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 |
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proofwiki-3500 | Countability Axioms Preserved under Open Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ has one of the following properties, then $T_B$ has the same property:
:First-Countability
:Second-Countability | === Proof for First-Countability ===
{{:First-Countability is Preserved under Open Continuous Surjection}} | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Surjection|surjective]] [[Definition:Open Mapping|open mapping]] which is also [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
If $... | === [[First-Countability is Preserved under Open Continuous Surjection|Proof for First-Countability]] ===
{{:First-Countability is Preserved under Open Continuous Surjection}} | Countability Axioms Preserved under Open Continuous Surjection | https://proofwiki.org/wiki/Countability_Axioms_Preserved_under_Open_Continuous_Surjection | https://proofwiki.org/wiki/Countability_Axioms_Preserved_under_Open_Continuous_Surjection | [
"Countability Axioms",
"Open Mappings",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Surjection",
"Definition:Open Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:First-Countable Space",
"Definition:Second-Countable Space"
] | [
"First-Countability is Preserved under Open Continuous Surjection"
] |
proofwiki-3501 | Connectedness of Points is Equivalence Relation | Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $a \sim b $ denote the [[Definition:Relation|relation]]:
:$a \sim b \iff a$ is [[Definition:Connected Points (Topology)|connected]] to $b$
where $a, b \in S$.
Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relat... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Connectedness of Points is Equivalence Relation | https://proofwiki.org/wiki/Connectedness_of_Points_is_Equivalence_Relation | https://proofwiki.org/wiki/Connectedness_of_Points_is_Equivalence_Relation | [
"Connected Points (Topology)",
"Examples of Equivalence Relations"
] | [
"Definition:Topological Space",
"Definition:Relation",
"Definition:Connected Points (Topology)",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-3502 | Atlas Belongs to Unique Differentiable Structure | Let $M$ be a locally Euclidean space of dimension $d$.
Let $\AA$ be an atlas on $M$.
Then there exists a unique differentiable structure $\FF$ on $M$ with $\AA \in \FF$. | Let $\FF$ be the equivalence class of $\AA$ under the relation of compatibility.
By Compatibility of Atlases is Equivalence Relation, this is indeed an equivalence relation.
By definition we have $\AA \in \FF$.
By Relation Partitions Set iff Equivalence, $\FF$ is an element of the partition of equivalence classes.
By d... | Let $M$ be a [[Definition:Locally Euclidean Space|locally Euclidean space of dimension $d$]].
Let $\AA$ be an [[Definition:Atlas|atlas]] on $M$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Differentiable Structure|differentiable structure]] $\FF$ on $M$ with $\AA \in \FF$. | Let $\FF$ be the [[Definition:Equivalence Class|equivalence class]] of $\AA$ under the [[Definition:Relation|relation]] of [[Definition:Compatible Atlases|compatibility]].
By [[Compatibility of Atlases is Equivalence Relation]], this is indeed an [[Definition:Equivalence Relation|equivalence relation]].
By definition... | Atlas Belongs to Unique Differentiable Structure | https://proofwiki.org/wiki/Atlas_Belongs_to_Unique_Differentiable_Structure | https://proofwiki.org/wiki/Atlas_Belongs_to_Unique_Differentiable_Structure | [
"Manifolds",
"Topological Manifolds"
] | [
"Definition:Locally Euclidean Space",
"Definition:Atlas",
"Definition:Unique",
"Definition:Differentiable Structure"
] | [
"Definition:Equivalence Class",
"Definition:Relation",
"Definition:Compatible Atlases",
"Compatibility of Atlases is Equivalence Relation",
"Definition:Equivalence Relation",
"Relation Partitions Set iff Equivalence",
"Definition:Element",
"Definition:Set Partition",
"Definition:Equivalence Class",
... |
proofwiki-3503 | Surgery for Rings | Let $R$ and $S$ be commutative rings with unity, and $\phi: R \to S$ a ring monomorphism.
Then there is a ring $T$ isomorphic to $S$ that contains $R$ as a subring. | Let $T$ be the disjoint union $T = R \cup \paren {S \setminus \Img \phi}$.
Define $\theta : T \to S$ as follows:
:If $x \in R$, then $\map \theta x = \map \phi x$
:If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$
We claim that $\theta$ is an isomorphism.
''Injectivity'': Let $\map \theta x = \map \the... | Let $R$ and $S$ be [[Definition:Commutative and Unitary Ring|commutative rings with unity]], and $\phi: R \to S$ a [[Definition:Ring Monomorphism|ring monomorphism]].
Then there is a ring $T$ [[Definition:Ring Isomorphism|isomorphic]] to $S$ that contains $R$ as a [[Definition:Subring|subring]]. | Let $T$ be the [[Definition:Disjoint Union (Set Theory)|disjoint union]] $T = R \cup \paren {S \setminus \Img \phi}$.
Define $\theta : T \to S$ as follows:
:If $x \in R$, then $\map \theta x = \map \phi x$
:If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$
We claim that $\theta$ is an isomorphism.
... | Surgery for Rings | https://proofwiki.org/wiki/Surgery_for_Rings | https://proofwiki.org/wiki/Surgery_for_Rings | [
"Ring Isomorphisms",
"Ring Monomorphisms"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Monomorphism",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Subring"
] | [
"Definition:Disjoint Union (Set Theory)",
"Definition:Identity Mapping",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Embedding Theorem",
"Category:Ring Isomorphisms",
"Category:Ring Monomorphisms"
] |
proofwiki-3504 | Clopen Set contains Components of All its Points | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be both closed and open in $T$.
Then $H$ contains the components of all of its points. | Let $H$ be a clopen set in $T$.
By definition, $H$ is open and so $H \in \tau$.
But as $H$ is also closed, by definition $\relcomp S H \in \tau$ where $\complement_S$ denotes complement relative to $S$.
Thus $H$ and $\relcomp S H$ are both open such that:
:$H \cap \relcomp S H = \O$ from Intersection with Relative Comp... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$ be [[Definition:Clopen Set|both closed and open]] in $T$.
Then $H$ contains the [[Definition:Component (Topology)|components]] of all of its [[Definition:Point of Set|points]]. | Let $H$ be a [[Definition:Clopen Set|clopen set]] in $T$.
By definition, $H$ is [[Definition:Open Set (Topology)|open]] and so $H \in \tau$.
But as $H$ is also [[Definition:Closed Set (Topology)|closed]], by definition $\relcomp S H \in \tau$ where $\complement_S$ denotes [[Definition:Relative Complement|complement r... | Clopen Set contains Components of All its Points | https://proofwiki.org/wiki/Clopen_Set_contains_Components_of_All_its_Points | https://proofwiki.org/wiki/Clopen_Set_contains_Components_of_All_its_Points | [
"Clopen Sets",
"Components (Topology)"
] | [
"Definition:Topological Space",
"Definition:Clopen Set",
"Definition:Component (Topology)",
"Definition:Element"
] | [
"Definition:Clopen Set",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Relative Complement",
"Definition:Open Set/Topology",
"Intersection with Relative Complement is Empty",
"Union with Relative Complement",
"Definition:Separation (Topology)",
"Definition:Closed Set/... |
proofwiki-3505 | Connectedness Between Two Points is an Equivalence Relation | Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff \exists S \subseteq T: S$ is connected between the two points $a$ and $b$
where $a, b \in X$.
Then $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $a \sim b $ denote the [[Definition:Relation|relation]]:
:$a \sim b \iff \exists S \subseteq T: S$ is [[Definition:Connected Between Two Points|connected between the two points ]] $a$ and $b$
where $a, b \in X$.
Then $\sim$ is a... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Connectedness Between Two Points is an Equivalence Relation | https://proofwiki.org/wiki/Connectedness_Between_Two_Points_is_an_Equivalence_Relation | https://proofwiki.org/wiki/Connectedness_Between_Two_Points_is_an_Equivalence_Relation | [
"Examples of Equivalence Relations",
"Connectedness Between Two Points"
] | [
"Definition:Topological Space",
"Definition:Relation",
"Definition:Connected Between Two Points",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-3506 | Quasicomponent is Intersection of Clopen Sets | Let $T = \struct {S, \tau}$ be a topological space.
Let $p \in S$.
Then the quasicomponent containing $p$ equals the intersection of all sets which are both open and closed which contain $p$. | Let $C$ be the quasicomponent of $p$.
Let $Q$ be the set of clopen sets containing $p$ that are not equal to $S$.
First let $x \in C$.
Let $U \in Q$.
By definition of $Q$:
:$p \in U$
Let $V = \relcomp S U$ be the complement of $U$ relative to $S$.
By Complement of Clopen Set is Clopen $V$ is also a clopen set of $T$.
B... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $p \in S$.
Then the [[Definition:Quasicomponent|quasicomponent]] containing $p$ equals the [[Definition:Set Intersection|intersection]] of all [[Definition:Clopen Set|sets which are both open and closed]] which contain $p$. | Let $C$ be the [[Definition:Quasicomponent|quasicomponent]] of $p$.
Let $Q$ be the [[Definition:Set|set]] of [[Definition:Clopen Set|clopen sets]] containing $p$ that are not equal to $S$.
First let $x \in C$.
Let $U \in Q$.
By definition of $Q$:
:$p \in U$
Let $V = \relcomp S U$ be the [[Definition:Relative Comp... | Quasicomponent is Intersection of Clopen Sets | https://proofwiki.org/wiki/Quasicomponent_is_Intersection_of_Clopen_Sets | https://proofwiki.org/wiki/Quasicomponent_is_Intersection_of_Clopen_Sets | [
"Clopen Sets",
"Connectedness Between Two Points"
] | [
"Definition:Topological Space",
"Definition:Quasicomponent",
"Definition:Set Intersection",
"Definition:Clopen Set"
] | [
"Definition:Quasicomponent",
"Definition:Set",
"Definition:Clopen Set",
"Definition:Relative Complement",
"Complement of Clopen Set is Clopen",
"Definition:Clopen Set",
"Clopen Set and Complement form Separation",
"Definition:Separation (Topology)",
"Definition:Quasicomponent",
"Definition:Separat... |
proofwiki-3507 | Topological Space with One Quasicomponent is Connected | Let $T = \struct {S, \tau}$ be a topological space which has one quasicomponent.
Then $T$ is connected. | Let $x \in S$.
By hypothesis, the quasicomponent of $x$ is $S$ itself.
Thus by definition of quasicomponent:
:$\forall y \in S: y \sim x$
where $\sim$ is the relation defined on $T$ as:
:$x \sim y \iff T$ is connected between the two points $x$ and $y$
Let $K = \ds \bigcap_{x \mathop \in U} U: U$ is clopen in $T$.
By Q... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which has one [[Definition:Quasicomponent|quasicomponent]].
Then $T$ is [[Definition:Connected Topological Space|connected]]. | Let $x \in S$.
[[Definition:By Hypothesis|By hypothesis]], the [[Definition:Quasicomponent|quasicomponent]] of $x$ is $S$ itself.
Thus by definition of [[Definition:Quasicomponent|quasicomponent]]:
:$\forall y \in S: y \sim x$
where $\sim$ is the [[Definition:Relation|relation]] defined on $T$ as:
:$x \sim y \iff ... | Topological Space with One Quasicomponent is Connected | https://proofwiki.org/wiki/Topological_Space_with_One_Quasicomponent_is_Connected | https://proofwiki.org/wiki/Topological_Space_with_One_Quasicomponent_is_Connected | [
"Connectedness Between Two Points",
"Quasicomponents",
"Connected Topological Spaces"
] | [
"Definition:Topological Space",
"Definition:Quasicomponent",
"Definition:Connected Topological Space"
] | [
"Definition:By Hypothesis",
"Definition:Quasicomponent",
"Definition:Quasicomponent",
"Definition:Relation",
"Definition:Connected Between Two Points",
"Definition:Clopen Set",
"Quasicomponent is Intersection of Clopen Sets",
"Definition:Non-Empty Set",
"Definition:Clopen Set",
"Definition:Connect... |
proofwiki-3508 | Localization of Ring Exists | Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.
Then there exists a localization $\struct {A_S, \iota}$ of $A$ at $S$. | Define a relation $\sim$ on the Cartesian product $A \times S$ by:
:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$ | Let $A$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $S \subseteq A$ be a [[Definition:Multiplicatively Closed Subset of Ring|multiplicatively closed subset]] with $0 \notin S$.
Then there exists a [[Definition:Localization of Ring|localization]] $\struct {A_S, \iota}$ of $A$ at ... | Define a [[Definition:Relation|relation]] $\sim$ on the [[Definition:Cartesian Product|Cartesian product]] $A \times S$ by:
:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$ | Localization of Ring Exists | https://proofwiki.org/wiki/Localization_of_Ring_Exists | https://proofwiki.org/wiki/Localization_of_Ring_Exists | [
"Localization of Rings",
"Localization of Ring Exists"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Multiplicatively Closed Subset of Ring",
"Definition:Localization of Ring"
] | [
"Definition:Relation",
"Definition:Cartesian Product"
] |
proofwiki-3509 | Transitivity of Integrality | Let $A \subseteq B \subseteq C$ be extensions of commutative rings with unity.
Suppose that $C$ is integral over $B$, and $B$ is integral over $A$.
Then $C$ is integral over $A$. | First, a lemma: | Let $A \subseteq B \subseteq C$ be [[Definition:Ring Extension|extensions]] of [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
Suppose that $C$ is [[Definition:Integral Ring Extension|integral]] over $B$, and $B$ is [[Definition:Integral Ring Extension|integral]] over $A$.
Then $C$ is [[Defi... | First, a lemma: | Transitivity of Integrality | https://proofwiki.org/wiki/Transitivity_of_Integrality | https://proofwiki.org/wiki/Transitivity_of_Integrality | [
"Algebraic Number Theory",
"Commutative Algebra",
"Transitivity of Integrality"
] | [
"Definition:Ring Extension",
"Definition:Commutative and Unitary Ring",
"Definition:Integral Ring Extension",
"Definition:Integral Ring Extension",
"Definition:Integral Ring Extension"
] | [] |
proofwiki-3510 | Integral Closure is Integrally Closed | Let $A \subseteq B$ be an extension of commutative rings with unity.
Let $C$ be the integral closure of $A$ in $B$.
Then $C$ is integrally closed. | Suppose $x \in B$ is integral over $C$.
Certainly $C$ is integral over $A$, so by Transitivity of Integrality, $C \sqbrk x$ is integral over $A$.
In particular, $x$ is integral over $A$, so $x \in C$.
{{Qed}}
Category:Algebraic Number Theory
Category:Commutative Algebra
3g683hc822lfpsjlmes125b6m8fhhme | Let $A \subseteq B$ be an [[Definition:Ring Extension|extension]] of [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
Let $C$ be the [[Definition:Integral Closure|integral closure]] of $A$ in $B$.
Then $C$ is [[Definition:Integrally Closed|integrally closed]]. | Suppose $x \in B$ is [[Definition:Integral Element of Ring Extension|integral]] over $C$.
Certainly $C$ is [[Definition:Integral Element of Ring Extension|integral]] over $A$, so by [[Transitivity of Integrality]], $C \sqbrk x$ is [[Definition:Integral Element of Ring Extension|integral]] over $A$.
In particular, $x$... | Integral Closure is Integrally Closed | https://proofwiki.org/wiki/Integral_Closure_is_Integrally_Closed | https://proofwiki.org/wiki/Integral_Closure_is_Integrally_Closed | [
"Algebraic Number Theory",
"Commutative Algebra"
] | [
"Definition:Ring Extension",
"Definition:Commutative and Unitary Ring",
"Definition:Integral Closure",
"Definition:Integrally Closed"
] | [
"Definition:Integral Element of Ring Extension",
"Definition:Integral Element of Ring Extension",
"Transitivity of Integrality",
"Definition:Integral Element of Ring Extension",
"Definition:Integral Element of Ring Extension",
"Category:Algebraic Number Theory",
"Category:Commutative Algebra"
] |
proofwiki-3511 | Universal Property for Field of Quotients | Let $\struct {D, +, \circ}$ be an integral domain.
Let $\struct {F, \oplus, \cdot}$ be a field of quotients of $D$.
Then $F$ satisfies the following universal property:
There exists a (ring) homomorphism $\iota : D \to F$ such that:
::for every field $\tilde F$ and
:and:
::for every (ring) homomorphism $\phi: D \to \ti... | {{ProofWanted}}
Category:Fields of Quotients
Category:Integral Domains
Category:Universal Properties
ci1x76tsuac3v1qn7xlmxos3tz8cv19 | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
Let $\struct {F, \oplus, \cdot}$ be a [[Definition:Field of Quotients|field of quotients]] of $D$.
Then $F$ satisfies the following [[Definition:Universal Property|universal property]]:
There exists a [[Definition:Ring Homomorphism|(r... | {{ProofWanted}}
[[Category:Fields of Quotients]]
[[Category:Integral Domains]]
[[Category:Universal Properties]]
ci1x76tsuac3v1qn7xlmxos3tz8cv19 | Universal Property for Field of Quotients | https://proofwiki.org/wiki/Universal_Property_for_Field_of_Quotients | https://proofwiki.org/wiki/Universal_Property_for_Field_of_Quotients | [
"Fields of Quotients",
"Integral Domains",
"Universal Properties"
] | [
"Definition:Integral Domain",
"Definition:Field of Quotients",
"Definition:Universal Property",
"Definition:Ring Homomorphism",
"Definition:Field (Abstract Algebra)",
"Definition:Ring Homomorphism",
"Definition:Field Homomorphism",
"Definition:Commutative Diagram"
] | [
"Category:Fields of Quotients",
"Category:Integral Domains",
"Category:Universal Properties"
] |
proofwiki-3512 | Constant Mapping is Continuous | Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.
Let $b \in B$ be any point in $B$.
Let $f_b: A \to B$ be the constant mapping defined by:
:$\forall x \in A: \map {f_b} x = b$
Then $f_b$ is continuous. | We have by definition that:
:$\forall x \in A: \map {f_b} x = b$
So:
$\map {f^{-1} } b = A$
For $c \in B: c \ne B$ we have that:
$\map {f^{-1} } c = \O$
Let $U \in \tau_B$ such that $b \in U$.
Then $f^{-1} \sqbrk U = A$
Let $V \in \tau_B$ such that $b \notin V$.
Then $f^{-1} \sqbrk V = \O$
From the definition of topolo... | Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $b \in B$ be any point in $B$.
Let $f_b: A \to B$ be the [[Definition:Constant Mapping|constant mapping]] defined by:
:$\forall x \in A: \map {f_b} x = b$
Then $f_b$ is [[Definition:Continuous... | We have by definition that:
:$\forall x \in A: \map {f_b} x = b$
So:
$\map {f^{-1} } b = A$
For $c \in B: c \ne B$ we have that:
$\map {f^{-1} } c = \O$
Let $U \in \tau_B$ such that $b \in U$.
Then $f^{-1} \sqbrk U = A$
Let $V \in \tau_B$ such that $b \notin V$.
Then $f^{-1} \sqbrk V = \O$
From the definiti... | Constant Mapping is Continuous | https://proofwiki.org/wiki/Constant_Mapping_is_Continuous | https://proofwiki.org/wiki/Constant_Mapping_is_Continuous | [
"Continuous Mappings",
"Constant Mappings"
] | [
"Definition:Topological Space",
"Definition:Constant Mapping",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Topology",
"Definition:Open Set/Topology",
"Empty Set is Element of Topology",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-3513 | Path-Connectedness is Equivalence Relation | Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is path-connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $a \sim b $ denote the [[Definition:Relation|relation]]:
:$a \sim b \iff a$ is [[Definition:Path-Connected Points|path-connected]] to $b$
where $a, b \in S$.
Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relati... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Path-Connectedness is Equivalence Relation | https://proofwiki.org/wiki/Path-Connectedness_is_Equivalence_Relation | https://proofwiki.org/wiki/Path-Connectedness_is_Equivalence_Relation | [
"Path-Connected Spaces",
"Examples of Equivalence Relations"
] | [
"Definition:Topological Space",
"Definition:Relation",
"Definition:Path-Connected/Points",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-3514 | Injective Path-Connectedness is Equivalence Relation | Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is injectively path-connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $a \sim b $ denote the [[Definition:Relation|relation]]:
:$a \sim b \iff a$ is [[Definition:Injectively Path-Connected Points|injectively path-connected]] to $b$
where $a, b \in S$.
Then $\sim$ is an [[Definition:Equivalence Rel... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Injective Path-Connectedness is Equivalence Relation | https://proofwiki.org/wiki/Injective_Path-Connectedness_is_Equivalence_Relation | https://proofwiki.org/wiki/Injective_Path-Connectedness_is_Equivalence_Relation | [
"Examples of Equivalence Relations",
"Injective Path-Connectedness"
] | [
"Definition:Topological Space",
"Definition:Relation",
"Definition:Injectively Path-Connected/Points",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-3515 | Joining Injective Paths makes Another Injective Path | Let $T$ be a topological space.
Let $\mathbb I \subseteq \R$ be the closed unit interval $\closedint 0 1$.
Let $a, b, c$ be three distinct points of $T$.
Let $f, g: \mathbb I \to T$ be injective paths in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
Let $h: \mathbb I \to T$ be the mapping given by:
:<nowiki>$\m... | From Injective Path in Topological Space is Path, $f$ and $g$ are also paths in $T$.
So by Joining Paths makes Another Path it follows that $h$ is a path in $T$.
Now if $\Img f \cap \Img g = b$ it can be seen that:
:<nowiki>$\forall x \in \Img h : x = \begin{cases}
\map f y & \text {for some $y \in \closedint 0 {\dfrac... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $\mathbb I \subseteq \R$ be the [[Definition:Closed Unit Interval|closed unit interval]] $\closedint 0 1$.
Let $a, b, c$ be three distinct points of $T$.
Let $f, g: \mathbb I \to T$ be [[Definition:Injective Path|injective paths]] in $T$ from $a$ t... | From [[Injective Path in Topological Space is Path]], $f$ and $g$ are also [[Definition:Path (Topology)|paths]] in $T$.
So by [[Joining Paths makes Another Path]] it follows that $h$ is a [[Definition:Path (Topology)|path]] in $T$.
Now if $\Img f \cap \Img g = b$ it can be seen that:
:<nowiki>$\forall x \in \Img h : ... | Joining Injective Paths makes Another Injective Path | https://proofwiki.org/wiki/Joining_Injective_Paths_makes_Another_Injective_Path | https://proofwiki.org/wiki/Joining_Injective_Paths_makes_Another_Injective_Path | [
"Injective Paths"
] | [
"Definition:Topological Space",
"Definition:Real Interval/Unit Interval/Closed",
"Definition:Injective Path",
"Definition:Mapping",
"Definition:Injective Path",
"Definition:Restriction/Mapping",
"Definition:Injective Path"
] | [
"Injective Path in Topological Space is Path",
"Definition:Path (Topology)",
"Joining Paths makes Another Path",
"Definition:Path (Topology)",
"Definition:Injection",
"Definition:Injective Path"
] |
proofwiki-3516 | Injective Path in Topological Space is Path | Let $T = \struct {X, \tau}$ be a topological space.
Let $f$ be an injective path in $T$.
Then $f$ is a path in $T$. | By definition, an injective path from $a$ to $b$ is a path $f: I \to T$ such that $f$ is injective.
Hence the result, by definition.
{{qed}} | Let $T = \struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $f$ be an [[Definition:Injective Path|injective path]] in $T$.
Then $f$ is a [[Definition:Path (Topology)|path]] in $T$. | By definition, an [[Definition:Injective Path|injective path]] from $a$ to $b$ is a [[Definition:Path (Topology)|path]] $f: I \to T$ such that $f$ is [[Definition:Injection|injective]].
Hence the result, by definition.
{{qed}} | Injective Path in Topological Space is Path | https://proofwiki.org/wiki/Injective_Path_in_Topological_Space_is_Path | https://proofwiki.org/wiki/Injective_Path_in_Topological_Space_is_Path | [
"Injective Paths",
"Paths (Topology)",
"Injective Path-Connectedness",
"Path-Connectedness"
] | [
"Definition:Topological Space",
"Definition:Injective Path",
"Definition:Path (Topology)"
] | [
"Definition:Injective Path",
"Definition:Path (Topology)",
"Definition:Injection"
] |
proofwiki-3517 | Relationship between Component Types | Let $T = \struct {S, \tau}$ be a topological space.
Let $p \in S$.
Let:
:$A$ be the arc component of $p$
:$J$ be the injective path component of $p$
:$P$ be the path component of $p$
:$C$ be the component of $p$
:$Q$ be the quasicomponent of $p$.
Then:
:$A \subseteq P \subseteq C \subseteq Q$
In general, the inclusions... | Let $f \in A$.
By Arc in Topological Space is Injective Path we have that $f \in J$.
That is, $A \subseteq J$.
{{qed|lemma}}
Let $f \in J$.
By Injective Path in Topological Space is Path we have that $f \in P$.
That is, $J \subseteq P$.
{{qed|lemma}}
Let $f \in P$.
From Path-Connected Space is Connected we have directl... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $p \in S$.
Let:
:$A$ be the [[Definition:Arc Component|arc component]] of $p$
:$J$ be the [[Definition:Injective Path Component|injective path component]] of $p$
:$P$ be the [[Definition:Path Component|path component]] of $p$
:$... | Let $f \in A$.
By [[Arc in Topological Space is Injective Path]] we have that $f \in J$.
That is, $A \subseteq J$.
{{qed|lemma}}
Let $f \in J$.
By [[Injective Path in Topological Space is Path]] we have that $f \in P$.
That is, $J \subseteq P$.
{{qed|lemma}}
Let $f \in P$.
From [[Path-Connected Space is Connec... | Relationship between Component Types | https://proofwiki.org/wiki/Relationship_between_Component_Types | https://proofwiki.org/wiki/Relationship_between_Component_Types | [
"Path Components",
"Injective Path Components",
"Arc Components",
"Quasicomponents",
"Components (Topology)",
"Topological Connectedness"
] | [
"Definition:Topological Space",
"Definition:Arc Component",
"Definition:Injective Path Component",
"Definition:Path Component",
"Definition:Component (Topology)",
"Definition:Quasicomponent",
"Definition:Subset"
] | [
"Arc in Topological Space is Injective Path",
"Injective Path in Topological Space is Path",
"Path-Connected Space is Connected",
"Connected Space is Connected Between Two Points",
"Injective Path Component is not necessarily Arc Component",
"Path Component is not necessarily Injective Path Component",
... |
proofwiki-3518 | Increasing Sequence in Ordered Set Terminates iff Maximal Element | Let $\struct {P, \le}$ be an ordered set.
{{TFAE}}
:$(1): \quad$ Every increasing sequence $x_1 \le x_2 \le x_3 \le \cdots$ with $x_i \in P$ eventually terminates: there is $n \in \N$ such that $x_n = x_{n + 1} = \cdots$.
:$(2): \quad$ Every non-empty subset of $P$ has a maximal element. | === $(1)$ implies $(2)$ ===
Suppose $(1)$ holds.
Pick $\O \ne S \subseteq P$.
Let $x_1 \in S$ be arbitrary.
Given $x_k \in S$, pick $x_{k + 1} \in S$ strictly bigger than $x_k$.
By hypothesis the process must eventually terminate, say $x_n$ is the last element.
{{explain|Why does this terminate? I think Axiom:Axiom of ... | Let $\struct {P, \le}$ be an [[Definition:Ordering|ordered set]].
{{TFAE}}
:$(1): \quad$ Every increasing [[Definition:Sequence|sequence]] $x_1 \le x_2 \le x_3 \le \cdots$ with $x_i \in P$ eventually terminates: there is $n \in \N$ such that $x_n = x_{n + 1} = \cdots$.
:$(2): \quad$ Every non-empty subset of $P$ ha... | === $(1)$ implies $(2)$ ===
Suppose $(1)$ holds.
Pick $\O \ne S \subseteq P$.
Let $x_1 \in S$ be arbitrary.
Given $x_k \in S$, pick $x_{k + 1} \in S$ strictly bigger than $x_k$.
By hypothesis the process must eventually terminate, say $x_n$ is the last element.
{{explain|Why does this terminate? I think [[Axiom:Ax... | Increasing Sequence in Ordered Set Terminates iff Maximal Element | https://proofwiki.org/wiki/Increasing_Sequence_in_Ordered_Set_Terminates_iff_Maximal_Element | https://proofwiki.org/wiki/Increasing_Sequence_in_Ordered_Set_Terminates_iff_Maximal_Element | [
"Increasing Sequences"
] | [
"Definition:Ordering",
"Definition:Sequence",
"Definition:Maximal"
] | [
"Axiom:Axiom of Choice"
] |
proofwiki-3519 | Characterisation of UFDs | Let $A$ be an integral domain.
{{TFAE}}
:$(1): \quad A$ is a unique factorisation domain
:$(2): \quad A$ is a GCD domain satisfying the ascending chain condition on principal ideals.
:$(3): \quad A$ satisfies the ascending chain condition on principal ideals and every irreducible element of $A$ is a prime element of $... | {{tidy}}
{{MissingLinks}}
We use the notation $a \sim b$ to mean $a \mid b$ and $b \mid a$, or equivalently that $a = bu$ for some unit $u \in A^\times$. We say $a$ and $b$ are "associates".
$\left( 1 \rightarrow 2 \right)$ Let $a \sim p_1^{\alpha_1} \cdots p_k ^ {\alpha_k}$, $b \sim p_1^{\beta_1} \cdots p_k ^ {\beta_k... | Let $A$ be an [[Definition:Integral Domain|integral domain]].
{{TFAE}}
:$(1): \quad A$ is a [[Definition:Unique Factorization Domain|unique factorisation domain]]
:$(2): \quad A$ is a [[Definition:GCD Domain|GCD domain]] satisfying the [[Definition:Ascending Chain Condition|ascending chain condition]] on [[Definit... | {{tidy}}
{{MissingLinks}}
We use the notation $a \sim b$ to mean $a \mid b$ and $b \mid a$, or equivalently that $a = bu$ for some unit $u \in A^\times$. We say $a$ and $b$ are "associates".
$\left( 1 \rightarrow 2 \right)$ Let $a \sim p_1^{\alpha_1} \cdots p_k ^ {\alpha_k}$, $b \sim p_1^{\beta_1} \cdots p_k ^ {\bet... | Characterisation of UFDs | https://proofwiki.org/wiki/Characterisation_of_UFDs | https://proofwiki.org/wiki/Characterisation_of_UFDs | [
"Unique Factorization Domains",
"Factorization"
] | [
"Definition:Integral Domain",
"Definition:Unique Factorization Domain",
"Definition:GCD Domain",
"Definition:Ascending Chain Condition",
"Definition:Principal Ideal of Ring",
"Definition:Ascending Chain Condition",
"Definition:Principal Ideal of Ring",
"Definition:Irreducible Element of Ring",
"Defi... | [
"Definition:GCD Domain",
"Definition:Ascending Chain Condition",
"Definition:Principal Ideal of Ring",
"Definition:Ascending Chain Condition",
"Definition:Principal Ideal of Ring",
"Cancellation Law for Ring Product of Integral Domain",
"Category:Unique Factorization Domains",
"Category:Factorization"... |
proofwiki-3520 | Ultraconnected Space is Path-Connected | Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is path-connected. | Let $T = \struct {S, \tau}$ be ultraconnected.
Let $a, b \in S$.
Let $p \in \set a^- \cap \set b^-$ where $\set a^-$ is the closure of $\set a$.
Such a $p$ can be chosen, as $T$ being ultraconnected guarantees that $\set a^- \cap \set b^- \ne \O$.
Consider the mapping $f: \closedint 0 1 \to X$ such that:
:<nowiki>$\map... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Ultraconnected Space|ultraconnected]].
Then $T$ is [[Definition:Path-Connected Space|path-connected]]. | Let $T = \struct {S, \tau}$ be [[Definition:Ultraconnected Space|ultraconnected]].
Let $a, b \in S$.
Let $p \in \set a^- \cap \set b^-$ where $\set a^-$ is the [[Definition:Closure (Topology)|closure]] of $\set a$.
Such a $p$ can be chosen, as $T$ being [[Definition:Ultraconnected Space|ultraconnected]] guarantees t... | Ultraconnected Space is Path-Connected | https://proofwiki.org/wiki/Ultraconnected_Space_is_Path-Connected | https://proofwiki.org/wiki/Ultraconnected_Space_is_Path-Connected | [
"Ultraconnected Spaces",
"Path-Connected Spaces",
"Sequence of Implications of Connectedness Properties"
] | [
"Definition:Topological Space",
"Definition:Ultraconnected Space",
"Definition:Path-Connected/Topological Space"
] | [
"Definition:Ultraconnected Space",
"Definition:Closure (Topology)",
"Definition:Ultraconnected Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Path-Connected/Topological Space"
] |
proofwiki-3521 | Continuous Real-Valued Function on Irreducible Space is Constant | Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Let $f: S \to \R$ be a continuous real-valued function.
Then $f$ is constant, that is:
:$\exists a \in \R: \forall x \in S: \map f x = a$ | {{Recall|Irreducible Space|irreducible space}}
{{:Definition:Irreducible Space/Definition 3}}
For every $x \in \R$, define:
:$L_x := f^{-1} \sqbrk {\openint {-\infty} x}$
:$U_x := f^{-1} \sqbrk {\openint x \infty}$
By continuity of $f$, these are open in $T$.
They are also disjoint, because for each $s \in S$, $\map f ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
Let $f: S \to \R$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Real-Valued Function|real-valued function]].
Then $f$ is [[Definition:Const... | {{Recall|Irreducible Space|irreducible space}}
{{:Definition:Irreducible Space/Definition 3}}
For every $x \in \R$, define:
:$L_x := f^{-1} \sqbrk {\openint {-\infty} x}$
:$U_x := f^{-1} \sqbrk {\openint x \infty}$
By [[Definition:Everywhere Continuous Mapping (Topology)|continuity]] of $f$, these are [[Definition:O... | Continuous Real-Valued Function on Irreducible Space is Constant | https://proofwiki.org/wiki/Continuous_Real-Valued_Function_on_Irreducible_Space_is_Constant | https://proofwiki.org/wiki/Continuous_Real-Valued_Function_on_Irreducible_Space_is_Constant | [
"Irreducible Spaces",
"Constant Mappings"
] | [
"Definition:Topological Space",
"Definition:Irreducible Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Real-Valued Function",
"Definition:Constant Mapping"
] | [
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Set/Topology",
"Definition:Disjoint Sets",
"Definition:Irreducible Space",
"Definition:Empty Set",
"Definition:Contradiction",
"Definition:Constant Mapping"
] |
proofwiki-3522 | Irreducible Space is Pseudocompact | Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is pseudocompact. | We have that Continuous Real-Valued Function on Irreducible Space is Constant.
A constant mapping is trivially bounded.
Hence the result by definition of pseudocompact.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
Then $T$ is [[Definition:Pseudocompact Space|pseudocompact]]. | We have that [[Continuous Real-Valued Function on Irreducible Space is Constant]].
A [[Definition:Constant Mapping|constant mapping]] is trivially [[Definition:Bounded Mapping|bounded]].
Hence the result by definition of [[Definition:Pseudocompact Space|pseudocompact]].
{{qed}} | Irreducible Space is Pseudocompact | https://proofwiki.org/wiki/Irreducible_Space_is_Pseudocompact | https://proofwiki.org/wiki/Irreducible_Space_is_Pseudocompact | [
"Pseudocompact Spaces",
"Irreducible Spaces"
] | [
"Definition:Topological Space",
"Definition:Irreducible Space",
"Definition:Pseudocompact Space"
] | [
"Continuous Real-Valued Function on Irreducible Space is Constant",
"Definition:Constant Mapping",
"Definition:Bounded Mapping",
"Definition:Pseudocompact Space"
] |
proofwiki-3523 | Sequence of Implications of Connectedness Properties | Let $P_1$ and $P_2$ be connectedness properties and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satisfies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | ||
| align="center" | ||
| align="center" | Ultraconnected ||
|-
| ... | The relevant justifications are listed as follows:
:Ultraconnected Space is Connected
:Ultraconnected Space is Path-Connected
:Injectively Path-Connected Space is Path-Connected
:Path-Connected Space is Connected
:Irreducible Space is Connected
{{qed|lemma}}
Then we have:
:Connected Space is not necessarily Locally Con... | Let $P_1$ and $P_2$ be [[Definition:Topological Connectedness|connectedness properties]] and let:
:$P_1 \implies P_2$
mean:
:If a [[Definition:Topological Space|topological space]] $T$ satisfies [[Definition:Property|property]] $P_1$, then $T$ also satisfies [[Definition:Property|property]] $P_2$.
Then the following ... | The relevant justifications are listed as follows:
:[[Ultraconnected Space is Connected]]
:[[Ultraconnected Space is Path-Connected]]
:[[Injectively Path-Connected Space is Path-Connected]]
:[[Path-Connected Space is Connected]]
:[[Irreducible Space is Connected]]
{{qed|lemma}}
Then we have:
:[[Connected Space is n... | Sequence of Implications of Connectedness Properties | https://proofwiki.org/wiki/Sequence_of_Implications_of_Connectedness_Properties | https://proofwiki.org/wiki/Sequence_of_Implications_of_Connectedness_Properties | [
"Sequence of Implications of Connectedness Properties",
"Topological Connectedness"
] | [
"Definition:Topological Connectedness",
"Definition:Topological Space",
"Definition:Property",
"Definition:Property",
"Definition:Ultraconnected Space",
"Definition:Injectively Path-Connected/Topological Space",
"Definition:Path-Connected/Topological Space",
"Definition:Irreducible Space",
"Definiti... | [
"Ultraconnected Space is Connected",
"Ultraconnected Space is Path-Connected",
"Injectively Path-Connected Space is Path-Connected",
"Path-Connected Space is Connected",
"Irreducible Space is Connected",
"Connected Space is not necessarily Locally Connected",
"Connected Space is not necessarily Path-Con... |
proofwiki-3524 | Injectively Path-Connected Space is Path-Connected | Let $T = \struct {S, \tau}$ be a topological space which is injectively path-connected.
Then $T$ is path-connected. | Let $T = \struct {S, \tau}$ be injectively path-connected.
Then $\forall x, y \in S$, there exists a continuous injection $f: \closedint 0 1 \to S$, such that $\map f 0 = x$ and $\map f 1 = y$.
As $f$ is a continuous injection, it is also simply a continuous mapping.
The result follows from the definition of path-conne... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Injectively Path-Connected Space|injectively path-connected]].
Then $T$ is [[Definition:Path-Connected Space|path-connected]]. | Let $T = \struct {S, \tau}$ be [[Definition:Injectively Path-Connected Space|injectively path-connected]].
Then $\forall x, y \in S$, there exists a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Injection|injection]] $f: \closedint 0 1 \to S$, such that $\map f 0 = x$ and $\map f 1 = ... | Injectively Path-Connected Space is Path-Connected | https://proofwiki.org/wiki/Injectively_Path-Connected_Space_is_Path-Connected | https://proofwiki.org/wiki/Injectively_Path-Connected_Space_is_Path-Connected | [
"Injectively Path-Connected Spaces",
"Path-Connected Spaces",
"Sequence of Implications of Connectedness Properties"
] | [
"Definition:Topological Space",
"Definition:Injectively Path-Connected/Topological Space",
"Definition:Path-Connected/Topological Space"
] | [
"Definition:Injectively Path-Connected/Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Injection",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Injection",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Path-Connected/Topol... |
proofwiki-3525 | Irreducible Space is Connected | Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is connected. | Let $T = \struct {S, \tau}$ be irreducible.
Then:
:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
So trivially there are no two open sets that can form a separation of $T$.
The result follows from definition of connected.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
Then $T$ is [[Definition:Connected Topological Space|connected]]. | Let $T = \struct {S, \tau}$ be [[Definition:Irreducible Space|irreducible]].
Then:
:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
So trivially there are no two [[Definition:Open Set (Topology)|open sets]] that can form a [[Definition:Separation (Topology)|separation]] of $T$.
The result f... | Irreducible Space is Connected | https://proofwiki.org/wiki/Irreducible_Space_is_Connected | https://proofwiki.org/wiki/Irreducible_Space_is_Connected | [
"Irreducible Spaces",
"Connected Topological Spaces",
"Sequence of Implications of Connectedness Properties"
] | [
"Definition:Topological Space",
"Definition:Irreducible Space",
"Definition:Connected Topological Space"
] | [
"Definition:Irreducible Space",
"Definition:Open Set/Topology",
"Definition:Separation (Topology)",
"Definition:Connected Topological Space"
] |
proofwiki-3526 | Non-Trivial Ultraconnected Space is not T1 | Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
If $S$ has more than one element, then $T$ is not a $T_1$ space.
That is, if $T$ is a $T_1$ space with more than one element, it is not ultraconnected. | Let $T = \struct {S, \tau}$ be ultraconnected.
{{Recall|Ultraconnected Space|ultraconnected space|index = 2}}
{{:Definition:Ultraconnected Space/Definition 2}}
Let $a, b \in S$ such that $a \ne b$.
{{AimForCont}} $T$ is a $T_1$ space.
{{Recall|T1 Space|$T_1$ space|index = 3}}
{{:Definition:T1 Space/Definition 3}}
Hence... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Ultraconnected Space|ultraconnected]].
If $S$ has more than one [[Definition:Element|element]], then $T$ is not a [[Definition:T1 Space|$T_1$ space]].
That is, if $T$ is a [[Definition:T1 Space|$T_1$ space]] wi... | Let $T = \struct {S, \tau}$ be [[Definition:Ultraconnected Space|ultraconnected]].
{{Recall|Ultraconnected Space|ultraconnected space|index = 2}}
{{:Definition:Ultraconnected Space/Definition 2}}
Let $a, b \in S$ such that $a \ne b$.
{{AimForCont}} $T$ is a [[Definition:T1 Space|$T_1$ space]].
{{Recall|T1 Space|$T... | Non-Trivial Ultraconnected Space is not T1 | https://proofwiki.org/wiki/Non-Trivial_Ultraconnected_Space_is_not_T1 | https://proofwiki.org/wiki/Non-Trivial_Ultraconnected_Space_is_not_T1 | [
"T1 Spaces",
"Ultraconnected Spaces"
] | [
"Definition:Topological Space",
"Definition:Ultraconnected Space",
"Definition:Element",
"Definition:T1 Space",
"Definition:T1 Space",
"Definition:Element",
"Definition:Ultraconnected Space"
] | [
"Definition:Ultraconnected Space",
"Definition:T1 Space",
"Definition:Closed Set/Topology",
"Set is Closed iff Equals Topological Closure",
"Definition:Contradiction",
"Definition:Ultraconnected Space/Definition 2",
"Proof by Contradiction"
] |
proofwiki-3527 | Ultraconnected Space is T4 | Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is a $T_4$ space. | {{Recall|T4 Space|$T_4$ space|index = 1}}
{{:Definition:T4 Space/Definition 1}}
As no two closed sets of an ultraconnected space are actually disjoint, it follows that $T_4$-ness follows vacuously.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Ultraconnected Space|ultraconnected]].
Then $T$ is a [[Definition:T4 Space|$T_4$ space]]. | {{Recall|T4 Space|$T_4$ space|index = 1}}
{{:Definition:T4 Space/Definition 1}}
As no two [[Definition:Closed Set (Topology)|closed sets]] of an [[Definition:Ultraconnected Space|ultraconnected space]] are actually [[Definition:Disjoint Sets|disjoint]], it follows that [[Definition:T4 Space|$T_4$-ness]] follows [[Defi... | Ultraconnected Space is T4 | https://proofwiki.org/wiki/Ultraconnected_Space_is_T4 | https://proofwiki.org/wiki/Ultraconnected_Space_is_T4 | [
"T4 Spaces",
"Ultraconnected Spaces"
] | [
"Definition:Topological Space",
"Definition:Ultraconnected Space",
"Definition:T4 Space"
] | [
"Definition:Closed Set/Topology",
"Definition:Ultraconnected Space",
"Definition:Disjoint Sets",
"Definition:T4 Space",
"Definition:Vacuous Truth"
] |
proofwiki-3528 | Dot Product of Vector with Itself | :$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$ | Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf u
| r = u_1 u_1 + u_2 u_2 + \cdots + u_n u_n
| c = {{Defof|Dot Product}}
}}
{{eqn | r = {u_1}^2 + {u_2}^2 + \cdots + {u_n}^2
| c =
}}
{{eqn | r = \paren {\sqrt {\sum_{i \mathop = 1}^n {u_i}^2} }^... | :$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$ | Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf u
| r = u_1 u_1 + u_2 u_2 + \cdots + u_n u_n
| c = {{Defof|Dot Product}}
}}
{{eqn | r = {u_1}^2 + {u_2}^2 + \cdots + {u_n}^2
| c =
}}
{{eqn | r = \paren {\sqrt {\sum_{i \mathop = 1}^n {u_i}^2} }... | Dot Product of Vector with Itself/Proof 1 | https://proofwiki.org/wiki/Dot_Product_of_Vector_with_Itself | https://proofwiki.org/wiki/Dot_Product_of_Vector_with_Itself/Proof_1 | [
"Dot Product",
"Dot Product of Vector with Itself"
] | [] | [] |
proofwiki-3529 | Dot Product of Vector with Itself | :$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$ | {{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf u
| r = \norm {\mathbf u} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf u
| c = Cosine Formula for Dot Product
}}
{{eqn | r = \norm {\mathbf u}^2 \cos 0
| c = since the angle between a vector and itself is $0$
}}
{{eqn | r = \norm {\mathbf u}^2
... | :$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$ | {{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf u
| r = \norm {\mathbf u} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf u
| c = [[Cosine Formula for Dot Product]]
}}
{{eqn | r = \norm {\mathbf u}^2 \cos 0
| c = since the angle between a vector and itself is $0$
}}
{{eqn | r = \norm {\mathbf u}^2
... | Dot Product of Vector with Itself/Proof 2 | https://proofwiki.org/wiki/Dot_Product_of_Vector_with_Itself | https://proofwiki.org/wiki/Dot_Product_of_Vector_with_Itself/Proof_2 | [
"Dot Product",
"Dot Product of Vector with Itself"
] | [] | [
"Cosine Formula for Dot Product",
"Cosine of Zero is One"
] |
proofwiki-3530 | Cosine Formula for Dot Product | Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^n$.
The dot product of $\mathbf v$ and $\mathbf w$ can be calculated by:
:$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
where:
:$\norm {\, \cdot \,}$ denotes vector length and
:$\theta$ is the angle between $\mathbf v$ and $\mathb... | There are two cases:
:$(1): \quad \mathbf v$ and $\mathbf w$ are not scalar multiples of each other
:$(2): \quad \mathbf v$ and $\mathbf w$ are scalar multiples of each other.
=== Case 1 ===
:400px
Let the two vectors $\mathbf v$ and $\mathbf w$ not be scalar multiples of each other.
Then by the definition of angle be... | Let $\mathbf v, \mathbf w$ be two non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Real Euclidean Space)|vectors in $\R^n$]].
The [[Definition:Dot Product|dot product]] of $\mathbf v$ and $\mathbf w$ can be calculated by:
:$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
where:
... | There are two cases:
:$(1): \quad \mathbf v$ and $\mathbf w$ are not [[Definition:Scalar Multiplication on Vector Space|scalar multiples]] of each other
:$(2): \quad \mathbf v$ and $\mathbf w$ are [[Definition:Scalar Multiplication on Vector Space|scalar multiples]] of each other.
=== Case 1 ===
:[[File:AngleBetween... | Cosine Formula for Dot Product/Proof 1 | https://proofwiki.org/wiki/Cosine_Formula_for_Dot_Product | https://proofwiki.org/wiki/Cosine_Formula_for_Dot_Product/Proof_1 | [
"Cosine Formula for Dot Product",
"Dot Product"
] | [
"Definition:Zero Vector",
"Definition:Vector/Real Euclidean Space",
"Definition:Dot Product",
"Definition:Vector Length",
"Definition:Angle between Vectors"
] | [
"Definition:Scalar Multiplication/Vector Space",
"Definition:Scalar Multiplication/Vector Space",
"File:AngleBetweenTwoVectors.png",
"Definition:Scalar Multiplication/Vector Space",
"Definition:Angle between Vectors",
"Definition:Triangle (Geometry)",
"Angle Between Non-Zero Vectors Always Defined",
"... |
proofwiki-3531 | Cosine Formula for Dot Product | Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^n$.
The dot product of $\mathbf v$ and $\mathbf w$ can be calculated by:
:$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
where:
:$\norm {\, \cdot \,}$ denotes vector length and
:$\theta$ is the angle between $\mathbf v$ and $\mathb... | Let $\mathbf v$ and $\mathbf w$ be considered to be embedded in a Cartesian plane $\CC$.
By Dot Product is Invariant under Coordinate Rotation, we may rotate $\CC$ arbitrarily, and $\mathbf v \cdot \mathbf w$ will not change.
So, let us rotate $\CC$ to $\CC'$ such that the $x$-axis is parallel to $\mathbf v$.
Hence $\m... | Let $\mathbf v, \mathbf w$ be two non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Real Euclidean Space)|vectors in $\R^n$]].
The [[Definition:Dot Product|dot product]] of $\mathbf v$ and $\mathbf w$ can be calculated by:
:$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
where:
... | Let $\mathbf v$ and $\mathbf w$ be considered to be embedded in a [[Definition:Cartesian Plane|Cartesian plane]] $\CC$.
By [[Dot Product is Invariant under Coordinate Rotation]], we may [[Definition:Plane Rotation|rotate]] $\CC$ arbitrarily, and $\mathbf v \cdot \mathbf w$ will not change.
So, let us [[Definition:Pla... | Cosine Formula for Dot Product/Proof 2 | https://proofwiki.org/wiki/Cosine_Formula_for_Dot_Product | https://proofwiki.org/wiki/Cosine_Formula_for_Dot_Product/Proof_2 | [
"Cosine Formula for Dot Product",
"Dot Product"
] | [
"Definition:Zero Vector",
"Definition:Vector/Real Euclidean Space",
"Definition:Dot Product",
"Definition:Vector Length",
"Definition:Angle between Vectors"
] | [
"Definition:Cartesian Plane",
"Dot Product is Invariant under Coordinate Rotation",
"Definition:Rotation (Geometry)/Plane",
"Definition:Rotation (Geometry)/Plane",
"Definition:Axis/X-Axis",
"Definition:Parallel (Geometry)/Lines",
"Definition:Dot Product/General Context"
] |
proofwiki-3532 | Irreducible Space is Locally Connected | Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is locally connected. | {{Recall|Locally Connected Space|locally connected space}}
{{:Definition:Locally Connected Space/Definition 3}}
Let $T = \struct {S, \tau}$ be irreducible.
{{Recall|Irreducible Space|irreducible space}}
{{:Definition:Irreducible Space/Definition 3}}
So trivially there are no two open sets that can form a separation of ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
Then $T$ is [[Definition:Locally Connected Space|locally connected]]. | {{Recall|Locally Connected Space|locally connected space}}
{{:Definition:Locally Connected Space/Definition 3}}
Let $T = \struct {S, \tau}$ be [[Definition:Irreducible Space|irreducible]].
{{Recall|Irreducible Space|irreducible space}}
{{:Definition:Irreducible Space/Definition 3}}
So trivially there are no two [[De... | Irreducible Space is Locally Connected | https://proofwiki.org/wiki/Irreducible_Space_is_Locally_Connected | https://proofwiki.org/wiki/Irreducible_Space_is_Locally_Connected | [
"Irreducible Spaces",
"Locally Connected Spaces"
] | [
"Definition:Topological Space",
"Definition:Irreducible Space",
"Definition:Locally Connected Space"
] | [
"Definition:Irreducible Space",
"Definition:Open Set/Topology",
"Definition:Separation (Topology)",
"Definition:Basis (Topology)",
"Definition:Open Set/Topology",
"Definition:Basis (Topology)",
"Definition:Locally Connected Space"
] |
proofwiki-3533 | Locally Injectively Path-Connected Space is Locally Path-Connected | Let $T = \struct {S, \tau}$ be a topological space which is locally injectively path-connected.
Then $T$ is also locally path-connected. | Let $T = \struct {S, \tau}$ be locally injectively path-connected.
Then $T$ has a basis consisting entirely of injectively path-connected sets.
From Injectively Path-Connected Space is Path-Connected, this basis consisting entirely of path-connected sets.
The result follows from definition of locally path-connected.
{{... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Locally Injectively Path-Connected Space|locally injectively path-connected]].
Then $T$ is also [[Definition:Locally Path-Connected Space|locally path-connected]]. | Let $T = \struct {S, \tau}$ be [[Definition:Locally Injectively Path-Connected Space|locally injectively path-connected]].
Then $T$ has a [[Definition:Basis (Topology)|basis]] consisting entirely of [[Definition:Injectively Path-Connected Set|injectively path-connected sets]].
From [[Injectively Path-Connected Space ... | Locally Injectively Path-Connected Space is Locally Path-Connected | https://proofwiki.org/wiki/Locally_Injectively_Path-Connected_Space_is_Locally_Path-Connected | https://proofwiki.org/wiki/Locally_Injectively_Path-Connected_Space_is_Locally_Path-Connected | [
"Locally Injectively Path-Connected Spaces",
"Locally Path-Connected Spaces"
] | [
"Definition:Topological Space",
"Definition:Locally Injectively Path-Connected Space",
"Definition:Locally Path-Connected Space"
] | [
"Definition:Locally Injectively Path-Connected Space",
"Definition:Basis (Topology)",
"Definition:Injectively Path-Connected/Subset",
"Injectively Path-Connected Space is Path-Connected",
"Definition:Basis (Topology)",
"Definition:Path-Connected/Set",
"Definition:Locally Path-Connected Space"
] |
proofwiki-3534 | Locally Path-Connected Space is Locally Connected | Let $T = \struct {S, \tau}$ be a topological space which is locally path-connected.
Then $T$ is also locally connected. | Let $x \in S$ be any point of $T$.
Let $\BB$ be a local basis of path-connected sets for $x$.
From Path-Connected Space is Connected, $\BB$ is a local basis of connected sets.
Thus, $T$ is locally connected by definition.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Locally Path-Connected Space|locally path-connected]].
Then $T$ is also [[Definition:Locally Connected Space|locally connected]]. | Let $x \in S$ be any point of $T$.
Let $\BB$ be a [[Definition:Local Basis|local basis]] of [[Definition:Path-Connected Set|path-connected sets]] for $x$.
From [[Path-Connected Space is Connected]], $\BB$ is a [[Definition:Local Basis|local basis]] of [[Definition:Connected Set (Topology)|connected sets]].
Thus, $T$... | Locally Path-Connected Space is Locally Connected | https://proofwiki.org/wiki/Locally_Path-Connected_Space_is_Locally_Connected | https://proofwiki.org/wiki/Locally_Path-Connected_Space_is_Locally_Connected | [
"Locally Path-Connected Spaces",
"Locally Connected Spaces"
] | [
"Definition:Topological Space",
"Definition:Locally Path-Connected Space",
"Definition:Locally Connected Space"
] | [
"Definition:Local Basis",
"Definition:Path-Connected/Set",
"Path-Connected Space is Connected",
"Definition:Local Basis",
"Definition:Connected Set (Topology)",
"Definition:Locally Connected Space"
] |
proofwiki-3535 | Locally Euclidean Space is Locally Compact | Let $M$ be a locally Euclidean space of some dimension $d$.
Then $M$ is locally compact. | Let $m \in M$ be arbitrary.
By definition of locally Euclidean space, there exists an open neighborhood $U$ of $m$, homeomorphic to an open subset of $\R^d$.
By the definition of an open set, there is some open ball:
:$B = \map {B_\delta} {\map \phi m} = \set {x \in \R^d: \size {x - \map \phi m} < \delta}$
of radius $\... | Let $M$ be a [[Definition:Locally Euclidean Space|locally Euclidean space]] of some dimension $d$.
Then $M$ is [[Definition:Locally Compact Space|locally compact]]. | Let $m \in M$ be arbitrary.
By definition of [[Definition:Locally Euclidean Space|locally Euclidean space]], there exists an [[Definition:Open Neighborhood of Point|open neighborhood]] $U$ of $m$, [[Definition:Homeomorphic Metric Spaces|homeomorphic]] to an [[Definition:Open Set of Metric Space|open subset]] of $\R^d... | Locally Euclidean Space is Locally Compact/Proof 1 | https://proofwiki.org/wiki/Locally_Euclidean_Space_is_Locally_Compact | https://proofwiki.org/wiki/Locally_Euclidean_Space_is_Locally_Compact/Proof_1 | [
"Locally Compact Spaces",
"Locally Euclidean Spaces",
"Locally Euclidean Space is Locally Compact",
"Locally Compact Spaces"
] | [
"Definition:Locally Euclidean Space",
"Definition:Locally Compact Space"
] | [
"Definition:Locally Euclidean Space",
"Definition:Open Neighborhood/Point",
"Definition:Homeomorphism/Metric Spaces",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Metric Space",
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Closure of Open Ball in Metric Space",
"Topological Cl... |
proofwiki-3536 | Locally Euclidean Space is Locally Compact | Let $M$ be a locally Euclidean space of some dimension $d$.
Then $M$ is locally compact. | Let $m \in M$ be arbitrary.
From Locally Euclidean Space has Countable Neighborhood Basis Homeomorphic to Closed Balls:
:there exists a countable neighborhood basis $\family{K_n}_{n \in \N}$ of $m$ where each $N_n$ is the homeomorphic image of a closed ball of $\R^d$
From Closed Ball in Euclidean Space is Compact:
:$\f... | Let $M$ be a [[Definition:Locally Euclidean Space|locally Euclidean space]] of some dimension $d$.
Then $M$ is [[Definition:Locally Compact Space|locally compact]]. | Let $m \in M$ be arbitrary.
From [[Locally Euclidean Space has Countable Neighborhood Basis Homeomorphic to Closed Balls]]:
:there exists a [[Definition:Countable|countable]] [[Definition:Neighborhood Basis|neighborhood basis]] $\family{K_n}_{n \in \N}$ of $m$ where each $N_n$ is the [[Definition:Homeomorphism|homeom... | Locally Euclidean Space is Locally Compact/Proof 2 | https://proofwiki.org/wiki/Locally_Euclidean_Space_is_Locally_Compact | https://proofwiki.org/wiki/Locally_Euclidean_Space_is_Locally_Compact/Proof_2 | [
"Locally Compact Spaces",
"Locally Euclidean Spaces",
"Locally Euclidean Space is Locally Compact",
"Locally Compact Spaces"
] | [
"Definition:Locally Euclidean Space",
"Definition:Locally Compact Space"
] | [
"Locally Euclidean Space has Countable Neighborhood Basis Homeomorphic to Closed Balls",
"Definition:Countable Set",
"Definition:Neighborhood Basis",
"Definition:Homeomorphism",
"Definition:Image",
"Definition:Closed Ball/Metric Space",
"Closed Ball in Euclidean Space is Compact",
"Definition:Homeomor... |
proofwiki-3537 | Unique Factorization Domain is Integrally Closed | Let $A$ be a unique factorization domain (UFD).
Then $A$ is integrally closed. | Let $K$ be the field of quotients of $A$.
Let $x \in K$ be integral over $A$.
Let:
$x = a / b$
for $a, b \in A$ with $\gcd \set {a, b} \in A^\times$.
This makes sense because a UFD is GCD Domain.
There is an equation:
:$\paren {\dfrac a b}^n + a_{n - 1} \paren {\dfrac a b}^{n - 1} + \dotsb + a_0$
{{explain|This is not ... | Let $A$ be a [[Definition:Unique Factorization Domain|unique factorization domain (UFD)]].
Then $A$ is [[Definition:Integrally Closed Integral Domain|integrally closed]]. | Let $K$ be the [[Definition:Field of Quotients|field of quotients]] of $A$.
Let $x \in K$ be [[Definition:Integral Element of Ring Extension|integral]] over $A$.
Let:
$x = a / b$
for $a, b \in A$ with $\gcd \set {a, b} \in A^\times$.
This makes sense because a [[UFD is GCD Domain]].
There is an equation:
:$\paren ... | Unique Factorization Domain is Integrally Closed | https://proofwiki.org/wiki/Unique_Factorization_Domain_is_Integrally_Closed | https://proofwiki.org/wiki/Unique_Factorization_Domain_is_Integrally_Closed | [
"Algebraic Number Theory",
"Factorization",
"Unique Factorization Domains"
] | [
"Definition:Unique Factorization Domain",
"Definition:Integrally Closed Integral Domain"
] | [
"Definition:Field of Quotients",
"Definition:Integral Element of Ring Extension",
"UFD is GCD Domain",
"Definition:Unit of Ring",
"Definition:Contradiction",
"Definition:Unit of Ring",
"Definition:Integrally Closed Integral Domain",
"Definition:Integrally Closed Integral Domain",
"Definition:Integra... |
proofwiki-3538 | Totally Disconnected Space is T1 | Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is a $T_1$ space. | Let $T = \struct {S, \tau}$ be totally disconnected.
Then as its components are singletons, it follows that each of its points is closed.
From Equivalence of Definitions of $T_1$ Space, it follows that $T$ is a $T_1$ space.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Totally Disconnected Space|totally disconnected]].
Then $T$ is a [[Definition:T1 Space|$T_1$ space]]. | Let $T = \struct {S, \tau}$ be [[Definition:Totally Disconnected Space|totally disconnected]].
Then as its [[Definition:Component (Topology)|components]] are [[Definition:Singleton|singletons]], it follows that each of its [[Definition:Point of Set|points]] is [[Definition:Closed Point|closed]].
From [[Equivalence of... | Totally Disconnected Space is T1 | https://proofwiki.org/wiki/Totally_Disconnected_Space_is_T1 | https://proofwiki.org/wiki/Totally_Disconnected_Space_is_T1 | [
"Totally Disconnected Spaces",
"T1 Spaces",
"Sequence of Implications of Disconnectedness Properties"
] | [
"Definition:Topological Space",
"Definition:Totally Disconnected Space",
"Definition:T1 Space"
] | [
"Definition:Totally Disconnected Space",
"Definition:Component (Topology)",
"Definition:Singleton",
"Definition:Element",
"Definition:Closed Point",
"Equivalence of Definitions of T1 Space",
"Definition:T1 Space"
] |
proofwiki-3539 | Determinant of Linear Operator is Well Defined | Let $V$ be a nontrivial finite dimensional vector space over a field $K$.
Let $A: V \to V$ be a linear operator of $V$.
Then the determinant $\det A$ of $A$ is well defined, that is, does not depend on the choice of a basis of $V$. | Let $A_\BB$ and $A_\CC$ be the matrices of $A$ relative to $\BB$ and $\CC$ respectively.
Let $\det$ also denote the determinant of a matrix.
We are required to show that $\det A_\BB = \det A_\CC$.
Let $P$ be the change of basis matrix from $\BB$ to $\CC$.
By Change of Coordinate Vectors Under Linear Mapping and since $... | Let $V$ be a [[Definition:Trivial Vector Space|nontrivial]] [[Definition:Finite Dimensional Vector Space|finite dimensional]] [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $K$.
Let $A: V \to V$ be a [[Definition:Linear Operator|linear operator]] of $V$.
Then the [[Defi... | Let $A_\BB$ and $A_\CC$ be the [[Definition:Matrix|matrices]] of $A$ [[Definition:Relative Matrix of Linear Transformation|relative]] to $\BB$ and $\CC$ respectively.
Let $\det$ also denote the [[Definition:Determinant of Matrix|determinant]] of a [[Definition:Matrix|matrix]].
We are required to show that $\det A_\BB... | Determinant of Linear Operator is Well Defined | https://proofwiki.org/wiki/Determinant_of_Linear_Operator_is_Well_Defined | https://proofwiki.org/wiki/Determinant_of_Linear_Operator_is_Well_Defined | [
"Linear Operators",
"Determinants"
] | [
"Definition:Trivial Vector Space",
"Definition:Dimension of Vector Space/Finite",
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Linear Operator",
"Definition:Determinant/Linear Operator",
"Definition:Well-Defined/Mapping",
"Definition:Basis of Vector Space"
] | [
"Definition:Matrix",
"Definition:Relative Matrix of Linear Transformation",
"Definition:Determinant/Matrix",
"Definition:Matrix",
"Definition:Change of Basis Matrix",
"Change of Coordinate Vectors Under Linear Transformation",
"Definition:Linear Operator",
"File:Determinant Independent of Basis.png",
... |
proofwiki-3540 | Localization Preserves Integral Closure | Let $A \subseteq B$ be an extension of commutative rings with unity.
Let $C$ be the integral closure of $A$ in $B$.
Let $S \subseteq A$ be a multiplicatively closed subset.
Then $C_S$ is the integral closure of $A_S$ in $B_S$, where subscript $S$ indicates the localization at $S$. | First we show that $C_S$ is an integral extension of $A_S$.
Let $x \in C_S$, and $\iota$ be the canonical inclusion from a ring to its localization.
There exists $s \in S$ such that $sx \in \iota(C)$, say $sx = \iota(c)$.
Since $c \in C$ is integral, there is an equation:
:$c^n + a_{n-1}c^{n-1} + \cdots + a_1 c + a_0 =... | Let $A \subseteq B$ be an [[Definition:Ring Extension|extension]] of [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
Let $C$ be the [[Definition:Integral Closure|integral closure]] of $A$ in $B$.
Let $S \subseteq A$ be a [[Definition:Multiplicatively Closed Subset of Ring|multiplicatively cl... | First we show that $C_S$ is an integral extension of $A_S$.
Let $x \in C_S$, and $\iota$ be the canonical [[Definition:Inclusion Mapping|inclusion]] from a [[Definition:Ring (Abstract Algebra)|ring]] to its [[Definition:Localization of Ring|localization]].
There exists $s \in S$ such that $sx \in \iota(C)$, say $sx =... | Localization Preserves Integral Closure | https://proofwiki.org/wiki/Localization_Preserves_Integral_Closure | https://proofwiki.org/wiki/Localization_Preserves_Integral_Closure | [
"Algebraic Number Theory"
] | [
"Definition:Ring Extension",
"Definition:Commutative and Unitary Ring",
"Definition:Integral Closure",
"Definition:Multiplicatively Closed Subset of Ring",
"Definition:Integral Closure",
"Definition:Localization of Ring"
] | [
"Definition:Inclusion Mapping",
"Definition:Ring (Abstract Algebra)",
"Definition:Localization of Ring",
"Definition:Integral Element of Ring Extension",
"Definition:Ring Homomorphism",
"Definition:Monic Polynomial",
"Definition:Integral Element of Ring Extension",
"Category:Algebraic Number Theory"
] |
proofwiki-3541 | Integrally Closed is Local Property | Let $A$ be an integral domain.
For a prime ideal $\mathfrak p$ of $A$, let $A_{\mathfrak p}$ denote the localization at $S = A \divides \mathfrak p$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is integrally closed}}
{{item|(2):|$A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$}}
{{item|(3):|$A_{... | === $(1)$ implies $(2)$ ===
Let $\map Q R$ denote the field of quotients of an integral domain $R$.
We have by Localization Preserves Integral Closure that:
:$\map Q {A_{\mathfrak p}} = \map Q A$
Hence $A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.
{{questionable|What is the point of conside... | Let $A$ be an [[Definition:Integral Domain|integral domain]].
For a [[Definition:Prime Ideal of Ring|prime ideal]] $\mathfrak p$ of $A$, let $A_{\mathfrak p}$ denote the [[Definition:Localization of Ring|localization]] at $S = A \divides \mathfrak p$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is [[Definition:Integr... | === $(1)$ implies $(2)$ ===
Let $\map Q R$ denote the [[Definition:Field of Quotients|field of quotients]] of an [[Definition:Integral Domain|integral domain]] $R$.
We have by [[Localization Preserves Integral Closure]] that:
:$\map Q {A_{\mathfrak p}} = \map Q A$
Hence $A_{\mathfrak p}$ is [[Definition:Integrally C... | Integrally Closed is Local Property | https://proofwiki.org/wiki/Integrally_Closed_is_Local_Property | https://proofwiki.org/wiki/Integrally_Closed_is_Local_Property | [
"Algebraic Number Theory"
] | [
"Definition:Integral Domain",
"Definition:Prime Ideal of Ring",
"Definition:Localization of Ring",
"Definition:Integrally Closed",
"Definition:Integrally Closed",
"Definition:Prime Ideal of Ring",
"Definition:Integrally Closed",
"Definition:Maximal Ideal of Ring"
] | [
"Definition:Field of Quotients",
"Definition:Integral Domain",
"Localization Preserves Integral Closure",
"Definition:Integrally Closed",
"Definition:Prime Ideal of Ring",
"Localization Preserves Integral Closure"
] |
proofwiki-3542 | Totally Disconnected and Locally Connected Space is Discrete | Let $T = \struct {S, \tau}$ be a topological space which is both totally disconnected and locally connected.
Then $T$ is the discrete space on $S$. | So, let $T = \struct {S, \tau}$ be that topological space which is both totally disconnected and locally connected.
As $T$ is totally disconnected, every point is a component and therefore closed.
As $T$ is locally connected, there exists a basis $\BB$ of $T$ such that every element of $\BB$ is a component of $T$.
In o... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is both [[Definition:Totally Disconnected Space|totally disconnected]] and [[Definition:Locally Connected Space|locally connected]].
Then $T$ is the [[Definition:Discrete Space|discrete space]] on $S$. | So, let $T = \struct {S, \tau}$ be that [[Definition:Topological Space|topological space]] which is both [[Definition:Totally Disconnected Space|totally disconnected]] and [[Definition:Locally Connected Space|locally connected]].
As $T$ is [[Definition:Totally Disconnected Space|totally disconnected]], every [[Definit... | Totally Disconnected and Locally Connected Space is Discrete | https://proofwiki.org/wiki/Totally_Disconnected_and_Locally_Connected_Space_is_Discrete | https://proofwiki.org/wiki/Totally_Disconnected_and_Locally_Connected_Space_is_Discrete | [
"Discrete Topologies",
"Locally Connected Spaces",
"Totally Disconnected Spaces",
"Sequence of Implications of Disconnectedness Properties"
] | [
"Definition:Topological Space",
"Definition:Totally Disconnected Space",
"Definition:Locally Connected Space",
"Definition:Discrete Topology"
] | [
"Definition:Topological Space",
"Definition:Totally Disconnected Space",
"Definition:Locally Connected Space",
"Definition:Totally Disconnected Space",
"Definition:Element",
"Definition:Component (Topology)",
"Definition:Closed Point",
"Definition:Locally Connected Space",
"Definition:Basis (Topolog... |
proofwiki-3543 | Totally Disconnected but Connected Set must be Singleton | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be both totally disconnected and connected.
Then $H$ is a singleton. | If $H$ is totally disconnected then its individual points are separated.
If $H$ is connected it can not be represented as the union of two (or more) separated sets.
So $H$ can have only one point in it.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$ be both [[Definition:Totally Disconnected Space|totally disconnected]] and [[Definition:Connected Set (Topology)|connected]].
Then $H$ is a [[Definition:Singleton|singleton]]. | If $H$ is [[Definition:Totally Disconnected Space|totally disconnected]] then its individual points are [[Definition:Separated Points|separated]].
If $H$ is [[Definition:Connected Set (Topology)|connected]] it can not be represented as the [[Definition:Set Union|union]] of two (or more) [[Definition:Separated Sets|sep... | Totally Disconnected but Connected Set must be Singleton | https://proofwiki.org/wiki/Totally_Disconnected_but_Connected_Set_must_be_Singleton | https://proofwiki.org/wiki/Totally_Disconnected_but_Connected_Set_must_be_Singleton | [
"Totally Disconnected Spaces",
"Connected Sets (Topology)",
"Singletons"
] | [
"Definition:Topological Space",
"Definition:Totally Disconnected Space",
"Definition:Connected Set (Topology)",
"Definition:Singleton"
] | [
"Definition:Totally Disconnected Space",
"Definition:Separated Points",
"Definition:Connected Set (Topology)",
"Definition:Set Union",
"Definition:Separated Sets"
] |
proofwiki-3544 | Identity Morphism is Unique | Let $\mathbf C$ be a category.
Let $X$ be an object of $\mathbf C$.
Then the identity morphism $\operatorname{id}_X : X \to X$ is unique. | Let $\operatorname{id}_X^1$, $\operatorname{id}_X^2$ be two identity morphisms for $X$.
By definition, for any morphism $f : Y \to X$, we have:
:$\operatorname{id}_X^1 \circ f = f$
In particular, taking $Y = X$ and $f = \operatorname{id}_X^2$, we have:
:$\operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname... | Let $\mathbf C$ be a [[Definition:Category|category]].
Let $X$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Then the [[Definition:Identity Morphism|identity morphism]] $\operatorname{id}_X : X \to X$ is unique. | Let $\operatorname{id}_X^1$, $\operatorname{id}_X^2$ be two [[Definition:Identity Morphism|identity morphisms]] for $X$.
By definition, for any [[Definition:Morphism|morphism]] $f : Y \to X$, we have:
:$\operatorname{id}_X^1 \circ f = f$
In particular, taking $Y = X$ and $f = \operatorname{id}_X^2$, we have:
:$\opera... | Identity Morphism is Unique | https://proofwiki.org/wiki/Identity_Morphism_is_Unique | https://proofwiki.org/wiki/Identity_Morphism_is_Unique | [
"Morphisms"
] | [
"Definition:Category",
"Definition:Object (Category Theory)",
"Definition:Identity Morphism"
] | [
"Definition:Identity Morphism",
"Definition:Morphism",
"Definition:Morphism",
"Category:Morphisms"
] |
proofwiki-3545 | Product Category is Category | Let $\mathbf C$ and $\mathbf D$ be metacategories.
Then the product category $\mathbf C \times \mathbf D$ is a metacategory. | Let $\tuple {X, Y}$ and $\tuple {X', Y'}$ be objects of $\mathbf C \times \mathbf D$.
Let $\tuple {f, g}: \tuple {X, Y} \to \tuple {X', Y'}$ and $\tuple {h, k}: \tuple {X', Y'} \to \tuple {X, Y}$ be morphisms of $\mathbf C \times \mathbf D$.
Let $\operatorname {id}_X, \operatorname {id}_Y$ be the identity morphisms for... | Let $\mathbf C$ and $\mathbf D$ be [[Definition:Metacategory|metacategories]].
Then the [[Definition:Product Category|product category]] $\mathbf C \times \mathbf D$ is a [[Definition:Metacategory|metacategory]]. | Let $\tuple {X, Y}$ and $\tuple {X', Y'}$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C \times \mathbf D$.
Let $\tuple {f, g}: \tuple {X, Y} \to \tuple {X', Y'}$ and $\tuple {h, k}: \tuple {X', Y'} \to \tuple {X, Y}$ be [[Definition:Morphism (Category Theory)|morphisms]] of $\mathbf C \times \mathbf... | Product Category is Category | https://proofwiki.org/wiki/Product_Category_is_Category | https://proofwiki.org/wiki/Product_Category_is_Category | [
"Product Categories"
] | [
"Definition:Metacategory",
"Definition:Product Category",
"Definition:Metacategory"
] | [
"Definition:Object (Category Theory)",
"Definition:Morphism",
"Definition:Identity Morphism",
"Definition:Object (Category Theory)",
"Definition:Composition of Morphisms",
"Definition:Composition of Morphisms",
"Definition:Identity Morphism",
"Definition:Composable Morphisms",
"Definition:Compositio... |
proofwiki-3546 | Category of Sets is Category | Let $\mathbf{Set}$ be the category of sets.
Then $\mathbf{Set}$ is a metacategory. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
For any two mappings their composition (in the usual set theoretic sense) is again a mapping by Composite Mapping is Mapping.
For any set $X$, we have the identity mapping $\operatorname{id}_X$.
By Identity Mapping is Left Identity and Ident... | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Then $\mathbf{Set}$ is a [[Definition:Metacategory|metacategory]]. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a [[Definition:Metacategory|metacategory]].
For any two mappings their [[Definition:Composition of Mappings|composition]] (in the usual [[Definition:Set Theory|set theoretic]] sense) is again a mapping by [[Composite Mapping is Mapping]].
For any set $X... | Category of Sets is Category | https://proofwiki.org/wiki/Category_of_Sets_is_Category | https://proofwiki.org/wiki/Category_of_Sets_is_Category | [
"Examples of Categories",
"Category of Sets"
] | [
"Definition:Category of Sets",
"Definition:Metacategory"
] | [
"Definition:Metacategory",
"Definition:Composition of Mappings",
"Definition:Set Theory",
"Composite Mapping is Mapping",
"Definition:Identity Mapping",
"Identity Mapping is Left Identity",
"Identity Mapping is Right Identity",
"Definition:Identity Morphism",
"Composition of Mappings is Associative"... |
proofwiki-3547 | Projection from Product Category | Let $\mathbf C$ and $\mathbf D$ be categories.
Let $\mathbf C \times \mathbf D$ be the product category.
Then the projection functors:
:$\pi_{\mathbf C}: \mathbf C \times \mathbf D \to \mathbf C$
:$\pi_{\mathbf D}: \mathbf C \times \mathbf D \to \mathbf D$
are indeed functors.
Moreover, $\pi_{\mathbf C}$ and $\pi_{\mat... | {{ProofWanted}}
Category:Product Categories
Category:Projections
laq3yrt19kqib748s6ez9aq5igaw3z6 | Let $\mathbf C$ and $\mathbf D$ be [[Definition:Category|categories]].
Let $\mathbf C \times \mathbf D$ be the [[Definition:Product Category|product category]].
Then the [[Definition:Projection Functor|projection functors]]:
:$\pi_{\mathbf C}: \mathbf C \times \mathbf D \to \mathbf C$
:$\pi_{\mathbf D}: \mathbf C ... | {{ProofWanted}}
[[Category:Product Categories]]
[[Category:Projections]]
laq3yrt19kqib748s6ez9aq5igaw3z6 | Projection from Product Category | https://proofwiki.org/wiki/Projection_from_Product_Category | https://proofwiki.org/wiki/Projection_from_Product_Category | [
"Product Categories",
"Projections"
] | [
"Definition:Category",
"Definition:Product Category",
"Definition:Projection Functor",
"Definition:Functor",
"Product Category is Product in Category of Categories",
"Definition:Category",
"Definition:Functor",
"Definition:Unique",
"Definition:Functor",
"Definition:Commutative Diagram"
] | [
"Category:Product Categories",
"Category:Projections"
] |
proofwiki-3548 | Extremally Disconnected Hausdorff Space is Totally Separated | Let $T = \struct {S, \tau}$ be an extremally disconnected Hausdorff space.
Then $T$ is totally separated. | {{Recall|Totally Separated Space|totally separated topological space}}
{{:Definition:Totally Separated Space/Definition 1}}
Let $T = \struct {S, \tau}$ be an extremally disconnected Hausdorff space.
{{Recall|Extremally Disconnected Space|extremally disconnected topological space}}
{{:Definition:Extremally Disconnected ... | Let $T = \struct {S, \tau}$ be an [[Definition:Extremally Disconnected Hausdorff Space|extremally disconnected Hausdorff space]].
Then $T$ is [[Definition:Totally Separated Space|totally separated]]. | {{Recall|Totally Separated Space|totally separated topological space}}
{{:Definition:Totally Separated Space/Definition 1}}
Let $T = \struct {S, \tau}$ be an [[Definition:Extremally Disconnected Hausdorff Space|extremally disconnected Hausdorff space]].
{{Recall|Extremally Disconnected Space|extremally disconnected t... | Extremally Disconnected Hausdorff Space is Totally Separated | https://proofwiki.org/wiki/Extremally_Disconnected_Hausdorff_Space_is_Totally_Separated | https://proofwiki.org/wiki/Extremally_Disconnected_Hausdorff_Space_is_Totally_Separated | [
"Extremally Disconnected Hausdorff Spaces",
"Totally Separated Spaces",
"Sequence of Implications of Disconnectedness Properties"
] | [
"Definition:Extremally Disconnected Hausdorff Space",
"Definition:Totally Separated Space"
] | [
"Definition:Extremally Disconnected Hausdorff Space",
"Definition:Arbitrary",
"Definition:T2 Space",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Definition:Closure (Topology)",
"Topological Closure is Closed",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Defi... |
proofwiki-3549 | Zero Dimensional Space is T3 | Let $T = \struct {S, \tau}$ be a zero dimensional topological space.
Then $T$ is a $T_3$ space. | {{Recall|T3 Space|$T_3$ space}}
{{:Definition:T3 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a zero dimensional space.
{{Recall|Zero Dimensional Space|zero dimensional space}}
{{:Definition:Zero Dimensional Space}}
Let $F \subseteq S$ be closed in $T$.
Let also $y \notin F$.
Then by definition of closed, $\relc... | Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional topological space]].
Then $T$ is a [[Definition:T3 Space|$T_3$ space]]. | {{Recall|T3 Space|$T_3$ space}}
{{:Definition:T3 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional space]].
{{Recall|Zero Dimensional Space|zero dimensional space}}
{{:Definition:Zero Dimensional Space}}
Let $F \subseteq S$ be [[Definition:Closed Set (Topology... | Zero Dimensional Space is T3 | https://proofwiki.org/wiki/Zero_Dimensional_Space_is_T3 | https://proofwiki.org/wiki/Zero_Dimensional_Space_is_T3 | [
"Zero Dimensional Spaces",
"T3 Spaces",
"Sequence of Implications of Disconnectedness Properties"
] | [
"Definition:Zero Dimensional Space",
"Definition:T3 Space"
] | [
"Definition:Zero Dimensional Space",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Relative Complement",
"Definition:Zero Dimensional Space",
"Definition:Basis (Topology)",
"Definition:Clopen Set",
"Definition:Basis (Topology)",
"De... |
proofwiki-3550 | Zero Dimensional T0 Space is Totally Separated | Let $T = \struct {S, \tau}$ be a zero dimensional topological space which is also a $T_0$ space.
Then $T$ is totally separated. | {{Recall|Totally Separated Space|totally separated space}}
{{:Definition:Totally Separated Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a zero dimensional space which is also a $T_0$ space.
{{Recall|Zero Dimensional Space|zero dimensional space}}
{{:Definition:Zero Dimensional Space}}
{{Recall|T0 Space|$T_0$ spa... | Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional topological space]] which is also a [[Definition:T0 Space|$T_0$ space]].
Then $T$ is [[Definition:Totally Separated Space|totally separated]]. | {{Recall|Totally Separated Space|totally separated space}}
{{:Definition:Totally Separated Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional space]] which is also a [[Definition:T0 Space|$T_0$ space]].
{{Recall|Zero Dimensional Space|zero dimensional space}}
{{... | Zero Dimensional T0 Space is Totally Separated | https://proofwiki.org/wiki/Zero_Dimensional_T0_Space_is_Totally_Separated | https://proofwiki.org/wiki/Zero_Dimensional_T0_Space_is_Totally_Separated | [
"T0 Spaces",
"Zero Dimensional Spaces",
"Totally Separated Spaces",
"Sequence of Implications of Disconnectedness Properties"
] | [
"Definition:Zero Dimensional Space",
"Definition:T0 Space",
"Definition:Totally Separated Space"
] | [
"Definition:Zero Dimensional Space",
"Definition:T0 Space",
"Definition:Arbitrary",
"Definition:Basis (Topology)",
"Definition:Clopen Set",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Clopen Set and Complement form Separation",
"Definition:Separation (Topology)",
"Definition:A... |
proofwiki-3551 | Scattered T1 Space is Totally Disconnected | Let $T = \struct {S, \tau}$ be a scattered topological space which is also a $T_1$ space.
Then $T$ is totally disconnected. | Let $T = \struct {S, \tau}$ be a scattered space which is also a $T_1$ space.
We have that every Non-Trivial Connected Set in $T_1$ Space is Dense-in-itself.
As $T$ is scattered, every $H \subseteq S$ contains at least one point which is isolated in $H$.
So $H$ is not dense-in-itself and so if $H$ has more than one ele... | Let $T = \struct {S, \tau}$ be a [[Definition:Scattered Space|scattered topological space]] which is also a [[Definition:T1 Space|$T_1$ space]].
Then $T$ is [[Definition:Totally Disconnected Space|totally disconnected]]. | Let $T = \struct {S, \tau}$ be a [[Definition:Scattered Space|scattered space]] which is also a [[Definition:T1 Space|$T_1$ space]].
We have that every [[Non-Trivial Connected Set in T1 Space is Dense-in-itself|Non-Trivial Connected Set in $T_1$ Space is Dense-in-itself]].
As $T$ is [[Definition:Scattered Space|scatt... | Scattered T1 Space is Totally Disconnected | https://proofwiki.org/wiki/Scattered_T1_Space_is_Totally_Disconnected | https://proofwiki.org/wiki/Scattered_T1_Space_is_Totally_Disconnected | [
"Scattered Spaces",
"T1 Spaces",
"Totally Disconnected Spaces",
"Sequence of Implications of Disconnectedness Properties"
] | [
"Definition:Scattered Space",
"Definition:T1 Space",
"Definition:Totally Disconnected Space"
] | [
"Definition:Scattered Space",
"Definition:T1 Space",
"Non-Trivial Connected Set in T1 Space is Dense-in-itself",
"Definition:Scattered Space",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Dense-in-itself",
"Definition:Element",
"Definition:Connected Set (Topology)",
"Definition:Totally... |
proofwiki-3552 | Non-Trivial Connected Set in T1 Space is Dense-in-itself | Let $T = \struct {S, \tau}$ be a $T_1$ space.
Let $H \subseteq S$ be connected in $T$.
If $H$ has more than one element, then $H$ is dense-in-itself. | {{AimForCont}} $H$ is not dense-in-itself.
Then $\exists x \in H$ such that $x$ is isolated in $H$.
That is, $\exists U \in \tau: U \cap H = \set x$.
Since $H$ has more than one element, we can find $y \in H$ with $y \ne x$.
Since $T$ is a $T_1$ space:
:$\forall y \in H: y \ne x \implies \paren {\exists V_y \in \tau: y... | Let $T = \struct {S, \tau}$ be a [[Definition:T1 Space|$T_1$ space]].
Let $H \subseteq S$ be [[Definition:Connected Set (Topology)|connected]] in $T$.
If $H$ has more than one [[Definition:Element|element]], then $H$ is [[Definition:Dense-in-itself|dense-in-itself]]. | {{AimForCont}} $H$ is not [[Definition:Dense-in-itself|dense-in-itself]].
Then $\exists x \in H$ such that $x$ is [[Definition:Isolated Point of Subset|isolated]] in $H$.
That is, $\exists U \in \tau: U \cap H = \set x$.
Since $H$ has more than one [[Definition:Element|element]], we can find $y \in H$ with $y \ne x... | Non-Trivial Connected Set in T1 Space is Dense-in-itself | https://proofwiki.org/wiki/Non-Trivial_Connected_Set_in_T1_Space_is_Dense-in-itself | https://proofwiki.org/wiki/Non-Trivial_Connected_Set_in_T1_Space_is_Dense-in-itself | [
"T1 Spaces",
"Dense-in-itself",
"Connected Sets (Topology)"
] | [
"Definition:T1 Space",
"Definition:Connected Set (Topology)",
"Definition:Element",
"Definition:Dense-in-itself"
] | [
"Definition:Dense-in-itself",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Element",
"Definition:T1 Space",
"Definition:Non-Empty Set",
"Definition:Disconnected (Topology)/Set",
"Definition:Contradiction",
"Proof by Contradiction",
"Definition:Dense-in-itself"
] |
proofwiki-3553 | Sequence of Implications of Disconnectedness Properties | Let $P_1$ and $P_2$ be disconnectedness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satisfies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | Regular||
| align="center" | $\impliedby$ ||
| align="center" | Zer... | The relevant justifications are listed as follows:
:Discrete Space is $T_0$ and Discrete Space is Zero Dimensional
:Discrete Space is $T_1$ and Discrete Space is Scattered
:Zero Dimensional Space is $T_3$, and by definition of a regular space as being both $T_3$ and $T_0$
:Zero Dimensional $T_0$ Space is Totally Separa... | Let $P_1$ and $P_2$ be [[Definition:Disconnected Space|disconnectedness properties]].
Let:
:$P_1 \implies P_2$
mean:
:If a [[Definition:Topological Space|topological space]] $T$ satisfies [[Definition:Property|property]] $P_1$, then $T$ also satisfies [[Definition:Property|property]] $P_2$.
Then the following sequen... | The relevant justifications are listed as follows:
:[[Discrete Space satisfies all Separation Properties|Discrete Space is $T_0$]] and [[Discrete Space is Zero Dimensional]]
:[[Discrete Space satisfies all Separation Properties|Discrete Space is $T_1$]] and [[Discrete Space is Scattered|Discrete Space is Scattered]]
:[... | Sequence of Implications of Disconnectedness Properties | https://proofwiki.org/wiki/Sequence_of_Implications_of_Disconnectedness_Properties | https://proofwiki.org/wiki/Sequence_of_Implications_of_Disconnectedness_Properties | [
"Sequence of Implications of Disconnectedness Properties",
"Topological Connectedness"
] | [
"Definition:Disconnected (Topology)/Topological Space",
"Definition:Topological Space",
"Definition:Property",
"Definition:Property",
"Definition:Regular Space",
"Definition:Zero Dimensional Space",
"Definition:T0 Space",
"Definition:Discrete Topology",
"Definition:Scattered Space",
"Definition:T1... | [
"Discrete Space satisfies all Separation Properties",
"Discrete Space is Zero Dimensional",
"Discrete Space satisfies all Separation Properties",
"Discrete Space is Scattered",
"Zero Dimensional Space is T3",
"Definition:Regular Space",
"Definition:T3 Space",
"Definition:T0 Space",
"Zero Dimensional... |
proofwiki-3554 | Scattered Space is T0 | Let $T = \struct {S, \tau}$ be a scattered topological space.
Then $T$ is also a $T_0$ space. | {{Recall|T0 Space|$T_0$ space|index = 2}}
{{:Definition:T0 Space/Definition 2}}
Suppose $T$ is not a $T_0$ space.
Hence by definition there exist $x, y \in S$ such that $x$ and $y$ are both limit points of each other.
So by definition of isolated point, neither $x$ nor $y$ are isolated in $\set {x, y}$.
Thus we have fo... | Let $T = \struct {S, \tau}$ be a [[Definition:Scattered Space|scattered topological space]].
Then $T$ is also a [[Definition:T0 Space|$T_0$ space]]. | {{Recall|T0 Space|$T_0$ space|index = 2}}
{{:Definition:T0 Space/Definition 2}}
Suppose $T$ is not a [[Definition:T0 Space|$T_0$ space]].
Hence by definition there exist $x, y \in S$ such that $x$ and $y$ are both [[Definition:Limit Point of Point|limit points]] of each other.
So by definition of [[Definition:Isolat... | Scattered Space is T0 | https://proofwiki.org/wiki/Scattered_Space_is_T0 | https://proofwiki.org/wiki/Scattered_Space_is_T0 | [
"T0 Spaces",
"Scattered Spaces",
"Sequence of Implications of Disconnectedness Properties"
] | [
"Definition:Scattered Space",
"Definition:T0 Space"
] | [
"Definition:T0 Space",
"Definition:Limit Point/Topology/Point",
"Definition:Isolated Point of Subset/Definition 2",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Subset",
"Definition:Dense-in-itself",
"Definition:Scattered Space",
"Rule of Transposition"
] |
proofwiki-3555 | Set with Dispersion Point is Biconnected | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a connected set in $T$.
Let $p \in H$ be a dispersion point of $H$.
Then $H$ is biconnected. | {{Recall|Biconnected Set|biconnected set}}
{{:Definition:Biconnected Set}}
{{AimForCont}} $H$ is not biconnected.
Then by definition there exist disjoint non-degenerate connected sets $U, V$ such that $H = U \cup V$.
{{WLOG}}, let $p \in U$.
Then $V \subset H \setminus \set p$.
As $p$ is a dispersion point of $H$, $H \... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$ be a [[Definition:Connected Set (Topology)|connected set]] in $T$.
Let $p \in H$ be a [[Definition:Dispersion Point|dispersion point]] of $H$.
Then $H$ is [[Definition:Biconnected Set|biconnected]]. | {{Recall|Biconnected Set|biconnected set}}
{{:Definition:Biconnected Set}}
{{AimForCont}} $H$ is not [[Definition:Biconnected Set|biconnected]].
Then by definition there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Non-Degenerate Connected Set|non-degenerate connected sets]] $U, V$ such that $H = U \cup V... | Set with Dispersion Point is Biconnected | https://proofwiki.org/wiki/Set_with_Dispersion_Point_is_Biconnected | https://proofwiki.org/wiki/Set_with_Dispersion_Point_is_Biconnected | [
"Biconnected Sets",
"Dispersion Points"
] | [
"Definition:Topological Space",
"Definition:Connected Set (Topology)",
"Definition:Dispersion Point",
"Definition:Biconnected Set"
] | [
"Definition:Biconnected Set",
"Definition:Disjoint Sets",
"Definition:Degenerate Connected Set/Non-Degenerate",
"Definition:Dispersion Point",
"Definition:Totally Disconnected Space",
"Definition:Degenerate Connected Set/Non-Degenerate",
"Proof by Contradiction",
"Definition:Biconnected Set"
] |
proofwiki-3556 | Totally Disconnected Space is Punctiform | Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is punctiform. | Let $T = \struct {S, \tau}$ be totally disconnected.
Then by definition its components are singletons.
Thus by definition each of its connected sets are degenerate.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Totally Disconnected Space|totally disconnected]].
Then $T$ is [[Definition:Punctiform Space|punctiform]]. | Let $T = \struct {S, \tau}$ be [[Definition:Totally Disconnected Space|totally disconnected]].
Then by definition its [[Definition:Component (Topology)|components]] are [[Definition:Singleton|singletons]].
Thus by definition each of its [[Definition:Connected Set (Topology)|connected sets]] are [[Definition:Degenerat... | Totally Disconnected Space is Punctiform | https://proofwiki.org/wiki/Totally_Disconnected_Space_is_Punctiform | https://proofwiki.org/wiki/Totally_Disconnected_Space_is_Punctiform | [
"Totally Disconnected Spaces",
"Punctiform Spaces"
] | [
"Definition:Topological Space",
"Definition:Totally Disconnected Space",
"Definition:Punctiform Space"
] | [
"Definition:Totally Disconnected Space",
"Definition:Component (Topology)",
"Definition:Singleton",
"Definition:Connected Set (Topology)",
"Definition:Degenerate Connected Set"
] |
proofwiki-3557 | Distance in Pseudometric is Non-Negative | Let $X$ be a set on which a pseudometric $d: X \times X \to \R$ has been imposed.
Then:
:$\forall x, y \in X: \map d {x, y} \ge 0$ | By definition of pseudometric, we have that:
{{begin-axiom}}
{{axiom | n = \text M 1
| q = \forall x \in A
| m = \map d {x, x} = 0
}}
{{axiom | n = \text M 2
| q = \forall x, y, z \in A
| m = \map d {x, y} + \map d {y, z} \ge \map d {x, z}
}}
{{axiom | n = \text M 3
| q = \forall... | Let $X$ be a [[Definition:Set|set]] on which a [[Definition:Pseudometric|pseudometric]] $d: X \times X \to \R$ has been imposed.
Then:
:$\forall x, y \in X: \map d {x, y} \ge 0$ | By definition of [[Definition:Pseudometric|pseudometric]], we have that:
{{begin-axiom}}
{{axiom | n = \text M 1
| q = \forall x \in A
| m = \map d {x, x} = 0
}}
{{axiom | n = \text M 2
| q = \forall x, y, z \in A
| m = \map d {x, y} + \map d {y, z} \ge \map d {x, z}
}}
{{axiom | n = \t... | Distance in Pseudometric is Non-Negative | https://proofwiki.org/wiki/Distance_in_Pseudometric_is_Non-Negative | https://proofwiki.org/wiki/Distance_in_Pseudometric_is_Non-Negative | [
"Pseudometrics",
"Pseudometric Spaces"
] | [
"Definition:Set",
"Definition:Pseudometric"
] | [
"Definition:Pseudometric",
"Category:Pseudometrics",
"Category:Pseudometric Spaces"
] |
proofwiki-3558 | Metric Defines Norm iff it Preserves Linear Structure | Let $\struct {k, \norm {\,\cdot\,}_k}$ be a valued field.
Let $V$ be a vector space over the valued field $\struct {k, \norm {\,\cdot\,}_k}$.
Let $d: V \times V \to k$ be a metric on $V$.
Then the function $\norm v := \map d {v, 0}$ is a norm on $V$ {{iff}} for all $x, y, z \in V$, $\lambda \in k$:
:$(1): \quad \map d ... | Suppose first that $d$ satisfies the hypotheses $(1)$ and $(2)$.
From {{Metric-space-axiom|4}}:
:$\forall u, v \in V: \map d {u, v} \ge 0$
Hence:
:$\forall u \in V: \norm u = \map d {u, 0} \ge 0$
Moreover, from {{Metric-space-axiom|1}}:
:$\norm u = 0 \implies \map d {u, 0} = 0$
and hence:
:$u = 0$
Now let $\lambda \in ... | Let $\struct {k, \norm {\,\cdot\,}_k}$ be a [[Definition:Valued Field|valued field]].
Let $V$ be a [[Definition:Vector Space|vector space]] over the [[Definition:Valued Field|valued field]] $\struct {k, \norm {\,\cdot\,}_k}$.
Let $d: V \times V \to k$ be a [[Definition:Metric|metric]] on $V$.
Then the function $\no... | Suppose first that $d$ satisfies the [[Definition:Hypothesis|hypotheses]] $(1)$ and $(2)$.
From {{Metric-space-axiom|4}}:
:$\forall u, v \in V: \map d {u, v} \ge 0$
Hence:
:$\forall u \in V: \norm u = \map d {u, 0} \ge 0$
Moreover, from {{Metric-space-axiom|1}}:
:$\norm u = 0 \implies \map d {u, 0} = 0$
and hence:
:... | Metric Defines Norm iff it Preserves Linear Structure | https://proofwiki.org/wiki/Metric_Defines_Norm_iff_it_Preserves_Linear_Structure | https://proofwiki.org/wiki/Metric_Defines_Norm_iff_it_Preserves_Linear_Structure | [
"Metric Spaces"
] | [
"Definition:Valued Field",
"Definition:Vector Space",
"Definition:Valued Field",
"Definition:Metric Space/Metric",
"Definition:Norm/Vector Space",
"Definition:Homogeneous (Metric Spaces)",
"Definition:Translation Mapping",
"Definition:Invariant",
"Definition:Enlargement Property"
] | [
"Definition:Hypothesis",
"Category:Metric Spaces"
] |
proofwiki-3559 | Real Numbers form Perfect Set | Consider the set of real numbers $\R$ as a (complete) metric space with the usual (Euclidean) metric.
Then $\R$ forms a perfect set. | {{Recall|Perfect Set|perfect set}}
{{:Definition:Perfect Set/Definition 1}}
Let $x \in \R$.
Consider the sequence:
:$\sequence {y_k} = x + \dfrac 1 k$
Then as $\sequence {z_k} = \dfrac 1 k$ is a basic null sequence it follows that:
:$\ds \lim_{n \mathop \to \infty} \sequence {y_k} = x$
Thus we see that $x$ is a limit p... | Consider the [[Definition:Real Number|set of real numbers]] $\R$ as a [[Real Number Line is Complete Metric Space|(complete) metric space]] with the [[Definition:Euclidean Metric on Real Number Line|usual (Euclidean) metric]].
Then $\R$ forms a [[Definition:Perfect Set|perfect set]]. | {{Recall|Perfect Set|perfect set}}
{{:Definition:Perfect Set/Definition 1}}
Let $x \in \R$.
Consider the [[Definition:Sequence|sequence]]:
:$\sequence {y_k} = x + \dfrac 1 k$
Then as $\sequence {z_k} = \dfrac 1 k$ is a [[Definition:Basic Null Sequences|basic null sequence]] it follows that:
:$\ds \lim_{n \mathop \to... | Real Numbers form Perfect Set | https://proofwiki.org/wiki/Real_Numbers_form_Perfect_Set | https://proofwiki.org/wiki/Real_Numbers_form_Perfect_Set | [
"Real Numbers",
"Examples of Perfect Sets"
] | [
"Definition:Real Number",
"Real Number Line is Complete Metric Space",
"Definition:Euclidean Metric/Real Number Line",
"Definition:Perfect Set"
] | [
"Definition:Sequence",
"Definition:Basic Null Sequence",
"Definition:Limit Point/Topology/Set",
"Category:Real Numbers",
"Category:Examples of Perfect Sets"
] |
proofwiki-3560 | Closed Interval in Reals is Uncountable | Let $a, b$ be extended real numbers such that $a < b$.
Then the closed interval $\set {x \in \R : a \le x \le b} \subseteq \R$ is uncountable. | First suppose that $a, b \in \R$.
We have that the unit interval is uncountable.
{{MissingLinks|Unit Interval is Uncountable (page does not exist at time of editing)}}
Let $f: \closedint 0 1 \to \closedint a b$ such that $\map f x = a + \paren {b - a} x$.
Then if $\map f {x_1} = \map f {x_2}$, we have:
:$a + \paren {b ... | Let $a, b$ be [[Definition:Extended Real Number Line|extended real numbers]] such that $a < b$.
Then the [[Definition:Closed Real Interval|closed interval]] $\set {x \in \R : a \le x \le b} \subseteq \R$ is [[Definition:Uncountable Set|uncountable]]. | First suppose that $a, b \in \R$.
We have that the [[Real Numbers are Uncountably Infinite/Cantor's Diagonal Argument|unit interval is uncountable]].
{{MissingLinks|[[Unit Interval is Uncountable]] (page does not exist at time of editing)}}
Let $f: \closedint 0 1 \to \closedint a b$ such that $\map f x = a + \paren {b... | Closed Interval in Reals is Uncountable | https://proofwiki.org/wiki/Closed_Interval_in_Reals_is_Uncountable | https://proofwiki.org/wiki/Closed_Interval_in_Reals_is_Uncountable | [
"Uncountable Sets"
] | [
"Definition:Extended Real Number Line",
"Definition:Real Interval/Closed",
"Definition:Uncountable/Set"
] | [
"Real Numbers are Uncountably Infinite/Cantor's Diagonal Argument",
"Unit Interval is Uncountable",
"Definition:Injective",
"Definition:Countable Set",
"Definition:Injective",
"Composite of Injections is Injection",
"Definition:Countable Set",
"Definition:Real Interval/Closed",
"Definition:Uncountab... |
proofwiki-3561 | Metric Space is T5 | A metric space $M = \struct {A, d}$ is a $T_5$ space. | {{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
Let $S, T \subseteq A$ such that $S$ and $T$ are separated in $A$.
Then:
:each point $x \in S$ has an open $\epsilon$-ball $\map {B_{\epsilon_x} } x$ which is disjoint from $T$
:each point $y \in T$ has an open $\epsilon$-ball $\map {B_{\epsilon_y} }... | A [[Definition:Metric Space|metric space]] $M = \struct {A, d}$ is a [[Definition:T5 Space|$T_5$ space]]. | {{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
Let $S, T \subseteq A$ such that $S$ and $T$ are [[Definition:Separated Sets|separated]] in $A$.
Then:
:each point $x \in S$ has an [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] $\map {B_{\epsilon_x} } x$ which is [[Definition:Disjo... | Metric Space is T5 | https://proofwiki.org/wiki/Metric_Space_is_T5 | https://proofwiki.org/wiki/Metric_Space_is_T5 | [
"Metric Space fulfils all Separation Axioms",
"Examples of T5 Spaces"
] | [
"Definition:Metric Space",
"Definition:T5 Space"
] | [
"Definition:Separated Sets",
"Definition:Open Ball",
"Definition:Disjoint Sets",
"Definition:Open Ball",
"Definition:Disjoint Sets",
"Definition:Disjoint Sets",
"Definition:Open Neighborhood",
"Definition:T5 Space"
] |
proofwiki-3562 | Metric Space is Perfectly T4 | A metric space $M = \struct {A, d}$ is a perfectly $T_4$ space. | We have that a metric space is $T_4$.
We also have that every closed set in a metric space is a $G_\delta$ set.
Hence the result, by definition of a perfectly $T_4$ space.
{{qed}} | A [[Definition:Metric Space|metric space]] $M = \struct {A, d}$ is a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]]. | We have that a [[Metric Space is T4|metric space is $T_4$]].
We also have that every [[Closed Set in Metric Space is G-Delta|closed set in a metric space is a $G_\delta$ set]].
Hence the result, by definition of a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
{{qed}} | Metric Space is Perfectly T4 | https://proofwiki.org/wiki/Metric_Space_is_Perfectly_T4 | https://proofwiki.org/wiki/Metric_Space_is_Perfectly_T4 | [
"Metric Space fulfils all Separation Axioms",
"Examples of Perfectly T4 Spaces"
] | [
"Definition:Metric Space",
"Definition:Perfectly T4 Space"
] | [
"Metric Space is T4",
"Closed Set in Metric Space is G-Delta",
"Definition:Perfectly T4 Space"
] |
proofwiki-3563 | Metric Space is Fully T4 | A metric space $M = \struct {A, d}$ is a fully $T_4$ space. | {{Recall|Fully T4 Space|fully $T_4$ space}}
{{:Definition:Fully T4 Space}}
{{proof wanted}} | A [[Definition:Metric Space|metric space]] $M = \struct {A, d}$ is a [[Definition:Fully T4 Space|fully $T_4$ space]]. | {{Recall|Fully T4 Space|fully $T_4$ space}}
{{:Definition:Fully T4 Space}}
{{proof wanted}} | Metric Space is Fully T4 | https://proofwiki.org/wiki/Metric_Space_is_Fully_T4 | https://proofwiki.org/wiki/Metric_Space_is_Fully_T4 | [
"Metric Space fulfils all Separation Axioms",
"Examples of Fully T4 Spaces"
] | [
"Definition:Metric Space",
"Definition:Fully T4 Space"
] | [] |
proofwiki-3564 | Metric Space is Fully Normal | A metric space $M = \struct {A, d}$ is a fully normal space. | {{Recall|Fully Normal Space|fully normal space}}
{{:Definition:Fully Normal Space/Definition 2}}
From Metric Space is Fully $T_4$
:$M$ is a fully $T_4$ space.
From Metric Space is $T_1$
:$M$ is a $T_1$ space.
The result follows.
{{qed}} | A [[Definition:Metric Space|metric space]] $M = \struct {A, d}$ is a [[Definition:Fully Normal Space|fully normal space]]. | {{Recall|Fully Normal Space|fully normal space}}
{{:Definition:Fully Normal Space/Definition 2}}
From [[Metric Space is Fully T4|Metric Space is Fully $T_4$]]
:$M$ is a [[Definition:Fully T4 Space|fully $T_4$ space]].
From [[Metric Space is T1|Metric Space is $T_1$]]
:$M$ is a [[Definition:T1 Space| $T_1$ space]].
T... | Metric Space is Fully Normal | https://proofwiki.org/wiki/Metric_Space_is_Fully_Normal | https://proofwiki.org/wiki/Metric_Space_is_Fully_Normal | [
"Metric Space fulfils all Separation Axioms",
"Examples of Fully Normal Spaces"
] | [
"Definition:Metric Space",
"Definition:Fully Normal Space"
] | [
"Metric Space is Fully T4",
"Definition:Fully T4 Space",
"Metric Space is T1",
"Definition:T1 Space"
] |
proofwiki-3565 | Metric Space is Paracompact | Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a paracompact space. | We have that a Metric Space is Fully Normal.
Then we have that a Fully Normal Space is Paracompact.
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then $M$ is a [[Definition:Paracompact Space|paracompact space]]. | We have that a [[Metric Space is Fully Normal]].
Then we have that a [[Fully Normal Space is Paracompact]].
{{qed}} | Metric Space is Paracompact/Proof 1 | https://proofwiki.org/wiki/Metric_Space_is_Paracompact | https://proofwiki.org/wiki/Metric_Space_is_Paracompact/Proof_1 | [
"Metric Space is Paracompact",
"Metric Spaces",
"Paracompact Spaces"
] | [
"Definition:Metric Space",
"Definition:Paracompact Space"
] | [
"Metric Space is Fully Normal",
"Fully Normal Space is Paracompact"
] |
proofwiki-3566 | Metric Space is Paracompact | Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a paracompact space. | Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {C_\alpha}$ be an open cover of $M$ indexed by ordinals.
Let $\map {B_r} x$ be the open $r$-ball in $M$ around $x$.
For $n \in \N$, let $D_{\alpha n}$ denote the union of all the open $r$-balls $\map {B_{2^{-n} } } x$ such that:
:$(1): \quad \alpha$ is the smal... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then $M$ is a [[Definition:Paracompact Space|paracompact space]]. | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $\sequence {C_\alpha}$ be an [[Definition:Open Cover|open cover]] of $M$ [[Definition:Indexed Set|indexed]] by [[Definition:Ordinal|ordinals]].
Let $\map {B_r} x$ be the [[Definition:Open Ball of Metric Space|open $r$-ball]] in $M$ around $x$... | Metric Space is Paracompact/Proof 2 | https://proofwiki.org/wiki/Metric_Space_is_Paracompact | https://proofwiki.org/wiki/Metric_Space_is_Paracompact/Proof_2 | [
"Metric Space is Paracompact",
"Metric Spaces",
"Paracompact Spaces"
] | [
"Definition:Metric Space",
"Definition:Paracompact Space"
] | [
"Definition:Metric Space",
"Definition:Open Cover",
"Definition:Indexing Set/Indexed Set",
"Definition:Ordinal",
"Definition:Open Ball",
"Definition:Set Union",
"Definition:Open Ball",
"Definition:Ordinal",
"Definition:Locally Finite Cover",
"Definition:Open Refinement",
"Definition:Cover",
"D... |
proofwiki-3567 | Metric Space is First-Countable | Let $M = \struct {A, d}$ be a metric space.
Then $M$ is first-countable. | {{Recall|First-Countable Space|first-countable space}}
{{:Definition:First-Countable Space}}
Let $x \in A$ be arbitrary.
Let:
:$\BB = \set {\map {B_{1/n} } x: n \in \N_{>0} }$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$ in $M$.
By Surjection from Natural Numbers iff Countable, we have that $\BB$... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then $M$ is [[Definition:First-Countable Space|first-countable]]. | {{Recall|First-Countable Space|first-countable space}}
{{:Definition:First-Countable Space}}
Let $x \in A$ be [[Definition:Arbitrary|arbitrary]].
Let:
:$\BB = \set {\map {B_{1/n} } x: n \in \N_{>0} }$
where $\map {B_\epsilon} x$ denotes the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball of $x$ in $M$]].... | Metric Space is First-Countable | https://proofwiki.org/wiki/Metric_Space_is_First-Countable | https://proofwiki.org/wiki/Metric_Space_is_First-Countable | [
"Metric Spaces",
"First-Countable Spaces"
] | [
"Definition:Metric Space",
"Definition:First-Countable Space"
] | [
"Definition:Arbitrary",
"Definition:Open Ball",
"Surjection from Natural Numbers iff Countable",
"Definition:Countable Set",
"Definition:First-Countable Space",
"Definition:Local Basis",
"Open Ball is Open Set/Pseudometric Space",
"Definition:Element",
"Definition:Open Neighborhood/Point",
"Defini... |
proofwiki-3568 | Metric Space is Separable iff Second-Countable | A metric space is separable {{iff}} it is second-countable. | Follows directly from:
:Separable Metric Space is Second-Countable
:Second-Countable Space is Separable
{{qed}}
{{ACC|Second-Countable Space is Separable}} | A [[Definition:Metric Space|metric space]] is [[Definition:Separable Space|separable]] {{iff}} it is [[Definition:Second-Countable Space|second-countable]]. | Follows directly from:
:[[Separable Metric Space is Second-Countable]]
:[[Second-Countable Space is Separable]]
{{qed}}
{{ACC|Second-Countable Space is Separable}} | Metric Space is Separable iff Second-Countable | https://proofwiki.org/wiki/Metric_Space_is_Separable_iff_Second-Countable | https://proofwiki.org/wiki/Metric_Space_is_Separable_iff_Second-Countable | [
"Metric Spaces",
"Second-Countable Spaces",
"Separable Spaces",
"Sequence of Implications of Metric Space Compactness Properties"
] | [
"Definition:Metric Space",
"Definition:Separable Space",
"Definition:Second-Countable Space"
] | [
"Separable Metric Space is Second-Countable",
"Second-Countable Space is Separable"
] |
proofwiki-3569 | Metric Space is Lindelöf iff Second-Countable | Let $M = \struct {A, d}$ be a metric space.
Then $M$ is Lindelöf {{iff}} $M$ is second-countable. | === Sufficient Condition ===
We have from Second-Countable Space is Lindelöf that second-countability implies Lindelöf in all topological spaces, regardless of whether they are metric spaces or not.
{{qed|lemma}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then $M$ is [[Definition:Lindelöf Space|Lindelöf]] {{iff}} $M$ is [[Definition:Second-Countable Space|second-countable]]. | === Sufficient Condition ===
We have from [[Second-Countable Space is Lindelöf]] that [[Definition:Second-Countable Space|second-countability]] implies [[Definition:Lindelöf Space|Lindelöf]] in all [[Definition:Topological Space|topological spaces]], regardless of whether they are [[Definition:Metric Space|metric spac... | Metric Space is Lindelöf iff Second-Countable | https://proofwiki.org/wiki/Metric_Space_is_Lindelöf_iff_Second-Countable | https://proofwiki.org/wiki/Metric_Space_is_Lindelöf_iff_Second-Countable | [
"Metric Spaces",
"Lindelöf Spaces",
"Second-Countable Spaces",
"Sequence of Implications of Metric Space Compactness Properties"
] | [
"Definition:Metric Space",
"Definition:Lindelöf Space",
"Definition:Second-Countable Space"
] | [
"Second-Countable Space is Lindelöf",
"Definition:Second-Countable Space",
"Definition:Lindelöf Space",
"Definition:Topological Space",
"Definition:Metric Space",
"Definition:Lindelöf Space",
"Definition:Lindelöf Space",
"Definition:Metric Space",
"Definition:Second-Countable Space"
] |
proofwiki-3570 | Metric Space is Countably Compact iff Sequentially Compact | Let $M$ be a metric space.
Then $M$ is countably compact {{iff}} $M$ is sequentially compact. | This follows directly from the results:
:Countably Compact Metric Space is Sequentially Compact
:Sequentially Compact Space is Countably Compact
{{qed}}
{{ACC|Sequentially Compact Space is Countably Compact}} | Let $M$ be a [[Definition:Metric Space|metric space]].
Then $M$ is [[Definition:Countably Compact Space|countably compact]] {{iff}} $M$ is [[Definition:Sequentially Compact Space|sequentially compact]]. | This follows directly from the results:
:[[Countably Compact Metric Space is Sequentially Compact]]
:[[Sequentially Compact Space is Countably Compact]]
{{qed}}
{{ACC|Sequentially Compact Space is Countably Compact}} | Metric Space is Countably Compact iff Sequentially Compact | https://proofwiki.org/wiki/Metric_Space_is_Countably_Compact_iff_Sequentially_Compact | https://proofwiki.org/wiki/Metric_Space_is_Countably_Compact_iff_Sequentially_Compact | [
"Metric Spaces",
"Countably Compact Spaces",
"Sequentially Compact Spaces",
"Sequence of Implications of Metric Space Compactness Properties"
] | [
"Definition:Metric Space",
"Definition:Countably Compact Space",
"Definition:Sequentially Compact Space"
] | [
"Countably Compact Metric Space is Sequentially Compact",
"Sequentially Compact Space is Countably Compact"
] |
proofwiki-3571 | Metric Space is Weakly Countably Compact iff Countably Compact | A metric space $M = \struct {A, d}$ is weakly countably compact {{iff}} $M$ is countably compact. | From Metric Space is $T_1$, $M$ is a $T_1$ space.
The result follows from $T_1$ Space is Weakly Countably Compact iff Countably Compact.
{{qed}} | A [[Definition:Metric Space|metric space]] $M = \struct {A, d}$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} $M$ is [[Definition:Countably Compact Space|countably compact]]. | From [[Metric Space is T1|Metric Space is $T_1$]], $M$ is a [[Definition:T1 Space|$T_1$ space]].
The result follows from [[T1 Space is Weakly Countably Compact iff Countably Compact|$T_1$ Space is Weakly Countably Compact iff Countably Compact]].
{{qed}} | Metric Space is Weakly Countably Compact iff Countably Compact | https://proofwiki.org/wiki/Metric_Space_is_Weakly_Countably_Compact_iff_Countably_Compact | https://proofwiki.org/wiki/Metric_Space_is_Weakly_Countably_Compact_iff_Countably_Compact | [
"Metric Spaces",
"Countably Compact Spaces",
"Weakly Countably Compact Spaces",
"Sequence of Implications of Metric Space Compactness Properties"
] | [
"Definition:Metric Space",
"Definition:Weakly Countably Compact Space",
"Definition:Countably Compact Space"
] | [
"Metric Space is T1",
"Definition:T1 Space",
"T1 Space is Weakly Countably Compact iff Countably Compact"
] |
proofwiki-3572 | Countably Compact Metric Space is Compact | Let $M = \struct {A, d}$ be a metric space.
Let $M$ be countably compact.
Then $M$ is compact. | This follows directly from:
:Metric Space is First-Countable
:Countably Compact First-Countable Space is Sequentially Compact
:Sequentially Compact Metric Space is Second-Countable
:Second-Countable Space is Compact iff Countably Compact
{{qed}}
{{ACC|Sequentially Compact Metric Space is Second-Countable}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $M$ be [[Definition:Countably Compact Space|countably compact]].
Then $M$ is [[Definition:Compact Metric Space|compact]]. | This follows directly from:
:[[Metric Space is First-Countable]]
:[[Countably Compact First-Countable Space is Sequentially Compact]]
:[[Sequentially Compact Metric Space is Second-Countable]]
:[[Second-Countable Space is Compact iff Countably Compact]]
{{qed}}
{{ACC|Sequentially Compact Metric Space is Second-Countab... | Countably Compact Metric Space is Compact | https://proofwiki.org/wiki/Countably_Compact_Metric_Space_is_Compact | https://proofwiki.org/wiki/Countably_Compact_Metric_Space_is_Compact | [
"Countably Compact Metric Space is Compact",
"Countably Compact Spaces",
"Compact Metric Spaces",
"Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Countably Compact Space",
"Definition:Compact Space/Metric Space"
] | [
"Metric Space is First-Countable",
"Countably Compact First-Countable Space is Sequentially Compact",
"Sequentially Compact Metric Space is Second-Countable",
"Second-Countable Space is Compact iff Countably Compact"
] |
proofwiki-3573 | Metric Space is Weakly Locally Compact iff Strongly Locally Compact | Let $M = \struct {A, d}$ be a metric space.
Then $M$ is weakly locally compact {{iff}} $M$ is strongly locally compact. | Let $M = \struct {A, d}$ be a metric space. | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then $M$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]] {{iff}} $M$ is [[Definition:Strongly Locally Compact Space|strongly locally compact]]. | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]]. | Metric Space is Weakly Locally Compact iff Strongly Locally Compact | https://proofwiki.org/wiki/Metric_Space_is_Weakly_Locally_Compact_iff_Strongly_Locally_Compact | https://proofwiki.org/wiki/Metric_Space_is_Weakly_Locally_Compact_iff_Strongly_Locally_Compact | [
"Weakly Locally Compact Spaces",
"Strongly Locally Compact Spaces",
"Locally Compact Hausdorff Spaces",
"Metric Spaces",
"Sequence of Implications of Metric Space Compactness Properties"
] | [
"Definition:Metric Space",
"Definition:Weakly Locally Compact Space",
"Definition:Strongly Locally Compact Space"
] | [
"Definition:Metric Space"
] |
proofwiki-3574 | Sequence of Implications of Metric Space Compactness Properties | Let $P_1$ and $P_2$ be compactness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a metric space $M$ satisfies property $P_1$, then $M$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | Sequentially Compact ||
| align="center" | $\implies$ ||
| align="center" | W... | The relevant justifications are listed as follows:
:Metric Space is Compact iff Countably Compact.
:Metric Space is Countably Compact iff Sequentially Compact.
:Metric Space is Weakly Countably Compact iff Countably Compact.
:Compact Space is Weakly $\sigma$-Locally Compact.
:By definition, a weakly $\sigma$-locally co... | Let $P_1$ and $P_2$ be [[Definition:Compact Topological Space|compactness properties]].
Let:
:$P_1 \implies P_2$
mean:
:If a [[Definition:Metric Space|metric space]] $M$ satisfies [[Definition:Property|property]] $P_1$, then $M$ also satisfies [[Definition:Property|property]] $P_2$.
Then the following sequence of im... | The relevant justifications are listed as follows:
:[[Metric Space is Compact iff Countably Compact]].
:[[Metric Space is Countably Compact iff Sequentially Compact]].
:[[Metric Space is Weakly Countably Compact iff Countably Compact]].
:[[Compact Space is Weakly Sigma-Locally Compact|Compact Space is Weakly $\sigma$... | Sequence of Implications of Metric Space Compactness Properties | https://proofwiki.org/wiki/Sequence_of_Implications_of_Metric_Space_Compactness_Properties | https://proofwiki.org/wiki/Sequence_of_Implications_of_Metric_Space_Compactness_Properties | [
"Sequence of Implications of Metric Space Compactness Properties",
"Metric Spaces",
"Topological Compactness"
] | [
"Definition:Compact Topological Space",
"Definition:Metric Space",
"Definition:Property",
"Definition:Property",
"Definition:Sequentially Compact Space",
"Definition:Weakly Sigma-Locally Compact Space",
"Definition:Weakly Locally Compact Space",
"Definition:Countably Compact Space",
"Definition:Sigm... | [
"Metric Space is Compact iff Countably Compact",
"Metric Space is Countably Compact iff Sequentially Compact",
"Metric Space is Weakly Countably Compact iff Countably Compact",
"Compact Space is Weakly Sigma-Locally Compact",
"Definition:Weakly Sigma-Locally Compact Space",
"Definition:Weakly Locally Comp... |
proofwiki-3575 | Extremally Disconnected Metric Space is Discrete | Let $M = \struct {A, d}$ be a metric space which is extremally disconnected.
Then $M$ is the discrete topology. | Let $M = \struct {A, d}$ be extremally disconnected.
Let $p \in A$.
As $M$ is a metric space, $\set p$ can be expressed as:
:$\set p = \ds \bigcap_{n \mathop \in \N_{>0} } \paren {\map {B_{1 / n} } p}^-$
where:
:$\map {B_{1 / n} } p$ denotes the open $1 / n$-ball of $p$
:$\paren {\map {B_{1 / n} } p}^-$ denotes the clo... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Extremally Disconnected Space|extremally disconnected]].
Then $M$ is the [[Definition:Discrete Topology|discrete topology]]. | Let $M = \struct {A, d}$ be [[Definition:Extremally Disconnected Space|extremally disconnected]].
Let $p \in A$.
As $M$ is a [[Definition:Metric Space|metric space]], $\set p$ can be expressed as:
:$\set p = \ds \bigcap_{n \mathop \in \N_{>0} } \paren {\map {B_{1 / n} } p}^-$
where:
:$\map {B_{1 / n} } p$ denotes the... | Extremally Disconnected Metric Space is Discrete | https://proofwiki.org/wiki/Extremally_Disconnected_Metric_Space_is_Discrete | https://proofwiki.org/wiki/Extremally_Disconnected_Metric_Space_is_Discrete | [
"Metric Spaces",
"Extremally Disconnected Spaces",
"Discrete Topologies"
] | [
"Definition:Metric Space",
"Definition:Extremally Disconnected Space",
"Definition:Discrete Topology"
] | [
"Definition:Extremally Disconnected Space",
"Definition:Metric Space",
"Definition:Open Ball",
"Definition:Closure (Topology)",
"Definition:Set Intersection",
"Definition:Closure (Topology)",
"Definition:Open Ball",
"Definition:Natural Numbers",
"Definition:Set Complement",
"Definition:Open Set/To... |
proofwiki-3576 | Range of Characters | Let $G$ be a finite abelian group of order $m$.
Let $\chi: G \to \C^\times$ be a character on $G$.
Then for any $g \in G$, $\map \chi g$ is an $m$th root of unity.
If $e$ is the identity of $G$ then $\map \chi g = 1$. | The claim that $\map \chi e = 1$ is shown by Group Homomorphism Preserves Identity.
Let $g \in G$ be arbitrary.
Let $k$ be the order of $g$.
By Order of Element Divides Order of Finite Group:
:$\exists l \in \Z: m = k l$
Therefore:
:$g^m = \paren {g^k}^l = e^l = e$
By the homomorphism property:
:$1 = \map \chi e = \map... | Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] of [[Definition:Order of Structure|order]] $m$.
Let $\chi: G \to \C^\times$ be a [[Definition:Character (Number Theory)|character]] on $G$.
Then for any $g \in G$, $\map \chi g$ is an [[Definition:Complex Roots of Unity|$m$th r... | The claim that $\map \chi e = 1$ is shown by [[Group Homomorphism Preserves Identity]].
Let $g \in G$ be arbitrary.
Let $k$ be the [[Definition:Order of Group Element|order]] of $g$.
By [[Order of Element Divides Order of Finite Group]]:
:$\exists l \in \Z: m = k l$
Therefore:
:$g^m = \paren {g^k}^l = e^l = e$
By ... | Range of Characters | https://proofwiki.org/wiki/Range_of_Characters | https://proofwiki.org/wiki/Range_of_Characters | [
"Analytic Number Theory"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:Character (Number Theory)",
"Definition:Root of Unity/Complex",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Group Homomorphism Preserves Identity",
"Definition:Order of Group Element",
"Order of Element Divides Order of Finite Group",
"Definition:Group Homomorphism",
"Category:Analytic Number Theory"
] |
proofwiki-3577 | Indicator is Well-Defined | Let $G$ be a finite group, and $a \in G$.
Let $H$ be a subgroup of $G$.
Then the indicator of $a$ in $H$ is well defined. | If $a \in H$, then for $n = 1$, $a^n \in H$.
Suppose that $a \notin H$. By Identity of Subgroup, we have that the identity $e$ of $G$ is in $H$.
Moreover, by Element of Finite Group is of Finite Order, there is $n \in \N$ such that $a^n = e \in H$.
Therefore, for any $g \in G$, there is ''some'' $n \in \N$ such that $a... | Let $G$ be a [[Definition:Finite Group|finite group]], and $a \in G$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then the [[Definition:Indicator of Group Element|indicator]] of $a$ in $H$ is well defined. | If $a \in H$, then for $n = 1$, $a^n \in H$.
Suppose that $a \notin H$. By [[Identity of Subgroup]], we have that the [[Definition:Identity Element|identity]] $e$ of $G$ is in $H$.
Moreover, by [[Element of Finite Group is of Finite Order]], there is $n \in \N$ such that $a^n = e \in H$.
Therefore, for any $g \in G$... | Indicator is Well-Defined | https://proofwiki.org/wiki/Indicator_is_Well-Defined | https://proofwiki.org/wiki/Indicator_is_Well-Defined | [
"Subgroups"
] | [
"Definition:Finite Group",
"Definition:Subgroup",
"Definition:Indicator of Group Element"
] | [
"Identity of Subgroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Element of Finite Group is of Finite Order",
"Definition:Positive/Integer",
"Definition:Non-Empty Set",
"Well-Ordering Principle",
"Definition:Indicator of Group Element",
"Category:Subgroups"
] |
proofwiki-3578 | Compact Metric Space is Totally Bounded | Let $M = \struct {A, d}$ be a metric space which is compact.
Then $M$ is totally bounded. | Let $M = \struct {A, d}$ be compact.
Let $\epsilon > 0$.
Then the family $\set {\map {B_\epsilon} x: x \in A}$ of open $\epsilon$-balls forms an open cover of $A$.
By the definition of compact, every open cover for $A$ has a finite subcover.
That is, there are points $x_0, \ldots, x_n$ such that:
:$\ds A = \bigcup_{0 \... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Compact Metric Space|compact]].
Then $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]]. | Let $M = \struct {A, d}$ be [[Definition:Compact Metric Space|compact]].
Let $\epsilon > 0$.
Then the family $\set {\map {B_\epsilon} x: x \in A}$ of [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] forms an [[Definition:Open Cover|open cover]] of $A$.
By the definition of [[Definition:Compact Metric S... | Compact Metric Space is Totally Bounded | https://proofwiki.org/wiki/Compact_Metric_Space_is_Totally_Bounded | https://proofwiki.org/wiki/Compact_Metric_Space_is_Totally_Bounded | [
"Totally Bounded Metric Spaces",
"Compact Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Compact Space/Metric Space",
"Definition:Totally Bounded Metric Space"
] | [
"Definition:Compact Space/Metric Space",
"Definition:Open Ball",
"Definition:Open Cover",
"Definition:Compact Space/Metric Space",
"Definition:Open Cover",
"Definition:Subcover/Finite"
] |
proofwiki-3579 | Compact Metric Space is Totally Bounded | Let $M = \struct {A, d}$ be a metric space which is compact.
Then $M$ is totally bounded. | Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By definition, $M$ is totally bounded only if there exists a finite $\epsilon$-net for $M$.
{{AimForCont}} there exists no finite $\epsilon$-net for $M$.
The aim is to construct an infinite sequence $\sequence {x_n}_{n \ge 1}$ in $A$ that has no convergent ... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Compact Metric Space|compact]].
Then $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]]. | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
By definition, $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]] only if there exists a [[Definition:Finite Epsilon-Net|finite $\epsilon$-net]] for $M$.
{{AimForCont}} there exists no [[Defini... | Sequentially Compact Metric Space is Totally Bounded/Proof 1 | https://proofwiki.org/wiki/Compact_Metric_Space_is_Totally_Bounded | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Totally_Bounded/Proof_1 | [
"Totally Bounded Metric Spaces",
"Compact Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Compact Space/Metric Space",
"Definition:Totally Bounded Metric Space"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Totally Bounded Metric Space",
"Definition:Epsilon-Net/Finite Net",
"Definition:Epsilon-Net/Finite Net",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Subsequence",
"Definition:Natural Numbers... |
proofwiki-3580 | Compact Metric Space is Totally Bounded | Let $M = \struct {A, d}$ be a metric space which is compact.
Then $M$ is totally bounded. | We have {{hypothesis}} that $M$ is a sequentially compact space.
So {{afortiori}} every infinite sequence in $M$ has a subsequence which converges to a point in $A$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By definition, $M$ is totally bounded {{iff}} there exists a finite $\epsilon$-net for $M$.... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Compact Metric Space|compact]].
Then $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]]. | We have {{hypothesis}} that $M$ is a [[Definition:Sequentially Compact Space|sequentially compact space]].
So {{afortiori}} every [[Definition:Infinite Sequence|infinite sequence]] in $M$ has a [[Definition:Subsequence|subsequence]] which [[Definition:Convergent Sequence (Metric Space)|converges]] to a point in $A$.
... | Sequentially Compact Metric Space is Totally Bounded/Proof 2 | https://proofwiki.org/wiki/Compact_Metric_Space_is_Totally_Bounded | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Totally_Bounded/Proof_2 | [
"Totally Bounded Metric Spaces",
"Compact Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Compact Space/Metric Space",
"Definition:Totally Bounded Metric Space"
] | [
"Definition:Sequentially Compact Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Subsequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Strictly Positive/Real Number",
"Definition:Totally Bounded Metric Space",
"Definition:Epsilon-Net/Finite Net",
"Definition:Epsilon-Net... |
proofwiki-3581 | Subgroup Generated by Subgroup and Element | Let $G$ be a finite abelian group.
Let $H$ be a proper subgroup of $G$.
Let $a \in G \setminus H$.
Let $n$ be the indicator of $a$ in $H$.
Then:
:$K = \left\{{x a^k: x \in H, \ 0 \le k < n}\right\}$
is a subgroup of $G$ such that $H \subseteq K$, and each element of $K$ has a unique representation in this form.
Moreove... | === $K$ is Subgroup of $G$ ===
We first show that $K$ is a subgroup of $G$ using the Two-Step Subgroup Test.
$(1): \quad K \ne \varnothing$:
From Identity of Subgroup:
:$e \in H$
From Indicator is Well-Defined, $n > 0$ and so $n - 1 \ge 0$.
Thus $k = 0$ fulfils the condition that $0 \le k < n$, and so:
:$e = e a^0 \in ... | Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$.
Let $a \in G \setminus H$.
Let $n$ be the [[Definition:Indicator of Group Element|indicator]] of $a$ in $H$.
Then:
:$K = \left\{{x a^k: x \in H, \ 0 \le k... | === $K$ is [[Definition:Subgroup|Subgroup]] of $G$ ===
We first show that $K$ is a [[Definition:Subgroup|subgroup]] of $G$ using the [[Two-Step Subgroup Test]].
$(1): \quad K \ne \varnothing$:
From [[Identity of Subgroup]]:
:$e \in H$
From [[Indicator is Well-Defined]], $n > 0$ and so $n - 1 \ge 0$.
Thus $k = 0$ f... | Subgroup Generated by Subgroup and Element | https://proofwiki.org/wiki/Subgroup_Generated_by_Subgroup_and_Element | https://proofwiki.org/wiki/Subgroup_Generated_by_Subgroup_and_Element | [
"Subgroups"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Proper Subgroup",
"Definition:Indicator of Group Element",
"Definition:Subgroup",
"Definition:Element",
"Definition:Unique",
"Definition:Generator of Subgroup",
"Definition:Order of Structure"
] | [
"Definition:Subgroup",
"Definition:Subgroup",
"Two-Step Subgroup Test",
"Identity of Subgroup",
"Indicator is Well-Defined",
"Definition:Closure (Abstract Algebra)",
"Definition:Abelian Group",
"Definition:Commutative/Elements",
"Definition:Group",
"Definition:Indicator of Group Element",
"Defin... |
proofwiki-3582 | Number of Characters on Finite Abelian Group | Let $G$ be a finite abelian group.
Then the number of characters $G \to \C^\times$ is $\order G$. | === Lemma ===
Let $H \le G$ be a subgroup.
Let $\chi: H \to \C^\times$ be a character on $G$.
Let $a \in G \divides H$ where $\divides$ denotes divisibility.
Then:
:$\chi$ extends to $\index G H$ distinct characters on $G$
where $\index G H$ denotes the index of $H$ in $G$. | Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
Then the number of [[Definition:Character (Number Theory)|characters]] $G \to \C^\times$ is $\order G$. | === Lemma ===
Let $H \le G$ be a [[Definition:Subgroup|subgroup]].
Let $\chi: H \to \C^\times$ be a [[Definition:Character (Number Theory)|character]] on $G$.
Let $a \in G \divides H$ where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$\chi$ [[Definition:Extension of Mapping|extends]] ... | Number of Characters on Finite Abelian Group | https://proofwiki.org/wiki/Number_of_Characters_on_Finite_Abelian_Group | https://proofwiki.org/wiki/Number_of_Characters_on_Finite_Abelian_Group | [
"Analytic Number Theory"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Character (Number Theory)"
] | [
"Definition:Subgroup",
"Definition:Character (Number Theory)",
"Definition:Divisor (Algebra)/Integer",
"Definition:Extension of Mapping",
"Definition:Character (Number Theory)",
"Definition:Index of Subgroup",
"Definition:Extension of Mapping"
] |
proofwiki-3583 | Sequentially Compact Metric Space is Complete | Let $M = \struct {A, d}$ be a metric space which is sequentially compact.
Then $M$ is complete. | Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $A$.
As $M$ is sequentially compact, $\sequence {x_n}$ has a convergent subsequence.
By Convergent Subsequence of Cauchy Sequence in Metric Space, this implies that the entire sequence $\sequence {x_n}$ is convergent.
Hence, $M$ is complete.
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Sequentially Compact Space|sequentially compact]].
Then $M$ is [[Definition:Complete Metric Space|complete]]. | Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A$.
As $M$ is [[Definition:Sequentially Compact Space|sequentially compact]], $\sequence {x_n}$ has a [[Definition:Convergent Sequence (Metric Space)|convergent]] [[Definition:Subsequence|subsequence]].
By... | Sequentially Compact Metric Space is Complete | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Complete | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Complete | [
"Sequentially Compact Spaces",
"Complete Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Sequentially Compact Space",
"Definition:Complete Metric Space"
] | [
"Definition:Cauchy Sequence/Metric Space",
"Definition:Sequentially Compact Space",
"Definition:Convergent Sequence/Metric Space",
"Definition:Subsequence",
"Convergent Subsequence of Cauchy Sequence/Metric Space",
"Definition:Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Compl... |
proofwiki-3584 | Complete Metrizability is Weakly Hereditary | Let $T = \struct {S, \tau}$ be a topological space which is completely metrizable.
Let $V \subseteq S$ be a closed subspace of $T$.
Then $V$ is also completely metrizable.
That is, complete metrizability is a weakly hereditary property. | {{finish|Use Subspace of Complete Metric Space is Closed iff Complete}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Completely Metrizable Space|completely metrizable]].
Let $V \subseteq S$ be a [[Definition:Closed Set (Topology)|closed]] [[Definition:Topological Subspace|subspace]] of $T$.
Then $V$ is also [[Definition:Comple... | {{finish|Use [[Subspace of Complete Metric Space is Closed iff Complete]]}} | Complete Metrizability is Weakly Hereditary | https://proofwiki.org/wiki/Complete_Metrizability_is_Weakly_Hereditary | https://proofwiki.org/wiki/Complete_Metrizability_is_Weakly_Hereditary | [
"Completely Metrizable Spaces",
"Examples of Weakly Hereditary Properties"
] | [
"Definition:Topological Space",
"Definition:Completely Metrizable Space",
"Definition:Closed Set/Topology",
"Definition:Topological Subspace",
"Definition:Completely Metrizable Space",
"Definition:Completely Metrizable Space",
"Definition:Weakly Hereditary Property"
] | [
"Subspace of Complete Metric Space is Closed iff Complete"
] |
proofwiki-3585 | Metric Space Completeness is Preserved by Isometry | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
If $M_1$ is complete then so is $M_2$. | Let $\tau_1$ be the topology on $A_1$ induced by $d_1$.
Let $\tau_2$ be the topology on $A_2$ induced by $d_2$.
Let $\sequence {x_n}$ be a Cauchy sequence in $A_2$.
From Inverse of Isometry of Metric Spaces is Isometry, $\phi^{-1}$ is an isometry.
Since Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence ... | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $\phi: M_1 \to M_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
If $M_1$ is [[Definition:Complete Metric Space|complete]] then so is $M_2$. | Let $\tau_1$ be the [[Definition:Topology Induced by Metric|topology on $A_1$ induced by $d_1$]].
Let $\tau_2$ be the [[Definition:Topology Induced by Metric|topology on $A_2$ induced by $d_2$]].
Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A_2$.
From [[Inverse of Isom... | Metric Space Completeness is Preserved by Isometry/Proof 1 | https://proofwiki.org/wiki/Metric_Space_Completeness_is_Preserved_by_Isometry | https://proofwiki.org/wiki/Metric_Space_Completeness_is_Preserved_by_Isometry/Proof_1 | [
"Metric Space Completeness is Preserved by Isometry",
"Complete Metric Spaces",
"Isometries (Metric Spaces)"
] | [
"Definition:Metric Space",
"Definition:Isometry (Metric Spaces)",
"Definition:Complete Metric Space"
] | [
"Definition:Topology Induced by Metric",
"Definition:Topology Induced by Metric",
"Definition:Cauchy Sequence/Metric Space",
"Equivalence of Definitions of Isometry of Metric Spaces",
"Definition:Isometry (Metric Spaces)",
"Isometric Image of Cauchy Sequence is Cauchy Sequence",
"Definition:Cauchy Seque... |
proofwiki-3586 | Metric Space Completeness is Preserved by Isometry | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
If $M_1$ is complete then so is $M_2$. | Let $\epsilon \in \R_{>0}$.
Let $\sequence {b_n}$ be a Cauchy sequence in $A_2$.
Thus:
:$\exists N_1 \in \N: \map {d_2} {b_n, b_m} < \epsilon$
whenever $n, m \ge N_1$ and $b_n, b_m \in A_2$.
We have that $M_1$ is isometric to $M_2$.
Isometry is Equivalence Relation and so in particular symmetric.
Hence $M_2$ is isometr... | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $\phi: M_1 \to M_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
If $M_1$ is [[Definition:Complete Metric Space|complete]] then so is $M_2$. | Let $\epsilon \in \R_{>0}$.
Let $\sequence {b_n}$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A_2$.
Thus:
:$\exists N_1 \in \N: \map {d_2} {b_n, b_m} < \epsilon$
whenever $n, m \ge N_1$ and $b_n, b_m \in A_2$.
We have that $M_1$ is [[Definition:Isometry (Metric Spaces)|isometric]] to $M_2... | Metric Space Completeness is Preserved by Isometry/Proof 2 | https://proofwiki.org/wiki/Metric_Space_Completeness_is_Preserved_by_Isometry | https://proofwiki.org/wiki/Metric_Space_Completeness_is_Preserved_by_Isometry/Proof_2 | [
"Metric Space Completeness is Preserved by Isometry",
"Complete Metric Spaces",
"Isometries (Metric Spaces)"
] | [
"Definition:Metric Space",
"Definition:Isometry (Metric Spaces)",
"Definition:Complete Metric Space"
] | [
"Definition:Cauchy Sequence/Metric Space",
"Definition:Isometry (Metric Spaces)",
"Isometry is Equivalence Relation",
"Definition:Symmetric Relation",
"Definition:Isometry (Metric Spaces)",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Complete Metric Space",
"Definition:Convergent Sequence/M... |
proofwiki-3587 | Urysohn's Metrization Theorem | Let $T = \struct {S, \tau}$ be a topological space which is regular and second-countable.
Then $T$ is metrizable. | From Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube:
:$T$ is homeomorphic to a subspace of the Hilbert cube $\struct {I^\omega, d_2}$
where $d_2$ is the metric defined as:
:$\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \in \... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Regular Space|regular]] and [[Definition:Second-Countable Space|second-countable]].
Then $T$ is [[Definition:Metrizable Space|metrizable]]. | From [[Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube]]:
:$T$ is [[Definition:Homeomorphism (Topological Spaces)|homeomorphic]] to a [[Definition:Topological Subspace|subspace]] of the [[Definition:Hilbert Cube|Hilbert cube]] $\struct {I^\omega, d_2}$
where $d_2$ is the [[Definition:Metric|... | Urysohn's Metrization Theorem | https://proofwiki.org/wiki/Urysohn's_Metrization_Theorem | https://proofwiki.org/wiki/Urysohn's_Metrization_Theorem | [
"Metrization Theorems",
"Metrizable Spaces",
"Regular Spaces",
"Second-Countable Spaces"
] | [
"Definition:Topological Space",
"Definition:Regular Space",
"Definition:Second-Countable Space",
"Definition:Metrizable Space"
] | [
"Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Topological Subspace",
"Definition:Hilbert Cube",
"Definition:Metric Space/Metric",
"Definition:Metrizable Space",
"Definition:Topology Induced by Metric",
"Subspace... |
proofwiki-3588 | Uniformity iff Quasiuniformity has Symmetric Basis | Let $S$ be a set.
Let $\UU$ be a quasiuniformity on $S$.
Then $\UU$ is a uniformity {{iff}} $\UU$ has a symmetric filter basis. | Let $\UU$ be a quasiuniformity on $S$ which has a symmetric filter basis $\BB$.
From the definition of filter basis, all the elements of $\UU$ can be formed from intersections of elements of $\BB$.
But from Intersection of Symmetric Relations is Symmetric, it follows that all elements of $\UU$ are symmetric.
Now suppos... | Let $S$ be a [[Definition:Set|set]].
Let $\UU$ be a [[Definition:Quasiuniformity|quasiuniformity]] on $S$.
Then $\UU$ is a [[Definition:Uniformity|uniformity]] {{iff}} $\UU$ has a [[Definition:Symmetric Filter Basis|symmetric filter basis]]. | Let $\UU$ be a [[Definition:Quasiuniformity|quasiuniformity]] on $S$ which has a [[Definition:Symmetric Filter Basis|symmetric filter basis]] $\BB$.
From the definition of [[Definition:Filter Basis|filter basis]], all the elements of $\UU$ can be formed from [[Definition:Set Intersection|intersections]] of [[Definitio... | Uniformity iff Quasiuniformity has Symmetric Basis | https://proofwiki.org/wiki/Uniformity_iff_Quasiuniformity_has_Symmetric_Basis | https://proofwiki.org/wiki/Uniformity_iff_Quasiuniformity_has_Symmetric_Basis | [
"Uniformities"
] | [
"Definition:Set",
"Definition:Quasiuniformity",
"Definition:Uniformity",
"Definition:Symmetric Filter Basis"
] | [
"Definition:Quasiuniformity",
"Definition:Symmetric Filter Basis",
"Definition:Filter Basis",
"Definition:Set Intersection",
"Definition:Element",
"Intersection of Symmetric Relations is Symmetric",
"Definition:Element",
"Definition:Symmetric Entourage",
"Definition:Uniformity",
"Definition:Filter... |
proofwiki-3589 | Topological Space is Quasiuniformizable | Every topological space is quasiuniformizable. | Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be defined as:
:$\BB := \set {u_G: u_G = \paren {G \times G} \cup \paren {\paren {S \setminus G} \times G}, G \in \tau}$
Then $\BB$ is a filter sub-basis for a quasiuniformity on $S$ such that $\struct {\struct {S, \UU}, \tau}$ is a quasiuniform space.
{{fin... | Every [[Definition:Topological Space|topological space]] is [[Definition:Quasiuniformizable Space|quasiuniformizable]]. | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\BB$ be defined as:
:$\BB := \set {u_G: u_G = \paren {G \times G} \cup \paren {\paren {S \setminus G} \times G}, G \in \tau}$
Then $\BB$ is a [[Definition:Filter Sub-Basis|filter sub-basis]] for a [[Definition:Quasiuniformity|qu... | Topological Space is Quasiuniformizable | https://proofwiki.org/wiki/Topological_Space_is_Quasiuniformizable | https://proofwiki.org/wiki/Topological_Space_is_Quasiuniformizable | [
"Quasiuniformizable Spaces"
] | [
"Definition:Topological Space",
"Definition:Quasiuniformizable Space"
] | [
"Definition:Topological Space",
"Definition:Filter Sub-Basis",
"Definition:Quasiuniformity",
"Definition:Quasiuniform Space"
] |
proofwiki-3590 | Pseudometric Space generates Uniformity | Let $P = \struct {A, d}$ be a pseudometric space.
Let $\UU$ be the set of sets defined as:
:$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$
where:
:$\R_{>0}$ is the set of strictly positive real numbers
:$u_\epsilon$ is defined as:
::$u_\epsilon := \set {\tuple {x, y}: \map d {x, ... | We check whether the Uniformity Axioms are satisfied. | Let $P = \struct {A, d}$ be a [[Definition:Pseudometric Space|pseudometric space]].
Let $\UU$ be the [[Definition:Set of Sets|set of sets]] defined as:
:$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$
where:
:$\R_{>0}$ is the set of [[Definition:Strictly Positive Real Number|stri... | We check whether the [[Axiom:Uniformity Axioms|Uniformity Axioms]] are satisfied. | Pseudometric Space generates Uniformity | https://proofwiki.org/wiki/Pseudometric_Space_generates_Uniformity | https://proofwiki.org/wiki/Pseudometric_Space_generates_Uniformity | [
"Pseudometric Spaces",
"Uniformities"
] | [
"Definition:Pseudometric/Pseudometric Space",
"Definition:Set of Sets",
"Definition:Strictly Positive/Real Number",
"Definition:Uniformity",
"Definition:Uniform Space",
"Definition:Topology",
"Pseudometric induces Topology"
] | [
"Axiom:Uniformity Axioms"
] |
proofwiki-3591 | Metric Space generates Uniformity | Let $M = \struct {A, d}$ be a metric space.
Let $\UU$ be the set of sets defined as:
:$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$
where:
:$\R_{>0}$ is the set of strictly positive real numbers
:$u_\epsilon$ is defined as:
::$u_\epsilon := \set {\tuple {x, y}: \map d {x, y} < \... | From definition it is clear that a metric space is a pseudometric space.
The result then follows from Pseudometric Space generates Uniformity.
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $\UU$ be the [[Definition:Set of Sets|set of sets]] defined as:
:$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$
where:
:$\R_{>0}$ is the set of [[Definition:Strictly Positive|strictly positive]] [[Defini... | From definition it is clear that a [[Definition:Metric Space|metric space]] is a [[Definition:Pseudometric Space|pseudometric space]].
The result then follows from [[Pseudometric Space generates Uniformity]].
{{qed}} | Metric Space generates Uniformity | https://proofwiki.org/wiki/Metric_Space_generates_Uniformity | https://proofwiki.org/wiki/Metric_Space_generates_Uniformity | [
"Metric Spaces",
"Uniformities"
] | [
"Definition:Metric Space",
"Definition:Set of Sets",
"Definition:Strictly Positive",
"Definition:Real Number",
"Definition:Uniformity",
"Definition:Uniform Space",
"Definition:Topology",
"Metric Induces Topology"
] | [
"Definition:Metric Space",
"Definition:Pseudometric/Pseudometric Space",
"Pseudometric Space generates Uniformity"
] |
proofwiki-3592 | Multiplication is Superfunction | The function $f: \C \to \C$, defined as:
:$\map f z = z \times c$
is a superfunction for any complex number $c$. | Define $h: \C \to \C$ by $\map h z = z + c$.
Then:
{{begin-eqn}}
{{eqn | l = \map h {\map f z}
| r = \map h {z \times c}
| c =
}}
{{eqn | r = z \times c + c
| c =
}}
{{eqn | r = \paren {z + 1} \times c
| c =
}}
{{eqn | r = \map f {z + 1}
| c =
}}
{{end-eqn}}
Thus $\map f z = z \times c... | The [[Definition:Function|function]] $f: \C \to \C$, defined as:
:$\map f z = z \times c$
is a [[Definition:Superfunction|superfunction]] for any [[Definition:Complex Number|complex number]] $c$. | Define $h: \C \to \C$ by $\map h z = z + c$.
Then:
{{begin-eqn}}
{{eqn | l = \map h {\map f z}
| r = \map h {z \times c}
| c =
}}
{{eqn | r = z \times c + c
| c =
}}
{{eqn | r = \paren {z + 1} \times c
| c =
}}
{{eqn | r = \map f {z + 1}
| c =
}}
{{end-eqn}}
Thus $\map f z = z \time... | Multiplication is Superfunction | https://proofwiki.org/wiki/Multiplication_is_Superfunction | https://proofwiki.org/wiki/Multiplication_is_Superfunction | [
"Superfunctions"
] | [
"Definition:Function",
"Definition:Superfunction",
"Definition:Complex Number"
] | [
"Definition:Superfunction",
"Definition:Superfunction",
"Category:Superfunctions"
] |
proofwiki-3593 | Discrete Topology is Topology | :$\tau$ is a topology on $S$. | Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Then by definition $\tau = \powerset S$, that is, is the power set of $S$.
We confirm the criteria for $T$ to be a topology:
:$(1): \quad$ By definition of power set, $\O \in \powerset S$ and $S \in \powerset S$.
:$(2): \quad$ From Power Set with Union is Monoid... | :$\tau$ is a [[Definition:Topology|topology]] on $S$. | Let $T = \struct {S, \tau}$ be the [[Definition:Discrete Space|discrete space]] on $S$.
Then by definition $\tau = \powerset S$, that is, is the [[Definition:Power Set|power set]] of $S$.
We confirm the criteria for $T$ to be a [[Definition:Topology|topology]]:
:$(1): \quad$ By definition of [[Definition:Power Set|po... | Discrete Topology is Topology | https://proofwiki.org/wiki/Discrete_Topology_is_Topology | https://proofwiki.org/wiki/Discrete_Topology_is_Topology | [
"Discrete Topologies"
] | [
"Definition:Topology"
] | [
"Definition:Discrete Topology",
"Definition:Power Set",
"Definition:Topology",
"Definition:Power Set",
"Power Set with Union is Commutative Monoid",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Set Union",
"Power Set with Intersection is Commutative Monoid",
"Definition:C... |
proofwiki-3594 | Discrete Topology is Finest Topology | :$\tau$ is the finest topology on $S$. | Let $\phi$ be any topology on $S$.
Let $U \in \phi$.
Then, by the definition of topology, $U \subseteq S$.
Then, by the definition of discrete topological space, $U \in \tau$.
Hence by definition of subset, $\phi \subseteq \tau$.
Hence by definition of finer topology, $\tau$ is finer than $\phi$.
{{qed}} | :$\tau$ is the [[Definition:Finer Topology|finest topology]] on $S$. | Let $\phi$ be any [[Definition:Topology|topology]] on $S$.
Let $U \in \phi$.
Then, by the definition of [[Definition:Topology|topology]], $U \subseteq S$.
Then, by the definition of [[Definition:Discrete Topology|discrete topological space]], $U \in \tau$.
Hence by definition of [[Definition:Subset|subset]], $\phi ... | Discrete Topology is Finest Topology | https://proofwiki.org/wiki/Discrete_Topology_is_Finest_Topology | https://proofwiki.org/wiki/Discrete_Topology_is_Finest_Topology | [
"Discrete Topologies",
"Examples of Finer Topology"
] | [
"Definition:Finer Topology"
] | [
"Definition:Topology",
"Definition:Topology",
"Definition:Discrete Topology",
"Definition:Subset",
"Definition:Finer Topology",
"Definition:Finer Topology"
] |
proofwiki-3595 | Set in Discrete Topology is Clopen | :$\forall U \subseteq S: U$ is both closed and open in $\struct {S, \tau}$. | Let $U \subseteq S$.
By definition of discrete topological space, $U \in \tau$.
By definition of closed set, $\relcomp S U$ is closed in $T$, where $\relcomp S U$ is the relative complement of $U$ in $S$.
But from Set Difference is Subset:
:$\relcomp S U = S \setminus U \subseteq S$
and so:
:$\relcomp S U \in \tau$
Th... | :$\forall U \subseteq S: U$ is both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $\struct {S, \tau}$. | Let $U \subseteq S$.
By definition of [[Definition:Discrete Topology|discrete topological space]], $U \in \tau$.
By definition of [[Definition:Closed Set (Topology)|closed set]], $\relcomp S U$ is [[Definition:Closed Set (Topology)|closed]] in $T$, where $\relcomp S U$ is the [[Definition:Relative Complement|relative... | Set in Discrete Topology is Clopen | https://proofwiki.org/wiki/Set_in_Discrete_Topology_is_Clopen | https://proofwiki.org/wiki/Set_in_Discrete_Topology_is_Clopen | [
"Discrete Topologies",
"Examples of Clopen Sets"
] | [
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology"
] | [
"Definition:Discrete Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Relative Complement",
"Set Difference is Subset",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Relative Complement of Relative Complement",
"Definition:Closed Set/Top... |
proofwiki-3596 | Topological Space is Discrete iff All Points are Isolated | :$\tau$ is the discrete topology on $S$ {{iff}} all points in $S$ are isolated points of $T$. | === Necessary Condition ===
Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Then by definition $\tau = \powerset S$, that is, $\tau$ is the power set of $S$.
Let $x \in S$.
Then from Set in Discrete Topology is Clopen it follows that $\set x$ is open in $T$.
Thus by definition $x \in S$ is an isolated point o... | :$\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$ {{iff}} all [[Definition:Point of Set|points]] in $S$ are [[Definition:Isolated Point (Topology)|isolated points of $T$]]. | === Necessary Condition ===
Let $T = \struct {S, \tau}$ be the [[Definition:Discrete Space|discrete space]] on $S$.
Then by definition $\tau = \powerset S$, that is, $\tau$ is the [[Definition:Power Set|power set]] of $S$.
Let $x \in S$.
Then from [[Set in Discrete Topology is Clopen]] it follows that $\set x$ is ... | Topological Space is Discrete iff All Points are Isolated | https://proofwiki.org/wiki/Topological_Space_is_Discrete_iff_All_Points_are_Isolated | https://proofwiki.org/wiki/Topological_Space_is_Discrete_iff_All_Points_are_Isolated | [
"Topological Space is Discrete iff All Points are Isolated",
"Discrete Topologies",
"Examples of Isolated Points"
] | [
"Definition:Discrete Topology",
"Definition:Element",
"Definition:Isolated Point (Topology)"
] | [
"Definition:Discrete Topology",
"Definition:Power Set",
"Set in Discrete Topology is Clopen",
"Definition:Open Set/Topology",
"Definition:Isolated Point (Topology)",
"Definition:Isolated Point (Topology)",
"Definition:Discrete Topology",
"Definition:Open Set/Topology",
"Definition:Isolated Point (To... |
proofwiki-3597 | Point in Discrete Space is Adherent Point | Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $U \subseteq S$.
Then $x$ is an adherent point of $U$ {{iff}} $x \in U$. | Let $U \subseteq S$.
From Set in Discrete Topology is Clopen it follows that $U$ is open in $T$.
Let $x \in U$.
Then $U$ is an open neighborhood of $x$ containing (trivially) an element of $U$, that is, $x$.
So, by definition, $x$ is an adherent point of $U$.
Now suppose $x \notin U$.
Then $\set x$ is an open neighborh... | Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
Let $U \subseteq S$.
Then $x$ is an [[Definition:Adherent Point of Set|adherent point of $U$]] {{iff}} $x \in U$. | Let $U \subseteq S$.
From [[Set in Discrete Topology is Clopen]] it follows that $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
Let $x \in U$.
Then $U$ is an [[Definition:Open Neighborhood of Point|open neighborhood]] of $x$ containing (trivially) an element of $U$, that is, $x$.
So, by definition, $x$ is ... | Point in Discrete Space is Adherent Point | https://proofwiki.org/wiki/Point_in_Discrete_Space_is_Adherent_Point | https://proofwiki.org/wiki/Point_in_Discrete_Space_is_Adherent_Point | [
"Discrete Topologies",
"Adherent Points of Sets"
] | [
"Definition:Discrete Topology",
"Definition:Adherent Point of Set"
] | [
"Set in Discrete Topology is Clopen",
"Definition:Open Set/Topology",
"Definition:Open Neighborhood/Point",
"Definition:Adherent Point of Set",
"Definition:Open Neighborhood/Point",
"Definition:Adherent Point of Set"
] |
proofwiki-3598 | Interior Equals Closure of Subset of Discrete Space | Let $A \subseteq S$.
Then:
:$A = A^\circ = A^-$
where:
:$A^\circ$ is the interior of $A$
:$A^-$ is the closure of $A$. | Let $A \subseteq S$.
Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$.
From Closed Set Equals its Closure we have that $A = A^-$.
From Set Interior is Largest Open Set, we have that $A^\circ = A$.
{{qed}} | Let $A \subseteq S$.
Then:
:$A = A^\circ = A^-$
where:
:$A^\circ$ is the [[Definition:Interior (Topology)|interior]] of $A$
:$A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$. | Let $A \subseteq S$.
Then from [[Set in Discrete Topology is Clopen]] it follows that $A$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
From [[Closed Set Equals its Closure]] we have that $A = A^-$.
From [[Set Interior is Largest Open Set]], we have that $A^\... | Interior Equals Closure of Subset of Discrete Space | https://proofwiki.org/wiki/Interior_Equals_Closure_of_Subset_of_Discrete_Space | https://proofwiki.org/wiki/Interior_Equals_Closure_of_Subset_of_Discrete_Space | [
"Discrete Topologies",
"Set Interiors",
"Examples of Set Closures"
] | [
"Definition:Interior (Topology)",
"Definition:Closure (Topology)"
] | [
"Set in Discrete Topology is Clopen",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Set is Closed iff Equals Topological Closure",
"Equivalence of Definitions of Interior (Topology)"
] |
proofwiki-3599 | Boundary of Subset of Discrete Space is Empty Set | Let $A \subseteq S$.
Then:
:$\partial A = \O$
where:
:$\partial A$ is the boundary of $A$ in $T$. | Let $A \subseteq S$.
Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$.
The result follows from Set Clopen iff Boundary is Empty.
{{qed}} | Let $A \subseteq S$.
Then:
:$\partial A = \O$
where:
:$\partial A$ is the [[Definition:Boundary (Topology)|boundary]] of $A$ in $T$. | Let $A \subseteq S$.
Then from [[Set in Discrete Topology is Clopen]] it follows that $A$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
The result follows from [[Set Clopen iff Boundary is Empty]].
{{qed}} | Boundary of Subset of Discrete Space is Empty Set | https://proofwiki.org/wiki/Boundary_of_Subset_of_Discrete_Space_is_Empty_Set | https://proofwiki.org/wiki/Boundary_of_Subset_of_Discrete_Space_is_Empty_Set | [
"Discrete Topologies",
"Set Boundaries",
"Empty Set"
] | [
"Definition:Boundary (Topology)"
] | [
"Set in Discrete Topology is Clopen",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Set is Clopen iff Boundary is Empty"
] |
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