:Compact Topological Space"
] | [
"Subspace of Finite Complement Topology is Compact",
"Finite Topological Space is Compact",
"Definition:Indiscrete Topology",
"Definition:Doubleton",
"Definition:Compact Topological Space",
"Tychonoff's Theorem"
] |
proofwiki-3907 | Double Pointed Finite Complement Topology fulfils no Separation Axioms | Let $T = \struct {S, \tau}$ be a finite complement space on an infinite set $S$.
Let $T \times D$ be the double pointed topological space on $T$.
Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space:
=== Double Pointed Finite Complement Space is not $T_0$ ===
{{:Doub... | From Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is not a $T_0$ space.
:$T \times D$ is not a $T_1$ space.
:$T \times D$ is not a $T_2$ space.
Also from Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is a $T_3$ space {{iff}} $T$ is a $T_3$ space
:$T \times D$ is ... | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Space|finite complement space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Let $T \times D$ be the [[Definition:Double Pointed Space|double pointed topological space]] on $T$.
Then $T \times D$ is not a [[Definition:T0 Space|$T_0$ space]], [[De... | From [[Separation Axioms on Double Pointed Topology]], we have that:
:$T \times D$ is not a [[Definition:T0 Space|$T_0$ space]].
:$T \times D$ is not a [[Definition:T1 Space|$T_1$ space]].
:$T \times D$ is not a [[Definition:T2 Space|$T_2$ space]].
Also from [[Separation Axioms on Double Pointed Topology]], we have... | Double Pointed Finite Complement Topology fulfils no Separation Axioms | https://proofwiki.org/wiki/Double_Pointed_Finite_Complement_Topology_fulfils_no_Separation_Axioms | https://proofwiki.org/wiki/Double_Pointed_Finite_Complement_Topology_fulfils_no_Separation_Axioms | [
"Double Pointed Finite Complement Topology fulfils no Separation Axioms",
"Finite Complement Topologies",
"Double Pointed Topologies",
"Examples of Separation Axioms"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Double Pointed Topology",
"Definition:T0 Space",
"Definition:T1 Space",
"Definition:T2 Space",
"Definition:T3 Space",
"Definition:T4 Space",
"Definition:T5 Space",
"Double Pointed Finite Complement Space is not T0",
... | [
"Separation Axioms on Double Pointed Topology",
"Definition:T0 Space",
"Definition:T1 Space",
"Definition:T2 Space",
"Separation Axioms on Double Pointed Topology",
"Definition:T3 Space",
"Definition:T3 Space",
"Definition:T4 Space",
"Definition:T4 Space",
"Definition:T5 Space",
"Definition:T5 S... |
proofwiki-3908 | Vaughan's Identity | Let $\Lambda$ be von Mangoldt's function.
Let $\mu$ be the Möbius function.
Then for $y, z \ge 1$ and $n > z$:
:$\ds \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y}} \map \mu d \map \ln {\frac n d} - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z... | By Sum Over Divisors of von Mangoldt is Logarithm:
:$\ds \ln n = \sum_{d \mathop \divides n} \map \Lambda d$
Hence:
{{begin-eqn}}
{{eqn | l = \map \Lambda n
| r = \sum_{d \mathop \divides n} \map \mu d \map \ln {\frac n d}
| c = Möbius Inversion Formula
}}
{{eqn | r = \sum_{\substack {d \mathop \divides n \... | Let $\Lambda$ be [[Definition:Von Mangoldt Function|von Mangoldt's function]].
Let $\mu$ be the [[Definition:Möbius Function|Möbius function]].
Then for $y, z \ge 1$ and $n > z$:
:$\ds \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y}} \map \mu d \map \ln {\frac n d} - \mathop {\sum \sum}_{... | By [[Sum Over Divisors of von Mangoldt is Logarithm]]:
:$\ds \ln n = \sum_{d \mathop \divides n} \map \Lambda d$
Hence:
{{begin-eqn}}
{{eqn | l = \map \Lambda n
| r = \sum_{d \mathop \divides n} \map \mu d \map \ln {\frac n d}
| c = [[Möbius Inversion Formula]]
}}
{{eqn | r = \sum_{\substack {d \mathop \d... | Vaughan's Identity | https://proofwiki.org/wiki/Vaughan's_Identity | https://proofwiki.org/wiki/Vaughan's_Identity | [
"Analytic Number Theory"
] | [
"Definition:Von Mangoldt Function",
"Definition:Möbius Function",
"Definition:Divisor (Algebra)/Integer"
] | [
"Sum Over Divisors of von Mangoldt is Logarithm",
"Möbius Inversion Formula",
"Definition:Summation",
"Definition:Summation",
"Definition:Summation",
"Sum of Möbius Function over Divisors",
"Category:Analytic Number Theory"
] |
proofwiki-3909 | Sum Over Divisors of von Mangoldt is Logarithm | Let $\Lambda$ be von Mangoldt's function.
Then for $n \ge 1$:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$ | Let $n \ge 1$, and by the Fundamental Theorem of Arithmetic write $n = p_1^{e_1} \cdots p_k^{e_k}$ with $p_1, \ldots, p_k$ distinct primes and $e_1, \ldots, e_k > 0$.
Now $d \divides n$ if any only if $d = p_1^{f_1} \cdots p_k^{f_k}$ with $0 \le f_i \le e_i$ for $i = 1, \ldots, k$.
By the definition of $\Lambda$, for s... | Let $\Lambda$ be [[Definition:Von Mangoldt Function|von Mangoldt's function]].
Then for $n \ge 1$:
:$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$ | Let $n \ge 1$, and by the [[Fundamental Theorem of Arithmetic]] write $n = p_1^{e_1} \cdots p_k^{e_k}$ with $p_1, \ldots, p_k$ distinct [[Definition:Prime Number|primes]] and $e_1, \ldots, e_k > 0$.
Now $d \divides n$ if any only if $d = p_1^{f_1} \cdots p_k^{f_k}$ with $0 \le f_i \le e_i$ for $i = 1, \ldots, k$.
By ... | Sum Over Divisors of von Mangoldt is Logarithm | https://proofwiki.org/wiki/Sum_Over_Divisors_of_von_Mangoldt_is_Logarithm | https://proofwiki.org/wiki/Sum_Over_Divisors_of_von_Mangoldt_is_Logarithm | [
"Analytic Number Theory",
"Von Mangoldt Function"
] | [
"Definition:Von Mangoldt Function"
] | [
"Fundamental Theorem of Arithmetic",
"Definition:Prime Number",
"Definition:Unique",
"Sum of Logarithms",
"Category:Analytic Number Theory",
"Category:Von Mangoldt Function"
] |
proofwiki-3910 | Multiplicative Function that Converges to Zero on Prime Powers | Let $f$ be a multiplicative function such that:
:$\ds \lim_{p^k \mathop \to \infty} \map f {p^k} = 0$
where $p^k$ runs though ''all'' prime powers.
Then:
:$\ds \lim_{n \mathop \to \infty} \map f n = 0$
where $n$ runs through the integers. | By hypothesis, there exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > 1$.
Let $\ds A = \prod_{\size {\map f {p^k} } \mathop > 1} \size {\map f {p^k} }$.
Thus $A \ge 1$.
Let $0 < \dfrac \epsilon A$.
There exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > \dfrac... | Let $f$ be a [[Definition:Multiplicative Arithmetic Function|multiplicative function]] such that:
:$\ds \lim_{p^k \mathop \to \infty} \map f {p^k} = 0$
where $p^k$ runs though ''all'' [[Definition:Prime Power|prime powers]].
Then:
:$\ds \lim_{n \mathop \to \infty} \map f n = 0$
where $n$ runs through the [[Definitio... | By hypothesis, there exist only [[Definition:Finite Set|finitely many]] [[Definition:Prime Power|prime powers]] $p^k$ such that $\size {\map f {p^k} } > 1$.
Let $\ds A = \prod_{\size {\map f {p^k} } \mathop > 1} \size {\map f {p^k} }$.
Thus $A \ge 1$.
Let $0 < \dfrac \epsilon A$.
There exist only [[Definition:Fini... | Multiplicative Function that Converges to Zero on Prime Powers | https://proofwiki.org/wiki/Multiplicative_Function_that_Converges_to_Zero_on_Prime_Powers | https://proofwiki.org/wiki/Multiplicative_Function_that_Converges_to_Zero_on_Prime_Powers | [
"Analytic Number Theory",
"Multiplicative Functions"
] | [
"Definition:Multiplicative Arithmetic Function",
"Definition:Prime Power",
"Definition:Integer"
] | [
"Definition:Finite Set",
"Definition:Prime Power",
"Definition:Finite Set",
"Definition:Prime Power",
"Definition:Finite Set",
"Definition:Integer",
"Definition:Prime Power",
"Definition:Divisor (Algebra)/Integer",
"Definition:Sufficiently Large",
"Definition:Prime Power",
"Definition:Divisor (A... |
proofwiki-3911 | Finite T1 Space is Discrete | Let $S$ be a finite set.
Let $T = \struct {S, \tau}$ be a $T_1$ space.
Then $\tau$ is the discrete topology on $S$. | Let $T = \struct {S, \tau}$ be a $T_1$ space on a finite set $S$.
Let $U \subseteq S$ be any subset of $S$.
Let $H = \relcomp S U$ be the complement of $U$ relative to $S$.
Then by Relative Complement of Relative Complement we have that $U = \relcomp S H$.
We can write $H$ as:
:$\ds H = \bigcup_{x \mathop \in H} \set ... | Let $S$ be a [[Definition:Finite Set|finite set]].
Let $T = \struct {S, \tau}$ be a [[Definition:T1 Space|$T_1$ space]].
Then $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$. | Let $T = \struct {S, \tau}$ be a [[Definition:T1 Space|$T_1$ space]] on a [[Definition:Finite Set|finite set]] $S$.
Let $U \subseteq S$ be any [[Definition:Subset|subset]] of $S$.
Let $H = \relcomp S U$ be the [[Definition:Relative Complement|complement of $U$ relative to $S$]].
Then by [[Relative Complement of Re... | Finite T1 Space is Discrete/Proof 1 | https://proofwiki.org/wiki/Finite_T1_Space_is_Discrete | https://proofwiki.org/wiki/Finite_T1_Space_is_Discrete/Proof_1 | [
"Finite T1 Space is Discrete",
"Finite Topological Spaces",
"T1 Spaces",
"Discrete Topologies"
] | [
"Definition:Finite Set",
"Definition:T1 Space",
"Definition:Discrete Topology"
] | [
"Definition:T1 Space",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Relative Complement",
"Relative Complement of Relative Complement",
"Equivalence of Definitions of T1 Space",
"Definition:Closed Set/Topology",
"Definition:Finite Set",
"Definition:Set Union/Finite Union",
"Definition... |
proofwiki-3912 | Finite T1 Space is Discrete | Let $S$ be a finite set.
Let $T = \struct {S, \tau}$ be a $T_1$ space.
Then $\tau$ is the discrete topology on $S$. | We have from Finite Complement Topology is Minimal $T_1$ Topology that $\tau$ space is an expansion of a finite complement space.
But from the definition, any such finite complement space on a finite set is discrete.
But as the Discrete Topology is Finest Topology, $\tau$ must itself be the discrete topology on $S$.
{{... | Let $S$ be a [[Definition:Finite Set|finite set]].
Let $T = \struct {S, \tau}$ be a [[Definition:T1 Space|$T_1$ space]].
Then $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$. | We have from [[Finite Complement Topology is Minimal T1 Topology|Finite Complement Topology is Minimal $T_1$ Topology]] that $\tau$ space is an [[Definition:Expansion of Topology|expansion]] of a [[Definition:Finite Complement Space|finite complement space]].
But from the definition, any such [[Definition:Finite Compl... | Finite T1 Space is Discrete/Proof 2 | https://proofwiki.org/wiki/Finite_T1_Space_is_Discrete | https://proofwiki.org/wiki/Finite_T1_Space_is_Discrete/Proof_2 | [
"Finite T1 Space is Discrete",
"Finite Topological Spaces",
"T1 Spaces",
"Discrete Topologies"
] | [
"Definition:Finite Set",
"Definition:T1 Space",
"Definition:Discrete Topology"
] | [
"Finite Complement Topology is Minimal T1 Topology",
"Definition:Expansion of Topology",
"Definition:Finite Complement Topology",
"Definition:Finite Complement Topology",
"Definition:Finite Set",
"Definition:Discrete Topology",
"Discrete Topology is Finest Topology",
"Definition:Discrete Topology"
] |
proofwiki-3913 | Finite Complement Topology is Minimal T1 Topology | Let $T = \struct {S, \tau}$ be a finite complement space.
Let $\tau'$ be a topology on $S$ such that $T' = \struct {S, \tau'}$ is a $T_1$ space.
Then $\tau$ is comparable with $\tau'$ such that $\tau$ is coarser than $\tau'$.
That is, of all the topologies on $S$ fulfilling the $T_1$ separation axiom, the finite comple... | Let $T = \struct {S, \tau}$ be the finite complement space on $S$.
Let $U \in \tau$ be any open set of $T$.
Let $H = \relcomp S U$ be the complement of $U$ relative to $S$.
By definition of finite complement topology, $H \subseteq S$ is a finite subset of $S$.
Let $T' = \struct {S, \tau'}$ be any arbitrary $T_1$ space ... | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Space|finite complement space]].
Let $\tau'$ be a [[Definition:Topology|topology]] on $S$ such that $T' = \struct {S, \tau'}$ is a [[Definition:T1 Space|$T_1$ space]].
Then $\tau$ is [[Definition:Comparable Topologies|comparable]] with $\tau'$ such that... | Let $T = \struct {S, \tau}$ be the [[Definition:Finite Complement Space|finite complement space]] on $S$.
Let $U \in \tau$ be any [[Definition:Open Set (Topology)|open set]] of $T$.
Let $H = \relcomp S U$ be the [[Definition:Relative Complement|complement of $U$ relative to $S$]].
By definition of [[Definition:Fini... | Finite Complement Topology is Minimal T1 Topology | https://proofwiki.org/wiki/Finite_Complement_Topology_is_Minimal_T1_Topology | https://proofwiki.org/wiki/Finite_Complement_Topology_is_Minimal_T1_Topology | [
"Finite Complement Topologies",
"Examples of T1 Spaces",
"Examples of Coarser Topology"
] | [
"Definition:Finite Complement Topology",
"Definition:Topology",
"Definition:T1 Space",
"Definition:Comparable Topologies",
"Definition:Coarser Topology",
"Definition:Topology",
"Definition:T1 Space",
"Definition:Finite Complement Topology",
"Definition:Minimal/Set",
"Definition:Finite Complement T... | [
"Definition:Finite Complement Topology",
"Definition:Open Set/Topology",
"Definition:Relative Complement",
"Definition:Finite Complement Topology",
"Definition:Finite Set",
"Definition:Subset",
"Definition:T1 Space",
"Equivalence of Definitions of T1 Space",
"Definition:Closed Set/Topology",
"Defi... |
proofwiki-3914 | Finite Complement Space is Connected | Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a connected space. | We have that a Finite Complement Space is Irreducible.
The result follows from Irreducible Space is Connected.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:Connected Topological Space|connected space]]. | We have that a [[Finite Complement Space is Irreducible]].
The result follows from [[Irreducible Space is Connected]].
{{qed}} | Finite Complement Space is Connected | https://proofwiki.org/wiki/Finite_Complement_Space_is_Connected | https://proofwiki.org/wiki/Finite_Complement_Space_is_Connected | [
"Finite Complement Topologies",
"Examples of Connected Topological Spaces"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Connected Topological Space"
] | [
"Finite Complement Space is Irreducible",
"Irreducible Space is Connected"
] |
proofwiki-3915 | Finite Complement Space is Locally Connected | Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a locally connected space. | We have that a Finite Complement Space is Irreducible.
The result follows from Irreducible Space is Locally Connected.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:Locally Connected Space|locally connected space]]. | We have that a [[Finite Complement Space is Irreducible]].
The result follows from [[Irreducible Space is Locally Connected]].
{{qed}} | Finite Complement Space is Locally Connected | https://proofwiki.org/wiki/Finite_Complement_Space_is_Locally_Connected | https://proofwiki.org/wiki/Finite_Complement_Space_is_Locally_Connected | [
"Locally Connected Spaces",
"Finite Complement Topologies"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Locally Connected Space"
] | [
"Finite Complement Space is Irreducible",
"Irreducible Space is Locally Connected"
] |
proofwiki-3916 | Mapping Induces Partition on Domain | Let $f: S \to T$ be a mapping.
Let $F$ be defined as:
:$F = \set {\map {f^{-1} } x: x \in T}$
where $\map {f^{-1} } x$ is the preimage of $x$.
Then $F$ is a partition of $S$. | Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Then from Relation Induced by Mapping is Equivalence Relation, $\RR_f$ is an equivalence relation.
The result follows from Relation Partitions Set iff Equivalence.
{{qed}}
Category:Mapping Th... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $F$ be defined as:
:$F = \set {\map {f^{-1} } x: x \in T}$
where $\map {f^{-1} } x$ is the [[Definition:Preimage of Element under Mapping|preimage]] of $x$.
Then $F$ is a [[Definition:Partition (Set Theory)|partition]] of $S$. | Let $\RR_f \subseteq S \times S$ be the [[Definition:Equivalence Relation Induced by Mapping|relation induced by $f$]]:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Then from [[Relation Induced by Mapping is Equivalence Relation]], $\RR_f$ is an [[Definition:Equivalence Relation|equivalence relation... | Mapping Induces Partition on Domain | https://proofwiki.org/wiki/Mapping_Induces_Partition_on_Domain | https://proofwiki.org/wiki/Mapping_Induces_Partition_on_Domain | [
"Mapping Theory"
] | [
"Definition:Mapping",
"Definition:Preimage/Mapping/Element",
"Definition:Set Partition"
] | [
"Definition:Equivalence Relation Induced by Mapping",
"Relation Induced by Mapping is Equivalence Relation",
"Definition:Equivalence Relation",
"Relation Partitions Set iff Equivalence",
"Category:Mapping Theory"
] |
proofwiki-3917 | Countable Finite Complement Space is not Path-Connected | Let $T = \struct {S, \tau}$ be a countable finite complement topology.
Then $T$ is not path-connected. | {{AimForCont}} $f: \closedint 0 1 \to S$ is a path on $T$.
Then $f$ is by definition continuous.
By definition of the finite complement topology, for all $x \in S$, the set $\set x$ is closed.
Now consider the set:
:$F = \set {\map {f^{-1} } x: x \in S}$
From Continuity Defined from Closed Sets, each of the elements of... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Finite Complement Topology|countable finite complement topology]].
Then $T$ is not [[Definition:Path-Connected Space|path-connected]]. | {{AimForCont}} $f: \closedint 0 1 \to S$ is a [[Definition:Path (Topology)|path]] on $T$.
Then $f$ is by definition [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
By definition of the [[Definition:Finite Complement Topology|finite complement topology]], for all $x \in S$, the [[Definition:Set|se... | Countable Finite Complement Space is not Path-Connected | https://proofwiki.org/wiki/Countable_Finite_Complement_Space_is_not_Path-Connected | https://proofwiki.org/wiki/Countable_Finite_Complement_Space_is_not_Path-Connected | [
"Countable Finite Complement Topologies",
"Examples of Path-Connected Spaces"
] | [
"Definition:Finite Complement Topology/Countable",
"Definition:Path-Connected/Topological Space"
] | [
"Definition:Path (Topology)",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Finite Complement Topology",
"Definition:Set",
"Definition:Closed Set/Topology",
"Definition:Set",
"Continuity Defined from Closed Sets",
"Definition:Element",
"Definition:Closed Set/Topology",
"Mappin... |
proofwiki-3918 | Countable Finite Complement Space is not Locally Path-Connected | Let $T = \struct {S, \tau}$ be a countable finite complement topology.
Then $T$ is not locally path-connected. | Let $\BB$ be a basis for $T$.
Let $B \in \BB$.
{{AimForCont}} $f: \closedint 0 1 \to B$ is a path on $T$.
Then $f$ is by definition continuous.
Now consider the set:
:$F = \set {\map {f^{-1} } x: x \in S}$
From Continuity Defined from Closed Sets, each of the elements of $F$ is closed.
Also, from Mapping Induces Partit... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Finite Complement Topology|countable finite complement topology]].
Then $T$ is not [[Definition:Locally Path-Connected|locally path-connected]]. | Let $\BB$ be a [[Definition:Basis (Topology)|basis]] for $T$.
Let $B \in \BB$.
{{AimForCont}} $f: \closedint 0 1 \to B$ is a [[Definition:Path (Topology)|path]] on $T$.
Then $f$ is by definition [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
Now consider the set:
:$F = \set {\map {f^{-1} } x:... | Countable Finite Complement Space is not Locally Path-Connected | https://proofwiki.org/wiki/Countable_Finite_Complement_Space_is_not_Locally_Path-Connected | https://proofwiki.org/wiki/Countable_Finite_Complement_Space_is_not_Locally_Path-Connected | [
"Countable Finite Complement Topologies",
"Examples of Locally Path-Connected Spaces"
] | [
"Definition:Finite Complement Topology/Countable",
"Definition:Locally Path-Connected Space"
] | [
"Definition:Basis (Topology)",
"Definition:Path (Topology)",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Continuity Defined from Closed Sets",
"Definition:Element",
"Definition:Closed Set/Topology",
"Mapping Induces Partition on Domain",
"Definition:Element",
"Definition:Pairwise Disjoin... |
proofwiki-3919 | Injective Path-Connectedness in Uncountable Finite Complement Space | Let $T = \struct {S, \tau}$ be a finite complement topology on a uncountable set $S$.
If the Continuum Hypothesis is accepted as true, then:
:$T$ is injectively path-connected
:$T$ is locally injectively path-connected. | Let $a, b \in S$ such that $a \ne b$.
Let us assume the truth of the Continuum Hypothesis.
Then $a$ and $b$ are contained in a subset $X \subseteq S$ whose cardinality is the same as that of $\closedint 0 1$.
So by definition of cardinality we can set up a bijection $f: \closedint 0 1 \leftrightarrow X$ such that $\map... | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on a [[Definition:Uncountable Set|uncountable]] set $S$.
If the [[Continuum Hypothesis]] is accepted as [[Definition:True|true]], then:
:$T$ is [[Definition:Injectively Path-Connected Space|injectively path-connected... | Let $a, b \in S$ such that $a \ne b$.
Let us assume the truth of the [[Continuum Hypothesis]].
Then $a$ and $b$ are contained in a [[Definition:Subset|subset]] $X \subseteq S$ whose [[Definition:Cardinality|cardinality]] is the same as that of $\closedint 0 1$.
So by definition of [[Definition:Cardinality|cardinalit... | Injective Path-Connectedness in Uncountable Finite Complement Space | https://proofwiki.org/wiki/Injective_Path-Connectedness_in_Uncountable_Finite_Complement_Space | https://proofwiki.org/wiki/Injective_Path-Connectedness_in_Uncountable_Finite_Complement_Space | [
"Uncountable Finite Complement Topologies",
"Examples of Injectively Path-Connected Spaces",
"Examples of Locally Injectively Path-Connected Spaces",
"Continuum Hypothesis"
] | [
"Definition:Finite Complement Topology",
"Definition:Uncountable/Set",
"Continuum Hypothesis",
"Definition:True",
"Definition:Injectively Path-Connected/Topological Space",
"Definition:Locally Injectively Path-Connected Space"
] | [
"Continuum Hypothesis",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Bijection",
"Finite Complement Space is T1",
"Definition:T1 Space",
"Definition:Element",
"Definition:Closed Point",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Inj... |
proofwiki-3920 | Countable Complement Topology is Topology | Let $T = \struct {S, \tau}$ be a countable complement space.
Then $\tau$ is a topology on $T$. | By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially countable.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
| r = A \cap B
| c =
}}
{{eqn | ll= \leadsto
| l = \relcomp S H
| r = \relcomp S {A \... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Space|countable complement space]].
Then $\tau$ is a [[Definition:Topology|topology]] on $T$. | By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially [[Definition:Countable Set|countable]].
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
| r = A \cap B
| c =
}}
{{eqn | ll= \leadsto
| l = \relc... | Countable Complement Topology is Topology | https://proofwiki.org/wiki/Countable_Complement_Topology_is_Topology | https://proofwiki.org/wiki/Countable_Complement_Topology_is_Topology | [
"Countable Complement Topologies"
] | [
"Definition:Countable Complement Topology",
"Definition:Topology"
] | [
"Definition:Countable Set",
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection",
"Definition:Countable Set",
"Countable Union of Countable Sets is Countable",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Countable Set",
"Definition:Relative Complement",
... |
proofwiki-3921 | Countable Complement Topology is Expansion of Finite Complement Topology | Let $T = \struct {S, \tau}$ be the countable complement topology on an infinite set $S$.
Let $T' = \struct {S, \tau'}$ be the finite complement topology on the same infinite set $S$.
Then $\tau$ is an expansion of $\tau'$. | Let $U \in \tau', U \ne \O$.
Then $\relcomp S U$ is finite by definition of finite complement topology.
Then by definition of countable complement topology, we have that $U \in \tau$.
So $\tau' \subseteq \tau$ and hence the result by definition of expansion.
{{qed}} | Let $T = \struct {S, \tau}$ be the [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Let $T' = \struct {S, \tau'}$ be the [[Definition:Finite Complement Topology|finite complement topology]] on the same [[Definition:Infinite Set|infinite set]... | Let $U \in \tau', U \ne \O$.
Then $\relcomp S U$ is [[Definition:Finite Set|finite]] by definition of [[Definition:Finite Complement Topology|finite complement topology]].
Then by definition of [[Definition:Countable Complement Topology|countable complement topology]], we have that $U \in \tau$.
So $\tau' \subseteq... | Countable Complement Topology is Expansion of Finite Complement Topology | https://proofwiki.org/wiki/Countable_Complement_Topology_is_Expansion_of_Finite_Complement_Topology | https://proofwiki.org/wiki/Countable_Complement_Topology_is_Expansion_of_Finite_Complement_Topology | [
"Countable Complement Topologies",
"Finite Complement Topologies",
"Examples of Expansions of Topologies"
] | [
"Definition:Countable Complement Topology",
"Definition:Infinite Set",
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Expansion of Topology"
] | [
"Definition:Finite Set",
"Definition:Finite Complement Topology",
"Definition:Countable Complement Topology",
"Definition:Expansion of Topology"
] |
proofwiki-3922 | Countable Complement Space is T1 | Let $T = \struct {S, \tau}$ be a countable complement topology.
Then $T$ is a $T_1$ space. | We have that the Countable Complement Topology is Expansion of Finite Complement Topology.
We also have that a Finite Complement Space is $T_1$.
Then from Separation Properties Preserved under Expansion, we have that $T$ is a $T_1$ space.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]].
Then $T$ is a [[Definition:T1 Space|$T_1$ space]]. | We have that the [[Countable Complement Topology is Expansion of Finite Complement Topology]].
We also have that a [[Finite Complement Space is T1|Finite Complement Space is $T_1$]].
Then from [[Separation Properties Preserved under Expansion]], we have that $T$ is a [[Definition:T1 Space|$T_1$ space]].
{{qed}} | Countable Complement Space is T1 | https://proofwiki.org/wiki/Countable_Complement_Space_is_T1 | https://proofwiki.org/wiki/Countable_Complement_Space_is_T1 | [
"Countable Complement Topologies",
"Examples of T1 Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:T1 Space"
] | [
"Countable Complement Topology is Expansion of Finite Complement Topology",
"Finite Complement Space is T1",
"Separation Properties Preserved under Expansion",
"Definition:T1 Space"
] |
proofwiki-3923 | Uncountable Subset of Countable Complement Space Intersects Open Sets | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then the intersection of $H$ with an arbitrary non-empty open set of $T$ is uncountable. | Let $U \in \tau$ be any non-empty open set of $T$.
Then $\relcomp S U$ is countable.
Suppose $H \cap U = \O$.
Then from Intersection with Complement is Empty iff Subset it follows that $H \subseteq \relcomp S U$ and so $H$ is countable.
So if $H$ is uncountable it is bound to have a non-empty intersection with every op... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
Let $H \subseteq S$ be an [[Definition:Uncountable Set|uncountable]] [[Definition:Subset|subset]] of $S$.
Then the [[Definition:Set Intersection|inters... | Let $U \in \tau$ be any [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] of $T$.
Then $\relcomp S U$ is [[Definition:Countable Set|countable]].
Suppose $H \cap U = \O$.
Then from [[Intersection with Complement is Empty iff Subset]] it follows that $H \subseteq \relcomp S U$ and so $... | Uncountable Subset of Countable Complement Space Intersects Open Sets | https://proofwiki.org/wiki/Uncountable_Subset_of_Countable_Complement_Space_Intersects_Open_Sets | https://proofwiki.org/wiki/Uncountable_Subset_of_Countable_Complement_Space_Intersects_Open_Sets | [
"Countable Complement Topologies"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Uncountable/Set",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Arbitrary",
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Uncountable/Set"
] | [
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Countable Set",
"Intersection with Complement is Empty iff Subset",
"Definition:Countable Set",
"Definition:Uncountable/Set",
"Definition:Non-Empty Set",
"Definition:Set Intersection",
"Definition:Open Set/Topology",
"Definiti... |
proofwiki-3924 | Countable Complement Space is not T2 | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $T_2$ space. | We have:
:Countable Complement Space is Irreducible
:Irreducible Hausdorff Space is Singleton
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:T2 Space|$T_2$ space]]. | We have:
:[[Countable Complement Space is Irreducible]]
:[[Irreducible Hausdorff Space is Singleton]]
{{qed}} | Countable Complement Space is not T2 | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_T2 | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_T2 | [
"Countable Complement Topologies",
"Examples of Hausdorff Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:T2 Space"
] | [
"Countable Complement Space is Irreducible",
"Irreducible Hausdorff Space is Singleton"
] |
proofwiki-3925 | Countable Complement Space is Irreducible | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is an irreducible space. | Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$.
We have that both $U_1$ and $U_2$ are both uncountable if $S$ is.
From Uncountable Subset of Countable Complement Space Intersects Open Sets, they intersect each other.
Hence the result from definition of irreducible space.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is an [[Definition:Irreducible Space|irreducible space]]. | Let $U_1, U_2 \in \tau$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open sets]] of $T$.
We have that both $U_1$ and $U_2$ are both [[Definition:Uncountable Set|uncountable]] if $S$ is.
From [[Uncountable Subset of Countable Complement Space Intersects Open Sets]], they [[Definition:Set ... | Countable Complement Space is Irreducible | https://proofwiki.org/wiki/Countable_Complement_Space_is_Irreducible | https://proofwiki.org/wiki/Countable_Complement_Space_is_Irreducible | [
"Countable Complement Topologies",
"Examples of Irreducible Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Irreducible Space"
] | [
"Definition:Non-Empty Set",
"Definition:Open Set/Topology",
"Definition:Uncountable/Set",
"Uncountable Subset of Countable Complement Space Intersects Open Sets",
"Definition:Set Intersection",
"Definition:Irreducible Space"
] |
proofwiki-3926 | Higher-Aleph Complement Topology is Topology | Let $T = \struct {S, \tau}$ be an $\aleph_n$ complement space.
Then $\tau$ is a topology on $T$. | By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially finite.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
| r = A \cap B
| c =
}}
{{eqn | ll= \leadsto
| l = \relcomp S H
| r = \relcomp S {A \cap... | Let $T = \struct {S, \tau}$ be an [[Definition:Higher-Aleph Complement Space|$\aleph_n$ complement space]].
Then $\tau$ is a [[Definition:Topology|topology]] on $T$. | By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially [[Definition:Finite Set|finite]].
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
| r = A \cap B
| c =
}}
{{eqn | ll= \leadsto
| l = \relcomp S ... | Higher-Aleph Complement Topology is Topology | https://proofwiki.org/wiki/Higher-Aleph_Complement_Topology_is_Topology | https://proofwiki.org/wiki/Higher-Aleph_Complement_Topology_is_Topology | [
"Higher-Aleph Complement Topology"
] | [
"Definition:Higher-Aleph Complement Topology",
"Definition:Topology"
] | [
"Definition:Finite Set",
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection",
"Definition:Cardinality",
"Definition:Set Union",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Relative Complement",
"Definition:Cardinality",
"De Morgan's Laws (Set Theory)/Re... |
proofwiki-3927 | Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets | Let $\mathbb I = \closedint 0 1$ be the closed unit interval.
Then $\mathbb I$ cannot be expressed as the union of a countably infinite set of pairwise disjoint closed sets. | {{AimForCont}} $\ds \mathbb I = \bigcup_{i \mathop = 1}^\infty C_i$ where $\set {C_i}$ is a set of pairwise disjoint closed sets.
Let:
:$\ds B = \bigcup \partial C_i = \mathbb I \setminus \bigcup C_i^\circ$
where $\partial C_i$ is the boundary of $C_i$ and $C_i^\circ$ is the interior of $C_i$.
Let $J \subseteq \mathbb ... | Let $\mathbb I = \closedint 0 1$ be the [[Definition:Closed Unit Interval|closed unit interval]].
Then $\mathbb I$ cannot be expressed as the [[Definition:Set Union|union]] of a [[Definition:Countably Infinite Set|countably infinite set]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Closed Set (T... | {{AimForCont}} $\ds \mathbb I = \bigcup_{i \mathop = 1}^\infty C_i$ where $\set {C_i}$ is a [[Definition:Set|set]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Closed Set (Topology)|closed sets]].
Let:
:$\ds B = \bigcup \partial C_i = \mathbb I \setminus \bigcup C_i^\circ$
where $\partial C_i$ is... | Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets | https://proofwiki.org/wiki/Closed_Unit_Interval_is_not_Countably_Infinite_Union_of_Disjoint_Closed_Sets | https://proofwiki.org/wiki/Closed_Unit_Interval_is_not_Countably_Infinite_Union_of_Disjoint_Closed_Sets | [
"Closed Sets",
"Topology"
] | [
"Definition:Real Interval/Unit Interval/Closed",
"Definition:Set Union",
"Definition:Countably Infinite/Set",
"Definition:Pairwise Disjoint",
"Definition:Closed Set/Topology"
] | [
"Definition:Set",
"Definition:Pairwise Disjoint",
"Definition:Closed Set/Topology",
"Definition:Boundary (Topology)",
"Definition:Interior (Topology)",
"Definition:Subinterval",
"Definition:Meager Space/Non-Meager",
"Definition:Dense-in-itself",
"Definition:Real Interval/Open",
"Definition:Closed ... |
proofwiki-3928 | Compact Sets in Countable Complement Space | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then the compact sets of $T$ are exactly the finite subsets of $S$. | From Finite Topological Space is Compact, any finite subset of $S$ is compact.
{{AimForCont}} $H \subseteq S$ is an infinite compact set.
Take a (countably) infinite sequence $\sequence {a_n}_{n \mathop \ge 0}$ of distinct elements of $H$.
Consider the open sets:
:$V_m := S \setminus \set {a_{m + n} }_{n \mathop \ge 0}... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then the [[Definition:Compact Topological Subspace|compact sets]] of $T$ are exactly the [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] o... | From [[Finite Topological Space is Compact]], any [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$ is [[Definition:Compact Topological Subspace|compact]].
{{AimForCont}} $H \subseteq S$ is an [[Definition:Infinite Set|infinite]] [[Definition:Compact Topological Subspace|compact set]].
Take a [[De... | Compact Sets in Countable Complement Space | https://proofwiki.org/wiki/Compact_Sets_in_Countable_Complement_Space | https://proofwiki.org/wiki/Compact_Sets_in_Countable_Complement_Space | [
"Countable Complement Topologies",
"Examples of Compact Topological Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Compact Topological Space/Subspace",
"Definition:Finite Set",
"Definition:Subset"
] | [
"Finite Topological Space is Compact",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Compact Topological Space/Subspace",
"Definition:Infinite Set",
"Definition:Compact Topological Space/Subspace",
"Definition:Sequence/Infinite Sequence",
"Definition:Distinct/Plural",
"Definition:Open Se... |
proofwiki-3929 | Countable Complement Space is not Sigma-Compact | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $\sigma$-compact space. | From Compact Sets in Countable Complement Space, the only compact sets in $T$ are finite.
A countable union of finite sets can not be an uncountable set.
Hence the result by definition of $\sigma$-compact space.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:Sigma-Compact Space|$\sigma$-compact space]]. | From [[Compact Sets in Countable Complement Space]], the only [[Definition:Compact Topological Subspace|compact sets]] in $T$ are [[Definition:Finite Set|finite]].
A [[Definition:Countable Union|countable union]] of [[Definition:Finite Set|finite sets]] can not be an [[Definition:Uncountable Set|uncountable set]].
He... | Countable Complement Space is not Sigma-Compact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Sigma-Compact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Sigma-Compact | [
"Countable Complement Topologies",
"Examples of Sigma-Compact Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Sigma-Compact Space"
] | [
"Compact Sets in Countable Complement Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Finite Set",
"Definition:Set Union/Countable Union",
"Definition:Finite Set",
"Definition:Uncountable/Set",
"Definition:Sigma-Compact Space"
] |
proofwiki-3930 | Countable Complement Space is not Countably Compact | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a countably compact space. | Consider $U \in \tau$.
By definition of countable complement topology, $\relcomp S U$ is countably infinite.
Then for each $x \in \relcomp S U$, $\relcomp S U \setminus \set x$ is countably infinite.
Then for each $x \in \relcomp S U$, $\relcomp S {\relcomp S U \setminus \set x} = U \cup \set x$ is open.
Hence $\set {U... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:Countably Compact Space|countably compact space]]. | Consider $U \in \tau$.
By definition of [[Definition:Countable Complement Topology|countable complement topology]], $\relcomp S U$ is [[Definition:Countably Infinite Set|countably infinite]].
Then for each $x \in \relcomp S U$, $\relcomp S U \setminus \set x$ is [[Definition:Countably Infinite Set|countably infinite]... | Countable Complement Space is not Countably Compact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Countably_Compact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Countably_Compact | [
"Countable Complement Topologies",
"Examples of Countably Compact Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Countably Compact Space"
] | [
"Definition:Countable Complement Topology",
"Definition:Countably Infinite/Set",
"Definition:Countably Infinite/Set",
"Definition:Open Set/Topology",
"Definition:Open Cover",
"Definition:Cover of Set",
"Definition:Countable Set",
"Definition:Set Equivalence",
"Definition:Countably Compact Space",
... |
proofwiki-3931 | Countable Complement Space is Lindelöf | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a Lindelöf space. | By definition, $T$ is a Lindelöf space {{iff}} every open cover of $S$ has a countable subcover.
Let $\CC$ be an open cover of $T$.
Let $U \in \CC$ be an arbitrary set in $\CC$.
$U$ covers all but a countable number of points of $T$.
So for each of those points we pick an element of $\CC$ which covers each of those poi... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]]. | By definition, $T$ is a [[Definition:Lindelöf Space|Lindelöf space]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
Let $U \in \CC$ be an [[Definition:Arbitrary|arbitrary]] [[Definition:... | Countable Complement Space is Lindelöf | https://proofwiki.org/wiki/Countable_Complement_Space_is_Lindelöf | https://proofwiki.org/wiki/Countable_Complement_Space_is_Lindelöf | [
"Countable Complement Topologies",
"Examples of Lindelöf Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Lindelöf Space"
] | [
"Definition:Lindelöf Space",
"Definition:Open Cover",
"Definition:Subcover/Countable",
"Definition:Open Cover",
"Definition:Arbitrary",
"Definition:Set",
"Definition:Cover of Set",
"Definition:Countable Set",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Cover o... |
proofwiki-3932 | Countable Complement Space is not First-Countable | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a first-countable space. | {{AimForCont}} some $x \in S$ has a countable local basis.
That means:
:there exists a countable set of sets $\BB_x \subseteq \tau$
such that:
:$\forall B \in \BB_x: x \in B$
and such that:
:every open neighborhood of $x$ contains some $B \in \BB_x$.
So:
{{begin-eqn}}
{{eqn | l = \bigcap \BB_x
| r = \set x
... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:First-Countable Space|first-countable space]]. | {{AimForCont}} some $x \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
That means:
:there exists a [[Definition:Countable Set|countable]] [[Definition:Set of Sets|set of sets]] $\BB_x \subseteq \tau$
such that:
:$\forall B \in \BB_x: x \in B$
and such that:
:every [[Definit... | Countable Complement Space is not First-Countable | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_First-Countable | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_First-Countable | [
"Countable Complement Topologies",
"Examples of First-Countable Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:First-Countable Space"
] | [
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Open Neighborhood/Point",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Definition:Countable Set",
"Countable Union of Countable Sets is ... |
proofwiki-3933 | Closed Form for Number of Derangements on Finite Set | The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:
{{begin-eqn}}
{{eqn | l = D_n
| r = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }
| c =
}}
{{eqn | r = n! \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k} {k!}
| c =
}}
{{... | Let $s_i$ be the $i$th element of set $S$.
Begin by defining set $A_m$, which is all of the permutations of $S$ which fixes $S_m$.
Then the number of permutations, $W$, with ''at least'' one element fixed, $m$, is:
:$\ds W = \size {\bigcup_{m \mathop = 1}^n A_m}$
Applying the Inclusion-Exclusion Principle:
{{begin-eqn}... | The number of [[Definition:Derangement|derangements]] $D_n$ on a [[Definition:Finite Set|finite set]] $S$ of [[Definition:Cardinality|cardinality]] $n$ is:
{{begin-eqn}}
{{eqn | l = D_n
| r = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }
| c =
}}
{{eq... | Let $s_i$ be the $i$th element of set $S$.
Begin by defining set $A_m$, which is all of the [[Definition:Permutation|permutations]] of $S$ which [[Definition:Fixed Element under Permutation|fixes]] $S_m$.
Then the number of [[Definition:Permutation|permutations]], $W$, with ''at least'' one [[Definition:Element|eleme... | Closed Form for Number of Derangements on Finite Set/Proof 1 | https://proofwiki.org/wiki/Closed_Form_for_Number_of_Derangements_on_Finite_Set | https://proofwiki.org/wiki/Closed_Form_for_Number_of_Derangements_on_Finite_Set/Proof_1 | [
"Closed Form for Number of Derangements on Finite Set",
"Number of Derangements on Finite Set",
"Derangements",
"Counting Arguments",
"Closed Forms"
] | [
"Definition:Derangement",
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Subfactorial"
] | [
"Definition:Permutation",
"Definition:Fixed Element under Permutation",
"Definition:Permutation",
"Definition:Element",
"Definition:Fixed Element under Permutation",
"Inclusion-Exclusion Principle",
"Definition:Permutation",
"Definition:Fixed Element under Permutation",
"Definition:Permutation",
"... |
proofwiki-3934 | Closed Form for Number of Derangements on Finite Set | The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:
{{begin-eqn}}
{{eqn | l = D_n
| r = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }
| c =
}}
{{eqn | r = n! \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k} {k!}
| c =
}}
{{... | A derangement is a permutation that fixes no elements.
So, to compute the total number of derangements, we compute the number of permutations that fix elements, that is leave elements in the same place.
Let $\pi$ be a permutation of $S$.
Let $x, y \in S$.
Let $i$ be a position in a given permutation of set $S$, where $... | The number of [[Definition:Derangement|derangements]] $D_n$ on a [[Definition:Finite Set|finite set]] $S$ of [[Definition:Cardinality|cardinality]] $n$ is:
{{begin-eqn}}
{{eqn | l = D_n
| r = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }
| c =
}}
{{eq... | A [[Definition:Derangement|derangement]] is a [[Definition:Permutation|permutation]] that [[Definition:Fixed Element|fixes no elements]].
So, to compute the total number of [[Definition:Derangement|derangements]], we compute the number of [[Definition:Permutation|permutations]] that [[Definition:Fixed Element|fix elem... | Closed Form for Number of Derangements on Finite Set/Proof 2 | https://proofwiki.org/wiki/Closed_Form_for_Number_of_Derangements_on_Finite_Set | https://proofwiki.org/wiki/Closed_Form_for_Number_of_Derangements_on_Finite_Set/Proof_2 | [
"Closed Form for Number of Derangements on Finite Set",
"Number of Derangements on Finite Set",
"Derangements",
"Counting Arguments",
"Closed Forms"
] | [
"Definition:Derangement",
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Subfactorial"
] | [
"Definition:Derangement",
"Definition:Permutation",
"Definition:Fixed Element",
"Definition:Derangement",
"Definition:Permutation",
"Definition:Fixed Element",
"Definition:Element",
"Definition:Permutation",
"Definition:Set",
"Definition:Property",
"Definition:Permutation",
"Definition:Permuta... |
proofwiki-3935 | Limit Points of Countable Complement Space | Let $T = \struct {S, \tau}$ be a countable complement space.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then every point of $S$ is a limit point of $H$.
Let $H \subseteq S$ be an countable or finite subset of $S$.
Then $H$ contains all its limit points. | Let $U \in \tau$ be any open set of $T$.
Let $H \subseteq S$ be an uncountable.
From Uncountable Subset of Countable Complement Space Intersects Open Sets, we have that $U \cap H$ is uncountable {{iff}} $H$ is uncountable.
Let $x \in S$.
Then every open set $U$ in $T$ such that $x \in U$ also contains an uncountable nu... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Space|countable complement space]].
Let $H \subseteq S$ be an [[Definition:Uncountable Set|uncountable]] [[Definition:Subset|subset]] of $S$.
Then every point of $S$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
Let $H \subseteq S$ be... | Let $U \in \tau$ be any [[Definition:Open Set (Topology)|open set]] of $T$.
Let $H \subseteq S$ be an [[Definition:Uncountable Set|uncountable]].
From [[Uncountable Subset of Countable Complement Space Intersects Open Sets]], we have that $U \cap H$ is [[Definition:Uncountable Set|uncountable]] {{iff}} $H$ is [[Defin... | Limit Points of Countable Complement Space | https://proofwiki.org/wiki/Limit_Points_of_Countable_Complement_Space | https://proofwiki.org/wiki/Limit_Points_of_Countable_Complement_Space | [
"Countable Complement Topologies",
"Examples of Limit Points"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Subset",
"Definition:Limit Point/Topology/Set",
"Definition:Countable Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Limit Point/Topology/Set"
] | [
"Definition:Open Set/Topology",
"Definition:Uncountable/Set",
"Uncountable Subset of Countable Complement Space Intersects Open Sets",
"Definition:Uncountable/Set",
"Definition:Uncountable/Set",
"Definition:Open Set/Topology",
"Definition:Uncountable/Set",
"Definition:Limit Point/Topology/Set",
"Def... |
proofwiki-3936 | Countable Complement Space is not Separable | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a separable space. | Let $U$ be a countable subset of $S$.
By the definition of $T$, $U$ is closed.
As $U$ is countable but $S$ is uncountable:
:$U \subsetneq S$
and so:
:$U \ne S$
From Closed Set Equals its Closure:
:$U^- = U$
where $U^-$ is the closure of $U$.
Thus:
:$U^- \ne S$
So by definition, $U$ is not everywhere dense in $T$.
As $U... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:Separable Space|separable space]]. | Let $U$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$.
By the definition of $T$, $U$ is [[Definition:Closed Set (Topology)|closed]].
As $U$ is [[Definition:Countable Set|countable]] but $S$ is [[Definition:Uncountable Set|uncountable]]:
:$U \subsetneq S$
and so:
:$U \ne S$
From [[Cl... | Countable Complement Space is not Separable | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Separable | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Separable | [
"Countable Complement Topologies",
"Examples of Separable Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Separable Space"
] | [
"Definition:Countable Set",
"Definition:Subset",
"Definition:Closed Set/Topology",
"Definition:Countable Set",
"Definition:Uncountable/Set",
"Set is Closed iff Equals Topological Closure",
"Definition:Closure (Topology)",
"Definition:Everywhere Dense",
"Definition:Countable Set",
"Definition:Every... |
proofwiki-3937 | Countable Complement Space Satisfies Countable Chain Condition | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ satisfies the countable chain condition. | We have that a Countable Complement Space is Irreducible.
Therefore there is no disjoint set of open sets of $T$.
So the result holds vacuously.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ satisfies the [[Definition:Countable Chain Condition|countable chain condition]]. | We have that a [[Countable Complement Space is Irreducible]].
Therefore there is no [[Definition:Disjoint Sets|disjoint]] set of [[Definition:Open Set (Topology)|open sets]] of $T$.
So the result holds [[Definition:Vacuous Truth|vacuously]].
{{qed}} | Countable Complement Space Satisfies Countable Chain Condition | https://proofwiki.org/wiki/Countable_Complement_Space_Satisfies_Countable_Chain_Condition | https://proofwiki.org/wiki/Countable_Complement_Space_Satisfies_Countable_Chain_Condition | [
"Countable Complement Topologies",
"Examples of Countable Chain Condition"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Countable Chain Condition"
] | [
"Countable Complement Space is Irreducible",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Definition:Vacuous Truth"
] |
proofwiki-3938 | Countable Complement Space is Connected | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a connected space. | We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Connected.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is a [[Definition:Connected Topological Space|connected space]]. | We have that a [[Countable Complement Space is Irreducible]].
The result follows from [[Irreducible Space is Connected]].
{{qed}} | Countable Complement Space is Connected | https://proofwiki.org/wiki/Countable_Complement_Space_is_Connected | https://proofwiki.org/wiki/Countable_Complement_Space_is_Connected | [
"Countable Complement Topologies",
"Examples of Connected Topological Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Connected Topological Space"
] | [
"Countable Complement Space is Irreducible",
"Irreducible Space is Connected"
] |
proofwiki-3939 | Countable Complement Space is Locally Connected | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a locally connected space. | We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Locally Connected.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is a [[Definition:Locally Connected Space|locally connected space]]. | We have that a [[Countable Complement Space is Irreducible]].
The result follows from [[Irreducible Space is Locally Connected]].
{{qed}} | Countable Complement Space is Locally Connected | https://proofwiki.org/wiki/Countable_Complement_Space_is_Locally_Connected | https://proofwiki.org/wiki/Countable_Complement_Space_is_Locally_Connected | [
"Countable Complement Topologies",
"Examples of Locally Connected Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Locally Connected Space"
] | [
"Countable Complement Space is Irreducible",
"Irreducible Space is Locally Connected"
] |
proofwiki-3940 | Countable Complement Space is Pseudocompact | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is pseudocompact. | We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Pseudocompact.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is [[Definition:Pseudocompact Space|pseudocompact]]. | We have that a [[Countable Complement Space is Irreducible]].
The result follows from [[Irreducible Space is Pseudocompact]].
{{qed}} | Countable Complement Space is Pseudocompact | https://proofwiki.org/wiki/Countable_Complement_Space_is_Pseudocompact | https://proofwiki.org/wiki/Countable_Complement_Space_is_Pseudocompact | [
"Countable Complement Topologies",
"Examples of Pseudocompact Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Pseudocompact Space"
] | [
"Countable Complement Space is Irreducible",
"Irreducible Space is Pseudocompact"
] |
proofwiki-3941 | Countable Complement Space is not Countably Metacompact | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not countably metacompact. | From Uncountable Subset of Countable Complement Space Intersects Open Sets, the intersection of any open sets is uncountable.
So for any open cover $\CC$ of $T$, every point is in an infinite number of open sets of $T$.
So no refinement of any open cover of $T$ can be point finite.
Hence the result by definition of cou... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not [[Definition:Countably Metacompact Space|countably metacompact]]. | From [[Uncountable Subset of Countable Complement Space Intersects Open Sets]], the [[Definition:Set Intersection|intersection]] of any [[Definition:Open Set (Topology)|open sets]] is [[Definition:Uncountable Set|uncountable]].
So for any [[Definition:Open Cover|open cover]] $\CC$ of $T$, every point is in an [[Defini... | Countable Complement Space is not Countably Metacompact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Countably_Metacompact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Countably_Metacompact | [
"Countable Complement Topologies",
"Examples of Countably Metacompact Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Countably Metacompact Space"
] | [
"Uncountable Subset of Countable Complement Space Intersects Open Sets",
"Definition:Set Intersection",
"Definition:Open Set/Topology",
"Definition:Uncountable/Set",
"Definition:Open Cover",
"Definition:Infinite Set",
"Definition:Open Set/Topology",
"Definition:Refinement of Cover",
"Definition:Open... |
proofwiki-3942 | Countable Complement Space is not Weakly Countably Compact | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not weakly countably compact. | By definition, $T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$.
Let $H \subseteq S$ be a countable set.
From Limit Points of Countable Complement Space, it contains all its limit points.
Let $x \in H$.
$\relcomp S {\relcomp S H \cup \set x} = H \setminus \set x$ is countabl... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not [[Definition:Weakly Countably Compact Space|weakly countably compact]]. | By definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} every [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ has a [[Definition:Limit Point of Set|limit point]] in $S$.
Let $H \subseteq S$ be a [[Definition:Countable Set|countable set]].
From [[Li... | Countable Complement Space is not Weakly Countably Compact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Weakly_Countably_Compact | https://proofwiki.org/wiki/Countable_Complement_Space_is_not_Weakly_Countably_Compact | [
"Countable Complement Topologies",
"Examples of Weakly Countably Compact Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Weakly Countably Compact Space"
] | [
"Definition:Weakly Countably Compact Space",
"Definition:Infinite Set",
"Definition:Subset",
"Definition:Limit Point/Topology/Set",
"Definition:Countable Set",
"Limit Points of Countable Complement Space",
"Definition:Limit Point/Topology/Set",
"Definition:Countable Set",
"Definition:Open Neighborho... |
proofwiki-3943 | Saturated Implies Universal | Let $\kappa$ be an infinite cardinal.
Let $\MM$ be a model of the $\LL$-theory $T$.
If $\MM$ is $\kappa$-saturated, then it is $\kappa^+$-universal, where $\kappa^+$ is the successor cardinal of $\kappa$. | The idea of the proof is that $\MM$ being saturated means that when we want to define an elementary map $\NN \to \MM$, we can find an image $y \in \MM$ for an element $x \in \NN$ by realizing the type made up of the formulas that such a $y$ would need to satisfy.
Let $\NN$ be a model of $T$ with universe of cardinality... | Let $\kappa$ be an [[Definition:Infinite|infinite]] [[Definition:Cardinal|cardinal]].
Let $\MM$ be a [[Definition:Model (Logic)|model]] of the $\LL$-[[Definition:Theory|theory]] $T$.
If $\MM$ is $\kappa$-[[Definition:Saturated Model|saturated]], then it is $\kappa^+$-[[Definition:Universal Model|universal]], where $\... | The idea of the proof is that $\MM$ being saturated means that when we want to define an elementary map $\NN \to \MM$, we can find an image $y \in \MM$ for an element $x \in \NN$ by realizing the type made up of the formulas that such a $y$ would need to satisfy.
Let $\NN$ be a [[Definition:Model (Logic)|model]] of $... | Saturated Implies Universal | https://proofwiki.org/wiki/Saturated_Implies_Universal | https://proofwiki.org/wiki/Saturated_Implies_Universal | [
"Model Theory for Predicate Logic"
] | [
"Definition:Infinite",
"Definition:Cardinal",
"Definition:Model (Logic)",
"Definition:Theory",
"Definition:Saturated Model",
"Definition:Universal Model",
"Definition:Successor Cardinal"
] | [
"Definition:Model (Logic)",
"Definition:Elementary Embedding",
"Definition:Limit Ordinal",
"Definition:Successor Ordinal",
"Category:Model Theory for Predicate Logic"
] |
proofwiki-3944 | Double Pointed Countable Complement Topology fulfils no Separation Axioms | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space. | From Separation Axioms on Double Pointed Topology, we have that:
{{begin-itemize}}
{{item||$T \times D$ is not a $T_0$ space}}
{{item||$T \times D$ is not a $T_1$ space}}
{{item||$T \times D$ is not a $T_2$ space.}}
{{end-itemize}}
Also from Separation Axioms on Double Pointed Topology, we have that:
{{begin-itemize}}
... | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
Then $T \times D$ is not a [[Definition:T0 Space|$T_0$ s... | From [[Separation Axioms on Double Pointed Topology]], we have that:
{{begin-itemize}}
{{item||$T \times D$ is not a [[Definition:T0 Space|$T_0$ space]]}}
{{item||$T \times D$ is not a [[Definition:T1 Space|$T_1$ space]]}}
{{item||$T \times D$ is not a [[Definition:Hausdorff Space|$T_2$ space]].}}
{{end-itemize}}
Al... | Double Pointed Countable Complement Topology fulfils no Separation Axioms | https://proofwiki.org/wiki/Double_Pointed_Countable_Complement_Topology_fulfils_no_Separation_Axioms | https://proofwiki.org/wiki/Double_Pointed_Countable_Complement_Topology_fulfils_no_Separation_Axioms | [
"Double Pointed Topologies",
"Countable Complement Topologies",
"Examples of Separation Axioms"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Double Pointed Topology",
"Definition:T0 Space",
"Definition:T1 Space",
"Definition:T2 Space",
"Definition:T3 Space",
"Definition:T4 Space",
"Definition:T5 Space"
] | [
"Separation Axioms on Double Pointed Topology",
"Definition:T0 Space",
"Definition:T1 Space",
"Definition:T2 Space",
"Separation Axioms on Double Pointed Topology",
"Definition:T3 Space",
"Definition:T3 Space",
"Definition:T4 Space",
"Definition:T4 Space",
"Definition:T5 Space",
"Definition:T5 S... |
proofwiki-3945 | Double Pointed Countable Complement Topology is Weakly Countably Compact | Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is weakly countably compact. | By definition, $T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$.
Let $D = \set {0, 1}$.
Let $\tuple {p, 0}$ belong to some infinite $A \subseteq S$.
Then its twin $\tuple {p, 1}$ is a limit point of $A$.
Hence the result by definition of weakly countably compact.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
Then $T \times D$ is [[Definition:Weakly Countably Compa... | By definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} every [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ has a [[Definition:Limit Point of Set|limit point]] in $S$.
Let $D = \set {0, 1}$.
Let $\tuple {p, 0}$ belong to some [[Definition:Infinit... | Double Pointed Countable Complement Topology is Weakly Countably Compact | https://proofwiki.org/wiki/Double_Pointed_Countable_Complement_Topology_is_Weakly_Countably_Compact | https://proofwiki.org/wiki/Double_Pointed_Countable_Complement_Topology_is_Weakly_Countably_Compact | [
"Double Pointed Topologies",
"Countable Complement Topologies",
"Examples of Weakly Countably Compact Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Double Pointed Topology",
"Definition:Weakly Countably Compact Space"
] | [
"Definition:Weakly Countably Compact Space",
"Definition:Infinite Set",
"Definition:Subset",
"Definition:Limit Point/Topology/Set",
"Definition:Infinite Set",
"Definition:Limit Point/Topology/Set",
"Definition:Weakly Countably Compact Space"
] |
proofwiki-3946 | Open Extension of Double Pointed Countable Complement Topology is T4 | Let $T = \struct {S, \tau_S}$ be a countable complement topology on an uncountable set $S$.
Let $D = \struct {\set {0, 1}, \tau_D}$ be the indiscrete topology on two points.
Let $T \times D$ be the double pointed topology on $T$.
Let $\paren {T \times D}^*_{\bar p}$ be the open extension topology on $S \times \set {0, ... | From Open Extension Topology is T4 we have that $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space.
From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T \times D$ is not a $T_0$ space or a $T_5$ space.
From Condition for Open Extension Space to be $T_0$, it follows that $\paren... | Let $T = \struct {S, \tau_S}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Let $D = \struct {\set {0, 1}, \tau_D}$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on two points.
Let $T \times D$ be the [[Definit... | From [[Open Extension Topology is T4]] we have that $\paren {T \times D}^*_{\bar p}$ is a [[Definition:T4 Space|$T_4$ space]].
From [[Double Pointed Countable Complement Topology fulfils no Separation Axioms]], we have that $T \times D$ is not a [[Definition:T0 Space|$T_0$ space]] or a [[Definition:T5 Space|$T_5$ spa... | Open Extension of Double Pointed Countable Complement Topology is T4 | https://proofwiki.org/wiki/Open_Extension_of_Double_Pointed_Countable_Complement_Topology_is_T4 | https://proofwiki.org/wiki/Open_Extension_of_Double_Pointed_Countable_Complement_Topology_is_T4 | [
"Open Extension Topologies",
"Double Pointed Topologies",
"Countable Complement Topologies",
"Examples of T4 Spaces"
] | [
"Definition:Countable Complement Topology",
"Definition:Uncountable/Set",
"Definition:Indiscrete Topology",
"Definition:Double Pointed Topology",
"Definition:Open Extension Topology",
"Definition:T4 Space",
"Definition:Tychonoff Separation Axioms",
"Definition:T0 Space",
"Definition:T1 Space",
"De... | [
"Open Extension Topology is T4",
"Definition:T4 Space",
"Double Pointed Countable Complement Topology fulfils no Separation Axioms",
"Definition:T0 Space",
"Definition:T5 Space",
"Condition for Open Extension Space to be T0",
"Definition:T0 Space",
"Condition for Open Extension Space to be T5",
"Def... |
proofwiki-3947 | Saturated Models of same Cardinality are Isomorphic | Let $T$ be an $\LL$-theory.
Let $\kappa$ be an infinite cardinal.
If $\MM$ and $\NN$ are saturated models of $T$ and the cardinality of the universes of $\MM$ and $\NN$ are both $\kappa$, then $\MM$ and $\NN$ are isomorphic. | === Outline ===
The idea of the proof is that since the models are saturated, we can define an isomorphism $f: \NN \to \MM$ by picking the image $\map f x$ of an element $x \in \NN$ as something which realizes the type that $\map f x$ would need to satisfy in $\MM$.
This is done using a (transfinite) recursive construc... | Let $T$ be an $\LL$-[[Definition:Theory|theory]].
Let $\kappa$ be an [[Definition:Infinite|infinite]] [[Definition:Cardinal|cardinal]].
If $\MM$ and $\NN$ are [[Definition:Saturated Model|saturated models]] of $T$ and the [[Definition:Cardinality|cardinality]] of the universes of $\MM$ and $\NN$ are both $\kappa$, th... | === Outline ===
The idea of the proof is that since the models are saturated, we can define an isomorphism $f: \NN \to \MM$ by picking the image $\map f x$ of an element $x \in \NN$ as something which realizes the type that $\map f x$ would need to satisfy in $\MM$.
This is done using a (transfinite) recursive constr... | Saturated Models of same Cardinality are Isomorphic | https://proofwiki.org/wiki/Saturated_Models_of_same_Cardinality_are_Isomorphic | https://proofwiki.org/wiki/Saturated_Models_of_same_Cardinality_are_Isomorphic | [
"Model Theory for Predicate Logic"
] | [
"Definition:Theory",
"Definition:Infinite",
"Definition:Cardinal",
"Definition:Saturated Model",
"Definition:Cardinality",
"Definition:Embedding (Model Theory)"
] | [
"Saturated Implies Universal"
] |
proofwiki-3948 | Compact Complement Topology is Topology | Let $T = \struct {\R, \tau}$ be the compact complement space.
Then $\tau$ is a topology on $T$. | By definition, we have that $\O \in \tau$.
We also have that $\R \in \tau$ as $\relcomp \R \R = \O$ which is trivially compact.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
| r = A \cap B
| c =
}}
{{eqn | ll= \leadsto
| l = \relcomp \R H
| r = \relcomp \R {... | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Space|compact complement space]].
Then $\tau$ is a [[Definition:Topology|topology]] on $T$. | By definition, we have that $\O \in \tau$.
We also have that $\R \in \tau$ as $\relcomp \R \R = \O$ which is trivially [[Definition:Compact Topological Subspace|compact]].
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
| r = A \cap B
| c =
}}
{{eqn | ll= \leadsto
... | Compact Complement Topology is Topology | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Topology | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Topology | [
"Compact Complement Topology"
] | [
"Definition:Compact Complement Topology",
"Definition:Topology"
] | [
"Definition:Compact Topological Space/Subspace",
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection",
"Definition:Compact Topological Space/Subspace",
"Union of Compact Sets is Compact",
"Definition:Set Union",
"Definition:Compact Topological Space/Subspace",
"Definition:Compa... |
proofwiki-3949 | Infinite Ramsey's Theorem | Let $k, n \in \N$.
For any set $S$, let $S^{\paren n}$ denote the set $\set {\set {s_1, \ldots, s_n}: \text{each } s_i \in S}$ of cardinality $n$ subsets of $S$.
Let $X$ be an infinite set.
Then:
:for every partition $P$ of $X^{\paren n}$ into $k$ many components
:there is an infinite subset $Y \subseteq X$
such that:
... | We will prove the theorem for fixed $k$ by induction on $n$. | Let $k, n \in \N$.
For any set $S$, let $S^{\paren n}$ denote the set $\set {\set {s_1, \ldots, s_n}: \text{each } s_i \in S}$ of [[Definition:Cardinality|cardinality]] $n$ [[Definition:Subset|subsets]] of $S$.
Let $X$ be an [[Definition:Infinite Set|infinite set]].
Then:
:for every [[Definition:Partition (Set Theo... | We will prove the theorem for fixed $k$ by [[Proof by Mathematical Induction|induction]] on $n$. | Infinite Ramsey's Theorem | https://proofwiki.org/wiki/Infinite_Ramsey's_Theorem | https://proofwiki.org/wiki/Infinite_Ramsey's_Theorem | [
"Ramsey Theory"
] | [
"Definition:Cardinality",
"Definition:Subset",
"Definition:Infinite Set",
"Definition:Set Partition"
] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-3950 | Compact Complement Space is T1 | Let $T = \struct {S, \tau}$ be a compact complement space.
Then $T$ is a $T_1$ space. | We have that a Finite Topological Space is Compact.
So an arbitrary finite subspace of $T$ is compact.
Let $\tau^*$ be the set defined as:
:$\tau^* = \leftset {S \subseteq \R: S = \O \text { or } \relcomp \R S}$ is finite$\rightset {}$
where $\relcomp \R S$ denotes the complement of $S$ in $\R$.
Then $\tau^*$ is a subs... | Let $T = \struct {S, \tau}$ be a [[Definition:Compact Complement Space|compact complement space]].
Then $T$ is a [[Definition:T1 Space|$T_1$ space]]. | We have that a [[Finite Topological Space is Compact]].
So an [[Definition:Arbitrary|arbitrary]] [[Definition:Finite Set|finite]] [[Definition:Topological Subspace|subspace]] of $T$ is [[Definition:Compact Topological Subspace|compact]].
Let $\tau^*$ be the [[Definition:Set|set]] defined as:
:$\tau^* = \leftset {S \... | Compact Complement Space is T1 | https://proofwiki.org/wiki/Compact_Complement_Space_is_T1 | https://proofwiki.org/wiki/Compact_Complement_Space_is_T1 | [
"Compact Complement Topology",
"Examples of T1 Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:T1 Space"
] | [
"Finite Topological Space is Compact",
"Definition:Arbitrary",
"Definition:Finite Set",
"Definition:Topological Subspace",
"Definition:Compact Topological Space/Subspace",
"Definition:Set",
"Definition:Finite Set",
"Definition:Relative Complement",
"Definition:Subset",
"Definition:Compact Compleme... |
proofwiki-3951 | Compact Complement Topology is Irreducible | Let $T = \struct {\R, \tau}$ be the compact complement space on $\R$.
Then $T$ is an irreducible space. | Let $U_1, U_2 \in \tau$ be open in $T$.
Let $\relcomp \R {U_1} = V_1$ and $\relcomp \R {U_2} = V_2$.
By definition of compact complement topology, $V_1, V_2 \subseteq \R$ such that $V_1$ and $V_2$ are both compact.
$V_1$ and $V_2$ are both bounded by definition of compact in $\R$.
From Union of Two Compact Sets is Comp... | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Space|compact complement space]] on $\R$.
Then $T$ is an [[Definition:Irreducible Space|irreducible space]]. | Let $U_1, U_2 \in \tau$ be [[Definition:Open Set (Topology)|open]] in $T$.
Let $\relcomp \R {U_1} = V_1$ and $\relcomp \R {U_2} = V_2$.
By definition of [[Definition:Compact Complement Topology|compact complement topology]], $V_1, V_2 \subseteq \R$ such that $V_1$ and $V_2$ are both [[Definition:Compact Topological S... | Compact Complement Topology is Irreducible | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Irreducible | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Irreducible | [
"Compact Complement Topology",
"Examples of Irreducible Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:Irreducible Space"
] | [
"Definition:Open Set/Topology",
"Definition:Compact Complement Topology",
"Definition:Compact Topological Space/Subspace",
"Definition:Bounded Set/Real Numbers",
"Definition:Compact Topological Space/Subspace",
"Union of Two Compact Sets is Compact",
"Definition:Compact Topological Space/Subspace",
"D... |
proofwiki-3952 | Weierstrass-Casorati Theorem | Let $f$ be a holomorphic function defined on the open ball $\map B {a, r} \setminus \set a$.
Let $f$ have an essential singularity at $a$.
Then:
:$\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a dense subset of $\C$. | {{WLOG}}, suppose $a = 0$ and $r = 1$.
{{AimForCont}} $\exists s < 1$ such that:
:$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.
Then, by definition of dense subset:
:$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$
{{explain|This d... | Let $f$ be a [[Definition:Holomorphic Function|holomorphic function]] defined on the [[Definition:Open Ball|open ball]] $\map B {a, r} \setminus \set a$.
Let $f$ have an [[Definition:Essential Singularity|essential singularity]] at $a$.
Then:
:$\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a [[Definiti... | {{WLOG}}, suppose $a = 0$ and $r = 1$.
{{AimForCont}} $\exists s < 1$ such that:
:$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.
Then, by definition of [[Definition:Dense|dense subset]]:
:$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0}... | Weierstrass-Casorati Theorem | https://proofwiki.org/wiki/Weierstrass-Casorati_Theorem | https://proofwiki.org/wiki/Weierstrass-Casorati_Theorem | [
"Weierstrass-Casorati Theorem",
"Essential Singularities",
"Complex Analysis"
] | [
"Definition:Holomorphic Function",
"Definition:Open Ball",
"Definition:Essential Singularity",
"Definition:Dense",
"Definition:Subset"
] | [
"Definition:Dense",
"Definition:Analytic Function",
"Definition:Bounded Mapping",
"Analytic Continuation Principle",
"Definition:Singular Point/Complex",
"Definition:Removable Singularity/Complex Function",
"Definition:Power Series",
"Definition:Isolated Singularity/Pole",
"Definition:Essential Sing... |
proofwiki-3953 | Compact Complement Topology is Connected | Let $T = \struct {\R, \tau}$ be the compact complement topology.
Then $T$ is a connected space. | Follows from:
:Compact Complement Topology is Irreducible
:Irreducible Space is Connected
{{qed}} | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]].
Then $T$ is a [[Definition:Connected Topological Space|connected space]]. | Follows from:
:[[Compact Complement Topology is Irreducible]]
:[[Irreducible Space is Connected]]
{{qed}} | Compact Complement Topology is Connected | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Connected | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Connected | [
"Compact Complement Topology",
"Examples of Connected Topological Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:Connected Topological Space"
] | [
"Compact Complement Topology is Irreducible",
"Irreducible Space is Connected"
] |
proofwiki-3954 | Compact Complement Topology is Locally Connected | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a locally connected space. | Follows from:
:Compact Complement Topology is Irreducible
:Irreducible Space is Locally Connected
{{qed}} | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Then $T$ is a [[Definition:Locally Connected Space|locally connected space]]. | Follows from:
:[[Compact Complement Topology is Irreducible]]
:[[Irreducible Space is Locally Connected]]
{{qed}} | Compact Complement Topology is Locally Connected | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Locally_Connected | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Locally_Connected | [
"Compact Complement Topology",
"Examples of Locally Connected Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:Locally Connected Space"
] | [
"Compact Complement Topology is Irreducible",
"Irreducible Space is Locally Connected"
] |
proofwiki-3955 | Compact Complement Topology is not Ultraconnected | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not an ultraconnected space. | ;Proof by Counterexample
{{Recall|Ultraconnected Space|ultraconnected space}}
{{:Definition:Ultraconnected Space/Definition 1}}
By definition, closed sets of $T$ are compact sets of $T$.
So, for example, $\closedint 0 1$ and $\closedint 2 3$ are disjoint compact sets and therefore closed sets of $T$.
Hence the result b... | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Then $T$ is not an [[Definition:Ultraconnected Space|ultraconnected space]]. | ;[[Proof by Counterexample]]
{{Recall|Ultraconnected Space|ultraconnected space}}
{{:Definition:Ultraconnected Space/Definition 1}}
By definition, [[Definition:Closed Set (Topology)|closed sets]] of $T$ are [[Definition:Compact Topological Space|compact sets]] of $T$.
So, for example, $\closedint 0 1$ and $\closedi... | Compact Complement Topology is not Ultraconnected/Proof 1 | https://proofwiki.org/wiki/Compact_Complement_Topology_is_not_Ultraconnected | https://proofwiki.org/wiki/Compact_Complement_Topology_is_not_Ultraconnected/Proof_1 | [
"Compact Complement Topology is not Ultraconnected",
"Compact Complement Topology",
"Examples of Ultraconnected Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:Ultraconnected Space"
] | [
"Proof by Counterexample",
"Definition:Closed Set/Topology",
"Definition:Compact Topological Space",
"Definition:Disjoint Sets",
"Definition:Compact Topological Space",
"Definition:Closed Set/Topology",
"Definition:Ultraconnected Space"
] |
proofwiki-3956 | Compact Complement Topology is not Ultraconnected | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not an ultraconnected space. | We have:
: Compact Complement Space is $T_1$
: Non-Trivial Ultraconnected Space is not $T_1$
Hence the result.
{{qed}} | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Then $T$ is not an [[Definition:Ultraconnected Space|ultraconnected space]]. | We have:
: [[Compact Complement Space is T1|Compact Complement Space is $T_1$]]
: [[Non-Trivial Ultraconnected Space is not T1|Non-Trivial Ultraconnected Space is not $T_1$]]
Hence the result.
{{qed}} | Compact Complement Topology is not Ultraconnected/Proof 2 | https://proofwiki.org/wiki/Compact_Complement_Topology_is_not_Ultraconnected | https://proofwiki.org/wiki/Compact_Complement_Topology_is_not_Ultraconnected/Proof_2 | [
"Compact Complement Topology is not Ultraconnected",
"Compact Complement Topology",
"Examples of Ultraconnected Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:Ultraconnected Space"
] | [
"Compact Complement Space is T1",
"Non-Trivial Ultraconnected Space is not T1"
] |
proofwiki-3957 | Compact Complement Topology is Compact | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a compact space. | {{Recall|Compact Topological Space|compact space}}
{{:Definition:Compact Topological Space}}
Let $\CC$ be an arbitrary open cover of $\R$.
Let $U \in \CC$ be arbitrary.
By definition of the compact complement topology, $\R \setminus U$ is compact in the usual (Euclidean) topology.
We note that $\CC$ is an open cover of... | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Then $T$ is a [[Definition:Compact Topological Space|compact space]]. | {{Recall|Compact Topological Space|compact space}}
{{:Definition:Compact Topological Space}}
Let $\CC$ be an [[Definition:Arbitrary|arbitrary]] [[Definition:Open Cover|open cover]] of $\R$.
Let $U \in \CC$ be [[Definition:Arbitrary|arbitrary]].
By definition of the [[Definition:Compact Complement Topology|compact co... | Compact Complement Topology is Compact | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Compact | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Compact | [
"Compact Complement Topology",
"Examples of Compact Topological Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:Compact Topological Space"
] | [
"Definition:Arbitrary",
"Definition:Open Cover",
"Definition:Arbitrary",
"Definition:Compact Complement Topology",
"Definition:Compact Topological Space/Subspace",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Open Cover",
"Definition:Compact Topological Space/Subspace"... |
proofwiki-3958 | Countable Local Basis in Compact Complement Topology | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Let $p \in \R$.
Then sets of the form:
:$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
form a countable local basis for $p$. | Let $p \in \R$.
Let:
:$\BB_p = \set {\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to: n \in \N}$
Let $n \in \N$ and $P_n \in \BB$, so that:
:$P_n = \openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
Then:
:$\R \setminus P_n = \closedint {-n} ... | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Let $p \in \R$.
Then [[Definition:Set|sets]] of the form:
:$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
form a [[Definition:Countable Set|countable]] [[Defin... | Let $p \in \R$.
Let:
:$\BB_p = \set {\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to: n \in \N}$
Let $n \in \N$ and $P_n \in \BB$, so that:
:$P_n = \openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
Then:
:$\R \setminus P_n = \closedint {... | Countable Local Basis in Compact Complement Topology | https://proofwiki.org/wiki/Countable_Local_Basis_in_Compact_Complement_Topology | https://proofwiki.org/wiki/Countable_Local_Basis_in_Compact_Complement_Topology | [
"Compact Complement Topology",
"Examples of Local Bases"
] | [
"Definition:Compact Complement Topology",
"Definition:Set",
"Definition:Countable Set",
"Definition:Local Basis"
] | [
"Definition:Set Union",
"Definition:Compact Topological Space/Subspace",
"Definition:Compact Topological Space/Subspace",
"Definition:Open Neighborhood/Point",
"Definition:Compact Topological Space/Subspace",
"Definition:Bounded Set/Real Numbers",
"Definition:Local Basis",
"Definition:Countable Set"
] |
proofwiki-3959 | Compact Complement Topology is First-Countable | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a first-countable space. | Let $\CC$ be an open cover of $\R$.
Let $p \in \R$.
Consider the set:
:$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$
From Countable Local Basis in Compact Complement Topology, $\BB_p$ is a countable local basis for $T$.
Hence the result, by def... | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Then $T$ is a [[Definition:First-Countable Space|first-countable space]]. | Let $\CC$ be an [[Definition:Open Cover|open cover]] of $\R$.
Let $p \in \R$.
Consider the set:
:$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$
From [[Countable Local Basis in Compact Complement Topology]], $\BB_p$ is a [[Definition:Countabl... | Compact Complement Topology is First-Countable | https://proofwiki.org/wiki/Compact_Complement_Topology_is_First-Countable | https://proofwiki.org/wiki/Compact_Complement_Topology_is_First-Countable | [
"Compact Complement Topology",
"Examples of First-Countable Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:First-Countable Space"
] | [
"Definition:Open Cover",
"Countable Local Basis in Compact Complement Topology",
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:First-Countable Space"
] |
proofwiki-3960 | Compact Complement Topology is Sequentially Compact | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is a sequentially compact space. | We have:
:Compact Complement Topology is Compact
:Compact Space is Countably Compact
:Compact Complement Topology is First-Countable
:First-Countable Space is Sequentially Compact iff Countably Compact
{{qed}} | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Then $T$ is a [[Definition:Sequentially Compact Space|sequentially compact space]]. | We have:
:[[Compact Complement Topology is Compact]]
:[[Compact Space is Countably Compact]]
:[[Compact Complement Topology is First-Countable]]
:[[First-Countable Space is Sequentially Compact iff Countably Compact]]
{{qed}} | Compact Complement Topology is Sequentially Compact | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Sequentially_Compact | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Sequentially_Compact | [
"Compact Complement Topology",
"Examples of Sequentially Compact Spaces"
] | [
"Definition:Compact Complement Topology",
"Definition:Sequentially Compact Space"
] | [
"Compact Complement Topology is Compact",
"Compact Space is Countably Compact",
"Compact Complement Topology is First-Countable",
"First-Countable Space is Sequentially Compact iff Countably Compact"
] |
proofwiki-3961 | Compact Complement Topology is Coarser than Euclidean Topology | Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $\tau$ is coarser than the usual (Euclidean) topology in $\R$. | Let $U \in \tau$.
Then $V = \R \setminus U$ is compact.
So by definition $V$ is closed in the Euclidean topology.
That is, $U = \R \setminus V$ is open in the Euclidean topology.
That is, every open set in the compact complement topology is also open in the Euclidean topology.
Hence the result, by definition of coarser... | Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
Then $\tau$ is [[Definition:Coarser Topology|coarser]] than the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]] in $\R$. | Let $U \in \tau$.
Then $V = \R \setminus U$ is [[Definition:Compact Topological Subspace|compact]].
So by definition $V$ is [[Definition:Closed Set (Topology)|closed]] in the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]].
That is, $U = \R \setminus V$ is [[Definition:Open Set (Topology)|op... | Compact Complement Topology is Coarser than Euclidean Topology | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Coarser_than_Euclidean_Topology | https://proofwiki.org/wiki/Compact_Complement_Topology_is_Coarser_than_Euclidean_Topology | [
"Compact Complement Topology",
"Examples of Coarser Topology"
] | [
"Definition:Compact Complement Topology",
"Definition:Coarser Topology",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line"
] | [
"Definition:Compact Topological Space/Subspace",
"Definition:Closed Set/Topology",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Open Set/Topology",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Open Set/Topology",
"Definition:Compact Compl... |
proofwiki-3962 | Existence of Uncomputable Mappings | There exists a mapping that is not computable. | Let $\CC$ be the set of all computer programs.
Every computer program in $\CC$ is a finite string of symbols from some alphabet.
Thus, from Set of Finite Strings is Countably Infinite and Subset of Countable Set is Countable, $\CC$ is countable.
Let $\BB \subset \CC$ be the set of all computer programs that implement m... | There exists a [[Definition:Mapping|mapping]] that is not [[Definition:Computable Function|computable]]. | Let $\CC$ be the [[Definition:Set|set]] of all [[Definition:Computer Program|computer programs]].
Every [[Definition:Computer Program|computer program]] in $\CC$ is a [[Definition:Finite String|finite string]] of [[Definition:Symbol|symbols]] from some [[Definition:Alphabet of Formal Language|alphabet]].
Thus, from [... | Existence of Uncomputable Mappings/Proof | https://proofwiki.org/wiki/Existence_of_Uncomputable_Mappings | https://proofwiki.org/wiki/Existence_of_Uncomputable_Mappings/Proof | [
"Existence of Uncomputable Mappings",
"Effectively Computable Functions",
"Computer Science",
"Logic"
] | [
"Definition:Mapping",
"Definition:Effectively Computable Function"
] | [
"Definition:Set",
"Definition:Computer Program",
"Definition:Computer Program",
"Definition:String/Finite",
"Definition:Symbol",
"Definition:Formal Language/Alphabet",
"Set of Finite Strings is Countably Infinite",
"Subset of Countable Set is Countable",
"Definition:Countable Set",
"Definition:Set... |
proofwiki-3963 | Theories with Infinite Models have Models with Order Indiscernibles | Let $T$ be an $\LL$-theory with infinite models.
Let $\struct {I, <$ be an infinite strict linearly ordered set.
There is a model $\MM \models T$ containing an order indiscernible set $\set {x_i : i \in I}$. | We will construct the claimed model using the Compactness Theorem.
Let $LL^*$ be the language obtained by adding constant symbols $c_i$ to $\LL$ for each $i \in I$.
Let $T^*$ be the $\LL^*$-theory obtained by adding to $T$ the $\LL^*$-sentences:
:$c_i \ne c_j$ for each $i \ne j$ in $I$
and:
:$\map \phi {c_{i_1}, \dotsc... | Let $T$ be an $\LL$-[[Definition:Theory|theory]] with [[Definition:Infinite|infinite]] [[Definition:Model (Logic)|models]].
Let $\struct {I, <$ be an infinite [[Definition:Strict Total Ordering|strict linearly ordered set]].
There is a model $\MM \models T$ containing an [[Definition:Order Indiscernible|order indisc... | We will construct the claimed model using the [[Compactness Theorem]].
Let $LL^*$ be the language obtained by adding constant symbols $c_i$ to $\LL$ for each $i \in I$.
Let $T^*$ be the $\LL^*$-theory obtained by adding to $T$ the $\LL^*$-sentences:
:$c_i \ne c_j$ for each $i \ne j$ in $I$
and:
:$\map \phi {c_{i_1},... | Theories with Infinite Models have Models with Order Indiscernibles | https://proofwiki.org/wiki/Theories_with_Infinite_Models_have_Models_with_Order_Indiscernibles | https://proofwiki.org/wiki/Theories_with_Infinite_Models_have_Models_with_Order_Indiscernibles | [
"Model Theory for Predicate Logic"
] | [
"Definition:Theory",
"Definition:Infinite",
"Definition:Model (Logic)",
"Definition:Strict Total Ordering",
"Definition:Order Indiscernible"
] | [
"Compactness Theorem",
"Definition:Free Variable",
"Definition:Chain (Order Theory)",
"Infinite Ramsey's Theorem",
"Definition:Set Partition",
"Infinite Ramsey's Theorem",
"Definition:Increasing",
"Definition:Chain (Order Theory)",
"Compactness Theorem",
"Category:Model Theory for Predicate Logic"... |
proofwiki-3964 | Fort Topology is Topology | Let $T = \struct {S, \tau_p}$ be a Fort space.
Then $\tau_p$ is a topology on $T$. | We have that $p \in \relcomp S \O = S$, so $\O \in \tau_p$.
We have that $\relcomp S S = \O$ which is finite, so $S \in \tau_p$.
Now consider $A, B \in \tau_p$, and let $H = A \cap B$.
If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$.
Now suppose $p \in A$ and $p \in B$.
Then:
{{begin-eqn}... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]].
Then $\tau_p$ is a [[Definition:Topology|topology]] on $T$. | We have that $p \in \relcomp S \O = S$, so $\O \in \tau_p$.
We have that $\relcomp S S = \O$ which is [[Definition:Finite Set|finite]], so $S \in \tau_p$.
Now consider $A, B \in \tau_p$, and let $H = A \cap B$.
If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$.
Now suppose $p \in A$ ... | Fort Topology is Topology | https://proofwiki.org/wiki/Fort_Topology_is_Topology | https://proofwiki.org/wiki/Fort_Topology_is_Topology | [
"Fort Spaces"
] | [
"Definition:Fort Space",
"Definition:Topology"
] | [
"Definition:Finite Set",
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union",
"Definition:Open Set/Topology",
"Definition:Finite Set",
"Definition:Set Union",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Relative Complement",
"Definition:Finite Set",
"De Morgan... |
proofwiki-3965 | Fort Space is Excluded Point Space with Finite Complement Space | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $\tau_p$ is the minimal topology that is generated by the excluded point topology and the finite complement topology. | Let $T_1 = \struct {S, \tau_1}$ be the excluded point space on $S$ from $p$.
Let $T_2 = \struct {S, \tau_2}$ be the finite complement space on $S$.
By definition:
:$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$
:$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is finite$\rightset {} \cup \set \O$
By ... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $\tau_p$ is the [[Definition:Minimal Topology|minimal topology]] that is [[Definition:Topology Generated by Synthetic Sub-Basis|generated]] by the [[Definition:Excluded Point Topology|exclud... | Let $T_1 = \struct {S, \tau_1}$ be the [[Definition:Excluded Point Space|excluded point space]] on $S$ from $p$.
Let $T_2 = \struct {S, \tau_2}$ be the [[Definition:Finite Complement Space|finite complement space]] on $S$.
By definition:
:$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$
:$\tau_2 = \... | Fort Space is Excluded Point Space with Finite Complement Space | https://proofwiki.org/wiki/Fort_Space_is_Excluded_Point_Space_with_Finite_Complement_Space | https://proofwiki.org/wiki/Fort_Space_is_Excluded_Point_Space_with_Finite_Complement_Space | [
"Fort Spaces",
"Excluded Point Topologies",
"Finite Complement Topologies"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Minimal Topology",
"Definition:Topology Generated by Synthetic Sub-Basis",
"Definition:Excluded Point Topology",
"Definition:Finite Complement Topology"
] | [
"Definition:Excluded Point Topology",
"Definition:Finite Complement Topology",
"Definition:Finite Set",
"Definition:Fort Space",
"Union is Smallest Superset"
] |
proofwiki-3966 | Fort Space is T1 | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_1$ space. | From Fort Space is Excluded Point Space with Finite Complement Space, $T$ is an expansion of a finite complement space.
Then we have that a Finite Complement Space is $T_1$.
Then from $T_1$ Property is Preserved under Expansion, $T$ is a $T_1$ space.
{{qed}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:T1 Space|$T_1$ space]]. | From [[Fort Space is Excluded Point Space with Finite Complement Space]], $T$ is an [[Definition:Expansion of Topology|expansion]] of a [[Definition:Finite Complement Space|finite complement space]].
Then we have that a [[Finite Complement Space is T1|Finite Complement Space is $T_1$]].
Then from [[T1 Property is Pr... | Fort Space is T1 | https://proofwiki.org/wiki/Fort_Space_is_T1 | https://proofwiki.org/wiki/Fort_Space_is_T1 | [
"Fort Spaces",
"Examples of T1 Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:T1 Space"
] | [
"Fort Space is Excluded Point Space with Finite Complement Space",
"Definition:Expansion of Topology",
"Definition:Finite Complement Topology",
"Finite Complement Space is T1",
"T1 Property is Preserved under Expansion",
"Definition:T1 Space"
] |
proofwiki-3967 | Fort Space is T5 | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_5$ space. | Let $A, B \in \tau_p$ such that $A$ and $B$ are separated.
If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open.
Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated.
{{WLOG}}, suppose that $p \in A$.
Then $p \notin B$ so $B$ is open.
{{AimForCont}} $B$ is not... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:T5 Space|$T_5$ space]]. | Let $A, B \in \tau_p$ such that $A$ and $B$ are [[Definition:Separated Sets|separated]].
If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both [[Definition:Open Set (Topology)|open]].
Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be [[Definition:Separated Sets|separate... | Fort Space is T5 | https://proofwiki.org/wiki/Fort_Space_is_T5 | https://proofwiki.org/wiki/Fort_Space_is_T5 | [
"Fort Spaces",
"Examples of T5 Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:T5 Space"
] | [
"Definition:Separated Sets",
"Definition:Open Set/Topology",
"Definition:Separated Sets",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Infinite Set",
"Definition:Fort Space",
"Definition:Open Set/Topology",
"Definition:Cofinite Subset",
"Definition:Closure (Topolog... |
proofwiki-3968 | Fort Space is Completely Normal | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a completely normal space.
Consequently, $T$ satisfies all weaker separation axioms. | We have:
:Fort Space is $T_1$
:Fort Space is $T_5$
and so by definition $T$ is completely normal.
{{qed}}
See Sequence of Implications of Separation Axioms for confirmation of the statement about weaker separation axioms. | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:Completely Normal Space|completely normal space]].
Consequently, $T$ satisfies all weaker [[Definition:Separation Axioms|separation axioms]]. | We have:
:[[Fort Space is T1|Fort Space is $T_1$]]
:[[Fort Space is T5|Fort Space is $T_5$]]
and so by definition $T$ is [[Definition:Completely Normal Space|completely normal]].
{{qed}}
See [[Sequence of Implications of Separation Axioms]] for confirmation of the statement about weaker [[Definition:Separation Axioms... | Fort Space is Completely Normal | https://proofwiki.org/wiki/Fort_Space_is_Completely_Normal | https://proofwiki.org/wiki/Fort_Space_is_Completely_Normal | [
"Fort Spaces",
"Examples of Completely Normal Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Completely Normal Space",
"Definition:Tychonoff Separation Axioms"
] | [
"Fort Space is T1",
"Fort Space is T5",
"Definition:Completely Normal Space",
"Sequence of Implications of Separation Axioms",
"Definition:Tychonoff Separation Axioms"
] |
proofwiki-3969 | Exchange Principle | Let $D$ be a strongly minimal set in $\MM$.
Let $A$ be a subset of $D$.
Let $b, c \in D$.
If $b$ is algebraic over $A \cup \set c$ but not over $A$, then $c$ is algebraic over $A \cup \set b$. | Let $\map D x$ be a formula defining $D$, which exists since by definition, strongly minimal sets are definable.
To simplify exposition, we will assume below that all further mentioned formulas are $\LL$-formulas with parameters from $A$, in addition to whatever other parameters are supplied.
Suppose $b$ is algebraic ... | Let $D$ be a [[Definition:Minimal (Model Theory)|strongly minimal]] set in $\MM$.
Let $A$ be a [[Definition:Subset|subset]] of $D$.
Let $b, c \in D$.
If $b$ is [[Definition:Algebraic (Model Theory)|algebraic]] over $A \cup \set c$ but not over $A$, then $c$ is [[Definition:Algebraic (Model Theory)|algebraic]] over ... | Let $\map D x$ be a formula defining $D$, which exists since by definition, strongly minimal sets are definable.
To simplify exposition, we will assume below that all further mentioned formulas are $\LL$-formulas with parameters from $A$, in addition to whatever other parameters are supplied.
Suppose $b$ is algebra... | Exchange Principle | https://proofwiki.org/wiki/Exchange_Principle | https://proofwiki.org/wiki/Exchange_Principle | [
"Model Theory for Predicate Logic",
"Named Theorems"
] | [
"Definition:Minimal (Model Theory)",
"Definition:Subset",
"Definition:Algebraic (Model Theory)",
"Definition:Algebraic (Model Theory)"
] | [
"Definition:Free Variable",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Cofinite Subset",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Cofinite Subset",
"Definition:Finite Set",
"Category:Model Theory for Predicate Logic",
"Category:Named Theorems"
] |
proofwiki-3970 | Uncountable Fort Space is not Perfectly Normal | Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a perfectly normal space. | From Clopen Points in Fort Space, $\set p$ is closed in $T$.
Consider a countable intersection of open sets of $T$ which contain $p$.
By definition, all these are cofinite in $S$ and so uncountable.
So this intersection must itself contain all but a countable number of points of $S$.
So $\set p$ is not a $G_\delta$ set... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:Perfectly Normal Space|perfectly normal space]]. | From [[Clopen Points in Fort Space]], $\set p$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
Consider a [[Definition:Countable Intersection|countable intersection]] of [[Definition:Open Set (Topology)|open sets]] of $T$ which contain $p$.
By definition, all these are [[Definition:Cofinite Subset|cofinite in ... | Uncountable Fort Space is not Perfectly Normal | https://proofwiki.org/wiki/Uncountable_Fort_Space_is_not_Perfectly_Normal | https://proofwiki.org/wiki/Uncountable_Fort_Space_is_not_Perfectly_Normal | [
"Uncountable Fort Spaces",
"Examples of Perfectly Normal Spaces"
] | [
"Definition:Fort Space",
"Definition:Uncountable/Set",
"Definition:Perfectly Normal Space"
] | [
"Clopen Points in Fort Space",
"Definition:Closed Set/Topology",
"Definition:Set Intersection/Countable Intersection",
"Definition:Open Set/Topology",
"Definition:Cofinite Subset",
"Definition:Uncountable/Set",
"Definition:Countable Set",
"Definition:G-Delta Set",
"Definition:Perfectly Normal Space"... |
proofwiki-3971 | Countable Fort Space is Perfectly Normal | Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Then $T$ is a perfectly normal space. | We have from Closed Set of Countable Fort Space is $G_\delta$ that every closed set in $T$ is a $G_\delta$ set.
From Fort Space is $T_5$ and $T_5$ Space is $T_4$, we have that a Fort space is a $T_4$ space.
From Fort Space is $T_1$ it follows by definition that $T$ is a perfectly normal space.
{{qed}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on a [[Definition:Countably Infinite Set|countably infinite set]] $S$.
Then $T$ is a [[Definition:Perfectly Normal Space|perfectly normal space]]. | We have from [[Closed Set of Countable Fort Space is G-Delta|Closed Set of Countable Fort Space is $G_\delta$]] that every [[Definition:Closed Set (Topology)|closed set]] in $T$ is a [[Definition:G-Delta Set|$G_\delta$ set]].
From [[Fort Space is T5|Fort Space is $T_5$]] and [[T5 Space is T4|$T_5$ Space is $T_4$]], w... | Countable Fort Space is Perfectly Normal | https://proofwiki.org/wiki/Countable_Fort_Space_is_Perfectly_Normal | https://proofwiki.org/wiki/Countable_Fort_Space_is_Perfectly_Normal | [
"Countable Fort Spaces",
"Examples of Perfectly Normal Spaces"
] | [
"Definition:Fort Space",
"Definition:Countably Infinite/Set",
"Definition:Perfectly Normal Space"
] | [
"Closed Set of Countable Fort Space is G-Delta",
"Definition:Closed Set/Topology",
"Definition:G-Delta Set",
"Fort Space is T5",
"T5 Space is T4",
"Definition:Fort Space",
"Definition:T4 Space",
"Fort Space is T1",
"Definition:Perfectly Normal Space"
] |
proofwiki-3972 | Fort Space is Compact | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a compact space. | Let $\CC$ be an open cover of $T$.
Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is finite.
For each $x \in \relcomp S U$ there exists some $C_x \in \CC$ such that $x \in C$.
So $U$, together with each of those $C_x \in \CC$, is a finite subcover of $\CC$.
Hence the result by definition of compact ... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:Compact Topological Space|compact space]]. | Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is [[Definition:Finite Set|finite]].
For each $x \in \relcomp S U$ there exists some $C_x \in \CC$ such that $x \in C$.
So $U$, together with each of those $C_x \in \CC$, is a [[Definition:... | Fort Space is Compact | https://proofwiki.org/wiki/Fort_Space_is_Compact | https://proofwiki.org/wiki/Fort_Space_is_Compact | [
"Fort Spaces",
"Examples of Compact Topological Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Compact Topological Space"
] | [
"Definition:Open Cover",
"Definition:Finite Set",
"Definition:Subcover/Finite",
"Definition:Compact Topological Space"
] |
proofwiki-3973 | Fort Space is Sequentially Compact | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a sequentially compact space. | Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $T$.
Suppose $\sequence {x_n}$ takes an infinite number of distinct values in $S$.
Then there is an infinite subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ with distinct terms.
Let $U$ be a neighborhood of $p$.
Then $S \setminus U$ is a finite s... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:Sequentially Compact Space|sequentially compact space]]. | Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $T$.
Suppose $\sequence {x_n}$ takes an [[Definition:Infinite Set|infinite number]] of [[Definition:Distinct|distinct]] values in $S$.
Then there is an [[Definition:Infinite Sequence|infinite]] [[Definition:Subsequenc... | Fort Space is Sequentially Compact | https://proofwiki.org/wiki/Fort_Space_is_Sequentially_Compact | https://proofwiki.org/wiki/Fort_Space_is_Sequentially_Compact | [
"Fort Spaces",
"Examples of Sequentially Compact Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Sequentially Compact Space"
] | [
"Definition:Sequence/Infinite Sequence",
"Definition:Infinite Set",
"Definition:Distinct",
"Definition:Sequence/Infinite Sequence",
"Definition:Subsequence",
"Definition:Sequence of Distinct Terms",
"Definition:Neighborhood (Topology)/Point",
"Definition:Finite Set",
"Definition:Convergent Sequence/... |
proofwiki-3974 | Boubaker's Theorem/Proof of Uniqueness | Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Consider the following properties:
{{begin-eqn}}
{{eqn ... | Let:
:$\struct {R, +, \circ}$ be a commutative ring
:$\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$
:$X \in R$ be transcendental over $D$.
It has been demonstrated that the Boubaker Polynomials sub-sequence $\map {B_{4 n} } x$, defined in $D \sqbrk X$ as:
:$\ds \map... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $X \in R$ be [[Definition:Transcendental over ... | Let:
:$\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]]
:$\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$
:$X \in R$ be [[Definition:Transcendental over Integral... | Boubaker's Theorem/Proof of Uniqueness | https://proofwiki.org/wiki/Boubaker's_Theorem/Proof_of_Uniqueness | https://proofwiki.org/wiki/Boubaker's_Theorem/Proof_of_Uniqueness | [
"Boubaker Polynomials"
] | [
"Definition:Commutative Ring",
"Definition:Subdomain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Transcendental (Abstract Algebra)/Ring",
"Definition:Ring of Polynomial Forms",
"Definition:Positive/Integer",
"Definition:Null Polynomial/Ring",
"Definition:Subsequ... | [
"Definition:Commutative Ring",
"Definition:Subdomain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Transcendental (Abstract Algebra)/Ring",
"Definition:Boubaker Polynomials"
] |
proofwiki-3975 | Uncountable Fort Space is not First-Countable | Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a first-countable space. | Let $\UU$ be a countable set of open neighborhoods of $p$.
Let $U \in \UU$.
Then as $p \in U$:
:$p \notin \relcomp S U$
So for $U$ to be open in $T$, it must follow that $\relcomp S U$ is finite.
From De Morgan's Laws: Complement of Intersection we have:
:$\ds H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S ... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:First-Countable Space|first-countable space]]. | Let $\UU$ be a [[Definition:Countable Set|countable set]] of [[Definition:Open Neighborhood of Point|open neighborhoods]] of $p$.
Let $U \in \UU$.
Then as $p \in U$:
:$p \notin \relcomp S U$
So for $U$ to be [[Definition:Open Set (Topology)|open]] in $T$, it must follow that $\relcomp S U$ is [[Definition:Finite Set... | Uncountable Fort Space is not First-Countable | https://proofwiki.org/wiki/Uncountable_Fort_Space_is_not_First-Countable | https://proofwiki.org/wiki/Uncountable_Fort_Space_is_not_First-Countable | [
"Uncountable Fort Spaces",
"Examples of First-Countable Spaces"
] | [
"Definition:Fort Space",
"Definition:Uncountable/Set",
"Definition:First-Countable Space"
] | [
"Definition:Countable Set",
"Definition:Open Neighborhood/Point",
"Definition:Open Set/Topology",
"Definition:Finite Set",
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection",
"Countable Union of Countable Sets is Countable",
"Definition:Countable Set",
"Definition:Uncountab... |
proofwiki-3976 | Generating Function for Boubaker Polynomials | The Boubaker polynomials, defined as:
:$\map {B_n} x = \begin{cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end{cases}$
have as an ordinary generating function:
:$\ds \map {f_{B_n, \operatorname {ORD} } } {x, t} = \sum_{n \mathop = 0}^{\infty} \map {... | {{begin-eqn}}
{{eqn | l = \map f {x, t}
| r = 1 + x t + \paren {x^2 + 2} t^2 + \sum_{n \mathop = 3}^\infty \paren {x \map {B_{n - 1} } x t^n - \map {B_{n - 2} } x t^n}
| c = Recursive definition of $\map {B_n} x$ for $n > 2$
}}
{{eqn | r = 1 + x t + \paren {x^2 + 2} t^2 + x t \sum_{n \mathop = 3}^\infty \ma... | The [[Definition:Boubaker Polynomials|Boubaker polynomials]], defined as:
:$\map {B_n} x = \begin{cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end{cases}$
have as an [[Definition:Generating Function|ordinary generating function]]:
:$\ds \map {f_{B_... | {{begin-eqn}}
{{eqn | l = \map f {x, t}
| r = 1 + x t + \paren {x^2 + 2} t^2 + \sum_{n \mathop = 3}^\infty \paren {x \map {B_{n - 1} } x t^n - \map {B_{n - 2} } x t^n}
| c = Recursive definition of $\map {B_n} x$ for $n > 2$
}}
{{eqn | r = 1 + x t + \paren {x^2 + 2} t^2 + x t \sum_{n \mathop = 3}^\infty \ma... | Generating Function for Boubaker Polynomials | https://proofwiki.org/wiki/Generating_Function_for_Boubaker_Polynomials | https://proofwiki.org/wiki/Generating_Function_for_Boubaker_Polynomials | [
"Boubaker Polynomials",
"Examples of Generating Functions"
] | [
"Definition:Boubaker Polynomials",
"Definition:Generating Function"
] | [
"Category:Boubaker Polynomials",
"Category:Examples of Generating Functions"
] |
proofwiki-3977 | Exponential Generating Function for Boubaker Polynomials | The Boubaker polynomials, defined as:
:<nowiki>$\map {B_n} x = \begin {cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end {cases}$</nowiki>
have as an exponential generating function:
:$\ds \map {f_{B_n, \operatorname {EXP} } } {x, t} = \sum_{... | From the definition of the Boubaker polynomials, we have:
:$\begin {cases} \map {B_0} x = 1 \\ \map {B_1} x = x \end {cases}$
We have also, for $T_n$, the Chebyshev polynomials of the first kind and $U_n$, the Chebyshev polynomials of the second kind:
:$\begin {cases} \map {T_0} x = 1 \\ \map {T_1} x = x \e... | The [[Definition:Boubaker Polynomials|Boubaker polynomials]], defined as:
:<nowiki>$\map {B_n} x = \begin {cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end {cases}$</nowiki>
have as an [[Definition:Exponential Generating Function|exponenti... | From the definition of the [[Definition:Boubaker Polynomials|Boubaker polynomials]], we have:
:$\begin {cases} \map {B_0} x = 1 \\ \map {B_1} x = x \end {cases}$
We have also, for $T_n$, the [[Definition:Chebyshev Polynomial of the First Kind|Chebyshev polynomials of the first kind]] and $U_n$, the [[Definition:C... | Exponential Generating Function for Boubaker Polynomials | https://proofwiki.org/wiki/Exponential_Generating_Function_for_Boubaker_Polynomials | https://proofwiki.org/wiki/Exponential_Generating_Function_for_Boubaker_Polynomials | [
"Boubaker Polynomials",
"Examples of Exponential Generating Functions"
] | [
"Definition:Boubaker Polynomials",
"Definition:Exponential Generating Function"
] | [
"Definition:Boubaker Polynomials",
"Definition:Chebyshev Polynomials/First Kind",
"Definition:Chebyshev Polynomials/Second Kind",
"Definition:Chebyshev Polynomials/First Kind",
"Definition:Chebyshev Polynomials/Second Kind",
"Definition:Exponential Generating Function",
"Category:Boubaker Polynomials",
... |
proofwiki-3978 | Clopen Points in Fort Space | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Let $q \in S: q \ne p$.
Then $\set q$ is both open and closed in $T$.
$\set p$ itself, on the other hand, is closed but not open. | We have that $\set q$ is finite so $\relcomp S {\set q}$ is cofinite.
So $\relcomp S {\set q}$ is open and so $\set q$ is closed.
Then we have that $p \notin \set q$ so $\set q$ is open.
However, $p \notin \relcomp S {\set p}$ and $\relcomp S {\set p}$ is infinite.
So $\set p$ is not open in $S$.
But $\set p$ is finite... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Let $q \in S: q \ne p$.
Then $\set q$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
$\set p$ itself, on the other hand, is [[Definiti... | We have that $\set q$ is [[Definition:Finite Set|finite]] so $\relcomp S {\set q}$ is [[Definition:Cofinite Subset|cofinite]].
So $\relcomp S {\set q}$ is [[Definition:Open Set (Topology)|open]] and so $\set q$ is [[Definition:Closed Set (Topology)|closed]].
Then we have that $p \notin \set q$ so $\set q$ is [[Defin... | Clopen Points in Fort Space | https://proofwiki.org/wiki/Clopen_Points_in_Fort_Space | https://proofwiki.org/wiki/Clopen_Points_in_Fort_Space | [
"Fort Spaces",
"Examples of Clopen Sets"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology"
] | [
"Definition:Finite Set",
"Definition:Cofinite Subset",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Infinite Set",
"Definition:Open Set/Topology",
"Definition:Finite Set",
"Definition:Cofinite Subset",
"Definition:Closed Set/Topology... |
proofwiki-3979 | Fort Space is Totally Separated | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a totally separated space. | Let $a, b \in S$ such that $a \ne b$.
From Clopen Points in Fort Space, all points in $S$ apart from $p$ are both open and closed in $T$.
{{WLOG}}, suppose that $a \ne p$.
Then as $\set a$ is closed, $\relcomp S {\set a}$ is open in $T$.
So we have a partition $\set a \mid \relcomp S {\set a}$ of $S$ such that $a \in \... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:Totally Separated Space|totally separated space]]. | Let $a, b \in S$ such that $a \ne b$.
From [[Clopen Points in Fort Space]], all points in $S$ apart from $p$ are both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
{{WLOG}}, suppose that $a \ne p$.
Then as $\set a$ is [[Definition:Closed Set (Topology)|closed]], $\re... | Fort Space is Totally Separated | https://proofwiki.org/wiki/Fort_Space_is_Totally_Separated | https://proofwiki.org/wiki/Fort_Space_is_Totally_Separated | [
"Fort Spaces",
"Examples of Totally Separated Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Totally Separated Space"
] | [
"Clopen Points in Fort Space",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Separation (Topology)",
"Definition:Totally Separated Space"
] |
proofwiki-3980 | Fort Space is not Extremally Disconnected | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is not an extremally disconnected space. | {{Recall|Extremally Disconnected Space|extremally disconnected space}}
{{:Definition:Extremally Disconnected Space/Definition 1}}
Let $H \subseteq S$ be an infinite subset of $S$ such that $p \in \relcomp S H$ and $\relcomp S H$ is infinite.
Then $H$ is open in $T$, as $p \in \relcomp S H$.
From Limit Points in Fort Sp... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is not an [[Definition:Extremally Disconnected Space|extremally disconnected space]]. | {{Recall|Extremally Disconnected Space|extremally disconnected space}}
{{:Definition:Extremally Disconnected Space/Definition 1}}
Let $H \subseteq S$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ such that $p \in \relcomp S H$ and $\relcomp S H$ is [[Definition:Infinite Set|infinite]].... | Fort Space is not Extremally Disconnected | https://proofwiki.org/wiki/Fort_Space_is_not_Extremally_Disconnected | https://proofwiki.org/wiki/Fort_Space_is_not_Extremally_Disconnected | [
"Fort Spaces",
"Examples of Extremally Disconnected Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Extremally Disconnected Space"
] | [
"Definition:Infinite Set",
"Definition:Subset",
"Definition:Infinite Set",
"Definition:Open Set/Topology",
"Limit Points in Fort Space",
"Definition:Infinite Set",
"Definition:Limit Point/Topology/Set",
"Definition:Closure (Topology)",
"Definition:Infinite Set",
"Definition:Fort Space",
"Definit... |
proofwiki-3981 | Fort Space is Zero Dimensional | Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a zero dimensional space. | Let $q \in S$ such that $q \ne p$.
Then from Clopen Points in Fort Space, $\set q$ is clopen.
So $\forall q \in S, q \ne p: \set {\set q}$ is a local basis for $q$.
If we take the open neighborhoods of $p$ we get a local basis $\UU_p$ with the following property:
:Since $p \in U \in \UU_p$, its complement does not cont... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:Zero Dimensional Space|zero dimensional space]]. | Let $q \in S$ such that $q \ne p$.
Then from [[Clopen Points in Fort Space]], $\set q$ is [[Definition:Clopen Set|clopen]].
So $\forall q \in S, q \ne p: \set {\set q}$ is a [[Definition:Local Basis|local basis]] for $q$.
If we take the [[Definition:Open Neighborhood of Point|open neighborhoods]] of $p$ we get a [[... | Fort Space is Zero Dimensional | https://proofwiki.org/wiki/Fort_Space_is_Zero_Dimensional | https://proofwiki.org/wiki/Fort_Space_is_Zero_Dimensional | [
"Fort Spaces",
"Examples of Zero Dimensional Spaces"
] | [
"Definition:Fort Space",
"Definition:Infinite Set",
"Definition:Zero Dimensional Space"
] | [
"Clopen Points in Fort Space",
"Definition:Clopen Set",
"Definition:Local Basis",
"Definition:Open Neighborhood/Point",
"Definition:Local Basis",
"Definition:Set Complement",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Set Union",
"Definition:Local Basis",
"Defi... |
proofwiki-3982 | Formula and its Negation Cannot Both Cause Forking | Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A \subseteq B$ be subsets of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an $n$-type over $B$.
If $\pi$ does not fork over $A$, then for any formula $\map \phi {\bar x, \bar b}$, either $\pi \cup \set \phi$ or $\pi \cup... | We prove the contrapositive.
Suppose both $\pi \cup \set \phi$ and $\pi \cup \set {\neg \phi}$ fork over $A$.
By definition of forking:
:$\pi \cup \set \phi$ implies $\map {\phi_1} {\bar x, \bar c_1} \vee \cdots \vee \map {\phi_k} {\bar x, \bar c_k}$ and
:$\pi \cup \set {\neg \phi}$ implies $\map {\psi_1} {\bar x, \bar... | Let $T$ be a [[Definition:Theory|complete $\LL$-theory]].
Let $\mathfrak C$ be a [[Definition:Monster Model|monster model]] for $T$.
Let $A \subseteq B$ be [[Definition:Subset|subsets]] of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an [[Definition:Type|$n$-type]] over $B$.
If $\pi$ does not [[Defini... | [[Proof by Contraposition|We prove the contrapositive]].
Suppose both $\pi \cup \set \phi$ and $\pi \cup \set {\neg \phi}$ fork over $A$.
By definition of forking:
:$\pi \cup \set \phi$ implies $\map {\phi_1} {\bar x, \bar c_1} \vee \cdots \vee \map {\phi_k} {\bar x, \bar c_k}$ and
:$\pi \cup \set {\neg \phi}$ impl... | Formula and its Negation Cannot Both Cause Forking | https://proofwiki.org/wiki/Formula_and_its_Negation_Cannot_Both_Cause_Forking | https://proofwiki.org/wiki/Formula_and_its_Negation_Cannot_Both_Cause_Forking | [
"Model Theory for Predicate Logic"
] | [
"Definition:Theory",
"Definition:Monster Model",
"Definition:Subset",
"Definition:Type",
"Definition:Fork"
] | [
"Proof by Contraposition",
"Definition:Divide (Model Theory)",
"Category:Model Theory for Predicate Logic"
] |
proofwiki-3983 | Non-Forking Types have Non-Forking Completions | Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A\subseteq B$ be subsets of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an $n$-type over $B$.
If $\pi$ does not fork over $A$, then there is a complete $n$-type $p$ over $B$ such that $\pi \subseteq p$ and $p$ does not ... | Suppose $\pi$ does not fork over $A$.
We will use Zorn's Lemma to find a candidate for the needed complete type.
Consider the collection $\Pi$ of all non-forking sets $\pi'$ of $\LL$-formulas with parameters from $B$ such that $\pi'$ contains $\pi$.
Order $\Pi$ by subset inclusion.
Since a set forks iff a finite subset... | Let $T$ be a [[Definition:Theory|complete $\LL$-theory]].
Let $\mathfrak C$ be a [[Definition:Monster Model|monster model]] for $T$.
Let $A\subseteq B$ be [[Definition:Subset|subsets]] of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an [[Definition:Type|$n$-type]] over $B$.
If $\pi$ does not [[Definit... | Suppose $\pi$ does not fork over $A$.
We will use [[Zorn's Lemma]] to find a candidate for the needed complete type.
Consider the collection $\Pi$ of all non-forking sets $\pi'$ of $\LL$-formulas with parameters from $B$ such that $\pi'$ contains $\pi$.
Order $\Pi$ by subset inclusion.
Since [[Forking is Local|a ... | Non-Forking Types have Non-Forking Completions | https://proofwiki.org/wiki/Non-Forking_Types_have_Non-Forking_Completions | https://proofwiki.org/wiki/Non-Forking_Types_have_Non-Forking_Completions | [
"Model Theory for Predicate Logic"
] | [
"Definition:Theory",
"Definition:Monster Model",
"Definition:Subset",
"Definition:Type",
"Definition:Fork",
"Definition:Type"
] | [
"Zorn's Lemma",
"Forking is Local",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Chain (Order Theory)/Subset Relation",
"Zorn's Lemma",
"Formula and its Negation Cannot Both Cause Forking",
"Category:Model Theory for Predicate Logic"
] |
proofwiki-3984 | Forking is Local | Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A\subseteq B$ be subsets of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an $n$-type over $B$.
$\pi$ forks over $A$ {{iff}} a finite subset of $\pi$ forks over $A$. | Suppose $\pi$ forks over $A$.
:By definition, $\pi$ implies a disjunction of formulas which each divide over $A$.
:Since proofs are finite, this means that there is a finite subset of $\pi$ which implies this disjunction, completing this direction of the proof.
Suppose a finite subset of $\pi$ forks over $A$.
:By defin... | Let $T$ be a [[Definition:Theory|complete $\LL$-theory]].
Let $\mathfrak C$ be a [[Definition:Monster Model|monster model]] for $T$.
Let $A\subseteq B$ be [[Definition:Subset|subsets]] of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an [[Definition:Type|$n$-type]] over $B$.
$\pi$ [[Definition:Fork|for... | Suppose $\pi$ [[Definition:Fork|forks]] over $A$.
:By definition, $\pi$ implies a disjunction of formulas which each [[Definition:Divide (Model Theory)|divide]] over $A$.
:Since proofs are finite, this means that there is a [[Definition:Finite Subset|finite subset]] of $\pi$ which implies this disjunction, completing... | Forking is Local | https://proofwiki.org/wiki/Forking_is_Local | https://proofwiki.org/wiki/Forking_is_Local | [
"Model Theory for Predicate Logic"
] | [
"Definition:Theory",
"Definition:Monster Model",
"Definition:Subset",
"Definition:Type",
"Definition:Fork",
"Definition:Finite Subset",
"Definition:Fork"
] | [
"Definition:Fork",
"Definition:Divide (Model Theory)",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Definition:Fork",
"Definition:Finite Subset",
"Definition:Divide (Model Theory)",
"Category:Model Theory for Predicate Logic"
] |
proofwiki-3985 | Subset of Finite Set is Finite | Let $X$ be a finite set.
If $Y$ is a subset of $X$, then $Y$ is also finite. | From the definition, $X$ is finite {{iff}} $\exists n \in \N$ such that there exists a bijection:
:$f: X \leftrightarrow \N_n$
where $\N_n$ is the set of all elements of $\N$ less than $n$, that is:
:$\N_n = \set {0, 1, 2, \ldots, n - 1}$
The case in which $X$ is empty is trivial.
We begin proving the following particu... | Let $X$ be a [[Definition:Finite Set|finite set]].
If $Y$ is a [[Definition:Subset|subset]] of $X$, then $Y$ is also [[Definition:Finite Set|finite]]. | From the definition, $X$ is [[Definition:Finite Set|finite]] {{iff}} $\exists n \in \N$ such that there exists a [[Definition:Bijection|bijection]]:
:$f: X \leftrightarrow \N_n$
where $\N_n$ is the [[Definition:Initial Segment of Natural Numbers|set of all elements of $\N$ less than $n$]], that is:
:$\N_n = \set {0, 1,... | Subset of Finite Set is Finite | https://proofwiki.org/wiki/Subset_of_Finite_Set_is_Finite | https://proofwiki.org/wiki/Subset_of_Finite_Set_is_Finite | [
"Subsets",
"Finite Sets"
] | [
"Definition:Finite Set",
"Definition:Subset",
"Definition:Finite Set"
] | [
"Definition:Finite Set",
"Definition:Bijection",
"Definition:Initial Segment of Natural Numbers",
"Definition:Empty Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Bijection between Specific Elements",
"Definition:Bijection",
"Principle of Mathematical Induction",
"Definition:Bijection",
... |
proofwiki-3986 | Statements Equivalent to Non-Dividing Type | Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A$ be a subset of the universe of $\mathfrak C$.
The following are equivalent:
:$(1): \quad$ $\map {\mathrm {tp} } {\bar c / A, \bar b}$ does not divide over $A$.
:$(2): \quad$ For every $\set {\bar b_i : i \in I}$ containing $\bar b... | === $(1) \implies (2)$ ===
Let $\set {\bar b_i : i \in I}$ containing $\bar b$ be order indiscernible over $A$.
{{AimForCont}} $\ds \bigcup_{i \mathop \in I} \map {\mathrm {tp} } {\bar c / A, \bar b_i}$ is not satisfiable.
By the Compactness Theorem, some finite subset $\set {\map {\phi_1} {x, \bar b_{i_1} }, \dots, \m... | Let $T$ be a [[Definition:Theory|complete $\LL$-theory]].
Let $\mathfrak C$ be a [[Definition:Monster Model|monster model]] for $T$.
Let $A$ be a [[Definition:Subset|subset]] of the universe of $\mathfrak C$.
The following are equivalent:
:$(1): \quad$ [[Definition:Type|$\map {\mathrm {tp} } {\bar c / A, \bar b}$]]... | === $(1) \implies (2)$ ===
Let $\set {\bar b_i : i \in I}$ containing $\bar b$ be [[Definition:Order Indiscernible|order indiscernible]] over $A$.
{{AimForCont}} $\ds \bigcup_{i \mathop \in I} \map {\mathrm {tp} } {\bar c / A, \bar b_i}$ is not satisfiable.
By the [[Compactness Theorem]], some finite subset $\set {... | Statements Equivalent to Non-Dividing Type | https://proofwiki.org/wiki/Statements_Equivalent_to_Non-Dividing_Type | https://proofwiki.org/wiki/Statements_Equivalent_to_Non-Dividing_Type | [
"Model Theory for Predicate Logic"
] | [
"Definition:Theory",
"Definition:Monster Model",
"Definition:Subset",
"Definition:Type",
"Definition:Divide (Model Theory)",
"Definition:Order Indiscernible",
"Definition:Order Indiscernible",
"Definition:Embedding (Model Theory)"
] | [
"Definition:Order Indiscernible",
"Compactness Theorem",
"Compactness Theorem",
"Definition:Cardinality",
"Infinite Ramsey's Theorem",
"Compactness Theorem",
"Infinite Ramsey's Theorem"
] |
proofwiki-3987 | Fortissimo Topology is Topology | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $\tau_p$ is a topology on $T$. | === {{Open-set-axiom|3|nolink}} ===
We have that $p \in \relcomp S \O = S$ so $\O \in \tau_p$.
We have that $\relcomp S S = \O$, and Empty Set is Countable, so $S \in \tau_p$.
{{qed|lemma}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $\tau_p$ is a [[Definition:Topology|topology]] on $T$. | === {{Open-set-axiom|3|nolink}} ===
We have that $p \in \relcomp S \O = S$ so $\O \in \tau_p$.
We have that $\relcomp S S = \O$, and [[Empty Set is Countable]], so $S \in \tau_p$.
{{qed|lemma}} | Fortissimo Topology is Topology | https://proofwiki.org/wiki/Fortissimo_Topology_is_Topology | https://proofwiki.org/wiki/Fortissimo_Topology_is_Topology | [
"Fortissimo Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:Topology"
] | [
"Empty Set is Countable"
] |
proofwiki-3988 | Fortissimo Space is Completely Normal | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $T$ is a completely normal space. | We have:
:Fortissimo Space is $T_1$
:Fortissimo Space is $T_5$
and so by definition $T$ is completely normal.
{{qed}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $T$ is a [[Definition:Completely Normal Space|completely normal space]]. | We have:
:[[Fortissimo Space is T1|Fortissimo Space is $T_1$]]
:[[Fortissimo Space is T5|Fortissimo Space is $T_5$]]
and so by definition $T$ is [[Definition:Completely Normal Space|completely normal]].
{{qed}} | Fortissimo Space is Completely Normal | https://proofwiki.org/wiki/Fortissimo_Space_is_Completely_Normal | https://proofwiki.org/wiki/Fortissimo_Space_is_Completely_Normal | [
"Fortissimo Spaces",
"Examples of Completely Normal Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:Completely Normal Space"
] | [
"Fortissimo Space is T1",
"Fortissimo Space is T5",
"Definition:Completely Normal Space"
] |
proofwiki-3989 | Fortissimo Space is T1 | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $T$ is a $T_1$ space. | From Fortissimo Space is Excluded Point Space with Countable Complement Space, $T$ is an expansion of a countable complement space.
Then we have that a Countable Complement Space is $T_1$.
Then from $T_1$ Property is Preserved under Expansion, $T$ is a $T_1$ space.
{{qed}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $T$ is a [[Definition:T1 Space|$T_1$ space]]. | From [[Fortissimo Space is Excluded Point Space with Countable Complement Space]], $T$ is an [[Definition:Expansion of Topology|expansion]] of a [[Definition:Countable Complement Space|countable complement space]].
Then we have that a [[Countable Complement Space is T1|Countable Complement Space is $T_1$]].
Then fro... | Fortissimo Space is T1 | https://proofwiki.org/wiki/Fortissimo_Space_is_T1 | https://proofwiki.org/wiki/Fortissimo_Space_is_T1 | [
"Fortissimo Spaces",
"Examples of T1 Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:T1 Space"
] | [
"Fortissimo Space is Excluded Point Space with Countable Complement Space",
"Definition:Expansion of Topology",
"Definition:Countable Complement Topology",
"Countable Complement Space is T1",
"T1 Property is Preserved under Expansion",
"Definition:T1 Space"
] |
proofwiki-3990 | Fortissimo Space is Excluded Point Space with Countable Complement Space | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $\tau_p$ is the minimal topology that is generated by the excluded point topology and the countable complement topology. | Let $T_1 = \struct {S, \tau_1}$ be the excluded point space on $S$ from $p$.
Let $T_2 = \struct {S, \tau_2}$ be the countable complement space on $S$.
By definition:
:$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$
:$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is countable $\rightset{} \cup \set \... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $\tau_p$ is the [[Definition:Minimal Topology|minimal topology]] that is [[Definition:Topology Generated by Synthetic Sub-Basis|generated]] by the [[Definition:Excluded Point Topology|excluded point topology]] and the [[Definitio... | Let $T_1 = \struct {S, \tau_1}$ be the [[Definition:Excluded Point Space|excluded point space]] on $S$ from $p$.
Let $T_2 = \struct {S, \tau_2}$ be the [[Definition:Countable Complement Space|countable complement space]] on $S$.
By definition:
:$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$
:$\tau... | Fortissimo Space is Excluded Point Space with Countable Complement Space | https://proofwiki.org/wiki/Fortissimo_Space_is_Excluded_Point_Space_with_Countable_Complement_Space | https://proofwiki.org/wiki/Fortissimo_Space_is_Excluded_Point_Space_with_Countable_Complement_Space | [
"Fortissimo Spaces",
"Excluded Point Topologies",
"Countable Complement Topologies"
] | [
"Definition:Fortissimo Space",
"Definition:Minimal Topology",
"Definition:Topology Generated by Synthetic Sub-Basis",
"Definition:Excluded Point Topology",
"Definition:Countable Complement Topology"
] | [
"Definition:Excluded Point Topology",
"Definition:Countable Complement Topology",
"Definition:Countable Set",
"Definition:Fortissimo Space",
"Union is Smallest Superset"
] |
proofwiki-3991 | Fortissimo Space is T5 | Let $T = \struct {S, \tau_p}$ be a Fortissimo space on an infinite set $S$.
Then $T$ is a $T_5$ space. | Let $A, B \in \tau_p$ such that $A$ and $B$ are separated.
If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open.
Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated.
{{WLOG}}, suppose $p \in A$.
Then $p \notin B$ so $B$ is open.
Now suppose $B$ were not close... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then $T$ is a [[Definition:T5 Space|$T_5$ space]]. | Let $A, B \in \tau_p$ such that $A$ and $B$ are [[Definition:Separated Sets|separated]].
If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both [[Definition:Open Set (Topology)|open]].
Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be [[Definition:Separated Sets|separate... | Fortissimo Space is T5 | https://proofwiki.org/wiki/Fortissimo_Space_is_T5 | https://proofwiki.org/wiki/Fortissimo_Space_is_T5 | [
"Fortissimo Spaces",
"Examples of T5 Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:Infinite Set",
"Definition:T5 Space"
] | [
"Definition:Separated Sets",
"Definition:Open Set/Topology",
"Definition:Separated Sets",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Uncountable/Set",
"Definition:Fortissimo Space",
"Definition:Open Set/Topology",
"Definition:Co-Countable Set",
"Definition:Closur... |
proofwiki-3992 | Fortissimo Space is Lindelöf | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $T$ is a Lindelöf space. | Let $\CC$ be an open cover of $T$.
Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is countable.
So $U$, together with an open neighborhood of each of the elements of $\relcomp S U$, is a countable subcover of $\CC$.
Hence the result by definition of Lindelöf space.
{{qed}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]]. | Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is [[Definition:Countable Set|countable]].
So $U$, together with an [[Definition:Open Neighborhood of Point|open neighborhood]] of each of the [[Definition:Element|elements]] of $\relcomp S ... | Fortissimo Space is Lindelöf | https://proofwiki.org/wiki/Fortissimo_Space_is_Lindelöf | https://proofwiki.org/wiki/Fortissimo_Space_is_Lindelöf | [
"Fortissimo Spaces",
"Examples of Lindelöf Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:Lindelöf Space"
] | [
"Definition:Open Cover",
"Definition:Countable Set",
"Definition:Open Neighborhood/Point",
"Definition:Element",
"Definition:Subcover/Countable",
"Definition:Lindelöf Space"
] |
proofwiki-3993 | Fortissimo Space is not Sigma-Compact | Let $T = \struct {S, \tau}$ be a Fortissimo space.
Then $T$ is not a $\sigma$-compact space. | From Compact Sets in Fortissimo Space we have that the only compact sets in $T$ are finite.
A union of countably many finite sets can not be an uncountable set.
Hence the result by definition of $\sigma$-compact space.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $T$ is not a [[Definition:Sigma-Compact Space|$\sigma$-compact space]]. | From [[Compact Sets in Fortissimo Space]] we have that the only [[Definition:Compact Topological Subspace|compact sets]] in $T$ are [[Definition:Finite Set|finite]].
A [[Definition:Set Union|union]] of [[Definition:Countable Set|countably]] many [[Definition:Finite Set|finite sets]] can not be an [[Definition:Uncounta... | Fortissimo Space is not Sigma-Compact | https://proofwiki.org/wiki/Fortissimo_Space_is_not_Sigma-Compact | https://proofwiki.org/wiki/Fortissimo_Space_is_not_Sigma-Compact | [
"Fortissimo Spaces",
"Examples of Sigma-Compact Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:Sigma-Compact Space"
] | [
"Compact Sets in Fortissimo Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Finite Set",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Finite Set",
"Definition:Uncountable/Set",
"Definition:Sigma-Compact Space"
] |
proofwiki-3994 | Fortissimo Space is not Separable | Let $T = \struct {S, \tau_p}$ be a fortissimo space on an uncountable set $S$.
Then $T$ is not a separable space. | Let $U$ be a countable subset of $S$.
By the definition of the fortissimo space, $U$ is closed.
From Closed Set Equals its Closure, $U^- = U \ne S$.
Thus, by definition, $U$ is not everywhere dense in $T$.
Thus, there exists no countable subset of $S$ which is everywhere dense in $T$.
So, by definition, $T$ is not a se... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|fortissimo space]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
Then $T$ is not a [[Definition:Separable Space|separable space]]. | Let $U$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$.
By the definition of the [[Definition:Fortissimo Space|fortissimo space]], $U$ is [[Definition:Closed Set (Topology)|closed]].
From [[Closed Set Equals its Closure]], $U^- = U \ne S$.
Thus, by definition, $U$ is not [[Definition... | Fortissimo Space is not Separable | https://proofwiki.org/wiki/Fortissimo_Space_is_not_Separable | https://proofwiki.org/wiki/Fortissimo_Space_is_not_Separable | [
"Fortissimo Spaces",
"Examples of Separable Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:Uncountable/Set",
"Definition:Separable Space"
] | [
"Definition:Countable Set",
"Definition:Subset",
"Definition:Fortissimo Space",
"Definition:Closed Set/Topology",
"Set is Closed iff Equals Topological Closure",
"Definition:Everywhere Dense",
"Definition:Countable Set",
"Definition:Subset",
"Definition:Everywhere Dense",
"Definition:Separable Spa... |
proofwiki-3995 | Fortissimo Space is not First-Countable | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $T$ is not a first-countable space. | This proof follows the proof from Countable Complement Space is not First-Countable.
{{AimForCont}} that $p \in S$ has a countable local basis.
That means:
:there exists a countable set of sets $\BB_p \subseteq \tau$
such that:
:$\forall B \in \BB_p: p \in B$
and such that:
:every open neighborhood of $p$ contains some... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $T$ is not a [[Definition:First-Countable Space|first-countable space]]. | This proof follows the proof from [[Countable Complement Space is not First-Countable]].
{{AimForCont}} that $p \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
That means:
:there exists a [[Definition:Countable Set|countable]] [[Definition:Set of Sets|set of sets]] $\BB_p ... | Fortissimo Space is not First-Countable | https://proofwiki.org/wiki/Fortissimo_Space_is_not_First-Countable | https://proofwiki.org/wiki/Fortissimo_Space_is_not_First-Countable | [
"Fortissimo Spaces",
"Examples of First-Countable Spaces"
] | [
"Definition:Fortissimo Space",
"Definition:First-Countable Space"
] | [
"Countable Complement Space is not First-Countable",
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Open Neighborhood/Point",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Definition:F... |
proofwiki-3996 | Modified Fort Topology is Topology | Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $\tau_{a, b}$ is a topology on $S$. | Let $S = N \cup \set {a, b}$ where $N$ is infinite, $a \ne b$ and $a, b \notin N$.
{{Recall|Modified Fort Space|modified Fort space}}
{{:Definition:Modified Fort Space}} | Let $T = \struct {S, \tau_{a, b} }$ be a [[Definition:Modified Fort Space|modified Fort space]].
Then $\tau_{a, b}$ is a [[Definition:Topology|topology]] on $S$. | Let $S = N \cup \set {a, b}$ where $N$ is [[Definition:Infinite Set|infinite]], $a \ne b$ and $a, b \notin N$.
{{Recall|Modified Fort Space|modified Fort space}}
{{:Definition:Modified Fort Space}} | Modified Fort Topology is Topology | https://proofwiki.org/wiki/Modified_Fort_Topology_is_Topology | https://proofwiki.org/wiki/Modified_Fort_Topology_is_Topology | [
"Modified Fort Spaces"
] | [
"Definition:Modified Fort Space",
"Definition:Topology"
] | [
"Definition:Infinite Set"
] |
proofwiki-3997 | Subset of Naturals is Finite iff Bounded | Let $X$ be a subset of the natural numbers $\N$.
Then $X$ is finite {{iff}} it is bounded. | A subset of the natural numbers is also a subset of the real numbers $\R$.
By definition, a bounded subset of $\R$ is bounded below in $\R$ and bounded above in $\R$.
By the Well-Ordering Principle, $X$ is bounded below
Thus $X$ is bounded {{iff}} $X$ is bounded above.
That is, {{iff}}:
:$\exists p \in \N: \forall x \i... | Let $X$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] $\N$.
Then $X$ is [[Definition:Finite Set|finite]] {{iff}} it is [[Definition:Bounded Subset of Real Numbers|bounded]]. | A [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] is also a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers $\R$]].
By definition, a [[Definition:Bounded Subset of Real Numbers|bounded subset of $\R$]] is [[Definition:Bounded Below Set|bounded below in $\R$]... | Subset of Naturals is Finite iff Bounded | https://proofwiki.org/wiki/Subset_of_Naturals_is_Finite_iff_Bounded | https://proofwiki.org/wiki/Subset_of_Naturals_is_Finite_iff_Bounded | [
"Subsets",
"Natural Numbers"
] | [
"Definition:Subset",
"Definition:Natural Numbers",
"Definition:Finite Set",
"Definition:Bounded Set/Real Numbers"
] | [
"Definition:Subset",
"Definition:Natural Numbers",
"Definition:Subset",
"Definition:Real Number",
"Definition:Bounded Set/Real Numbers",
"Definition:Bounded Below Set",
"Definition:Bounded Above Set",
"Well-Ordering Principle",
"Definition:Bounded Below Set",
"Definition:Bounded Set/Real Numbers",... |
proofwiki-3998 | Arens-Fort Topology is Topology | Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $\tau$ is a topology on $T$. | === {{Open-set-axiom|3|nolink}} ===
We have that $\tuple {0, 0} \notin \O$ so $\O \in \tau$.
We have that $\forall m: \set {n: \tuple {m, n} \notin S} = \O$ which is finite, so $S \in \tau$.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
Then $\tau$ is a [[Definition:Topology|topology]] on $T$. | === {{Open-set-axiom|3|nolink}} ===
We have that $\tuple {0, 0} \notin \O$ so $\O \in \tau$.
We have that $\forall m: \set {n: \tuple {m, n} \notin S} = \O$ which is [[Definition:Finite Set|finite]], so $S \in \tau$.
{{qed|lemma}} | Arens-Fort Topology is Topology | https://proofwiki.org/wiki/Arens-Fort_Topology_is_Topology | https://proofwiki.org/wiki/Arens-Fort_Topology_is_Topology | [
"Arens-Fort Space"
] | [
"Definition:Arens-Fort Space",
"Definition:Topology"
] | [
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Finite Set"
] |
proofwiki-3999 | Arens-Fort Space is Expansion of Countable Fort Space | Let $T = \struct {S, \tau}$ be the Arens-Fort space, where $S = \Z_{\ge 0} \times \Z_{\ge 0}$.
Let $T_p = \struct {S, \tau_p}$ be the Fort space on $S$ where $p = \left({0, 0}\right)$.
Then $\tau$ is an expansion of $\tau_p$.
Furthermore, $S$ is countably infinite, so $T_p$ is a countable Fort space. | Let $H \in \tau_p$ where $p = \tuple {0, 0}$.
Then either:
:$(1): \quad \tuple {0, 0} \in \relcomp S H$
or:
:$(2): \quad H$ is cofinite in $S$, that is $\relcomp S H$ is finite.
Case $(1)$ means that $\tuple {0, 0} \notin H$ and so $H \in \tau$ by definition of the Arens-Fort space.
Suppose case $(2)$ applies.
Then for... | Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]], where $S = \Z_{\ge 0} \times \Z_{\ge 0}$.
Let $T_p = \struct {S, \tau_p}$ be the [[Definition:Fort Space|Fort space]] on $S$ where $p = \left({0, 0}\right)$.
Then $\tau$ is an [[Definition:Expansion of Topology|expansion]] of $\tau_... | Let $H \in \tau_p$ where $p = \tuple {0, 0}$.
Then either:
:$(1): \quad \tuple {0, 0} \in \relcomp S H$
or:
:$(2): \quad H$ is [[Definition:Cofinite Subset|cofinite]] in $S$, that is $\relcomp S H$ is [[Definition:Finite Set|finite]].
Case $(1)$ means that $\tuple {0, 0} \notin H$ and so $H \in \tau$ by definition o... | Arens-Fort Space is Expansion of Countable Fort Space | https://proofwiki.org/wiki/Arens-Fort_Space_is_Expansion_of_Countable_Fort_Space | https://proofwiki.org/wiki/Arens-Fort_Space_is_Expansion_of_Countable_Fort_Space | [
"Arens-Fort Space",
"Countable Fort Spaces",
"Examples of Expansions of Topologies"
] | [
"Definition:Arens-Fort Space",
"Definition:Fort Space",
"Definition:Expansion of Topology",
"Definition:Countably Infinite/Set",
"Definition:Fort Space/Countable"
] | [
"Definition:Cofinite Subset",
"Definition:Finite Set",
"Definition:Arens-Fort Space",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Infinite Set",
"Definition:Expansion of Topology",
"Arens-Fort Space is Countable",
"Definition:Countably Infinite/Set"
] |
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