id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-3300 | Quotient Ring by Null Ideal | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {\set {0_R}, +, \circ}$ be the null ideal of $\struct {R, +, \circ}$.
Let $\struct {R / \set {0_R}, +, \circ}$ be the quotient ring of $R$ defined by $\set {0_R}$.
Then $\struct {R / \set {0_R}, +, \circ}$ is isomorphic to $\struct {R, +, \circ}$. | Consider the additive group $\struct {R, +}$.
From Trivial Quotient Group is Quotient Group:
:$\struct {R, +} / \set {0_R} \cong \struct {R, +}$
{{finish}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $\struct {\set {0_R}, +, \circ}$ be the [[Definition:Null Ideal|null ideal]] of $\struct {R, +, \circ}$.
Let $\struct {R / \set {0_R}, +, \circ}$ be the [[Definition:Quotient Ring|quotient ri... | Consider the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$.
From [[Trivial Quotient Group is Quotient Group]]:
:$\struct {R, +} / \set {0_R} \cong \struct {R, +}$
{{finish}} | Quotient Ring by Null Ideal | https://proofwiki.org/wiki/Quotient_Ring_by_Null_Ideal | https://proofwiki.org/wiki/Quotient_Ring_by_Null_Ideal | [
"Ideal Theory",
"Quotient Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Null Ideal",
"Definition:Quotient Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Definition:Additive Group of Ring",
"Trivial Quotient Group is Quotient Group"
] |
proofwiki-3301 | Ring Homomorphism Preserves Negatives | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then:
:$\forall x \in R_1: \map \phi {-x} = -\paren {\map \phi x}$ | We have that Ring Homomorphism of Addition is Group Homomorphism.
The result follows from Group Homomorphism Preserves Inverses.
{{qed}} | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Then:
:$\forall x \in R_1: \map \phi {-x} = -\paren {\map \phi x}$ | We have that [[Ring Homomorphism of Addition is Group Homomorphism]].
The result follows from [[Group Homomorphism Preserves Inverses]].
{{qed}} | Ring Homomorphism Preserves Negatives | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Negatives | https://proofwiki.org/wiki/Ring_Homomorphism_Preserves_Negatives | [
"Ring Homomorphisms"
] | [
"Definition:Ring Homomorphism"
] | [
"Ring Homomorphism of Addition is Group Homomorphism",
"Group Homomorphism Preserves Inverses"
] |
proofwiki-3302 | Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function | Let $s \in \C: \map \Re s > 1$.
Let $x \in \R_{>0}$.
Then:
:$\ds \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s = -\frac 1 {s \paren {1 - s} } + \dfrac 1 2 \int_1^\infty \paren {x^{s / 2 - 1} + x^{-\paren {s + 1} / 2} } \paren {\map {\vartheta_3} {0, e^{-\pi x} } - 1} \rd x$
where:
:$\map \Gamma s$ is the gamma func... | === Lemma 1 ===
{{:Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 1}}{{qed|lemma}} | Let $s \in \C: \map \Re s > 1$.
Let $x \in \R_{>0}$.
Then:
:$\ds \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s = -\frac 1 {s \paren {1 - s} } + \dfrac 1 2 \int_1^\infty \paren {x^{s / 2 - 1} + x^{-\paren {s + 1} / 2} } \paren {\map {\vartheta_3} {0, e^{-\pi x} } - 1} \rd x$
where:
:$\map \Gamma s$ is the [[Defin... | === [[Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 1|Lemma 1]] ===
{{:Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 1}}{{qed|lemma}} | Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function | https://proofwiki.org/wiki/Integral_Representation_of_Riemann_Zeta_Function_in_terms_of_Jacobi_Theta_Function | https://proofwiki.org/wiki/Integral_Representation_of_Riemann_Zeta_Function_in_terms_of_Jacobi_Theta_Function | [
"Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function",
"Jacobi Theta Functions",
"Riemann Zeta Function",
"Gamma Function"
] | [
"Definition:Gamma Function",
"Definition:Riemann Zeta Function",
"Definition:Jacobi Theta Function/Third Type"
] | [
"Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function/Lemma 1"
] |
proofwiki-3303 | Mapping on Quadratic Integers over 2 to Conjugate is Automorphism | Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the mapping $\phi: \Z \sqbrk {\sqrt 2} \to \Z \sqbrk {\sqrt 2}$ defined as:
:$\forall x = a + b \sqrt... | We have Quadratic Integers over 2 form Subdomain of Reals.
First we note that:
:$\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$ | Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]] of [[Definition:Quadratic Integer|quadratic integers]] over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
Then the [[Definition:Mapping... | We have [[Quadratic Integers over 2 form Subdomain of Reals]].
First we note that:
:$\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$ | Mapping on Quadratic Integers over 2 to Conjugate is Automorphism | https://proofwiki.org/wiki/Mapping_on_Quadratic_Integers_over_2_to_Conjugate_is_Automorphism | https://proofwiki.org/wiki/Mapping_on_Quadratic_Integers_over_2_to_Conjugate_is_Automorphism | [
"Examples of Ring Automorphisms",
"Examples of Integral Domains",
"Quadratic Integers"
] | [
"Definition:Set",
"Definition:Algebraic Integer/Quadratic",
"Definition:Integer",
"Definition:Mapping",
"Definition:Ring Automorphism"
] | [
"Quadratic Integers over 2 form Subdomain of Reals"
] |
proofwiki-3304 | Ring Epimorphism from Integers to Integers Modulo m | Let $\struct {\Z, +, \times}$ be the ring of integers.
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Let $\phi: \struct {\Z, +, \times} \to \struct {\Z_m, +_m, \times_m}$ be the mapping defined as:
:$\forall x \in \Z: \map \phi x = \eqclass x m$
where $\eqclass x m$ is the residue class modulo... | Let $a, b \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {a + b}
| r = \eqclass {a + b} m
| c = Definition of $\phi$
}}
{{eqn | r = \eqclass a m +_m \eqclass b m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \map \phi a +_m \map \phi b
| c = Definition of $\phi$
}}
{{end-eqn}}
{{begin-eqn}}... | Let $\struct {\Z, +, \times}$ be the [[Definition:Ring of Integers|ring of integers]].
Let $\struct {\Z_m, +_m, \times_m}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Let $\phi: \struct {\Z, +, \times} \to \struct {\Z_m, +_m, \times_m}$ be the [[Definition:Mapping|mapping]] defined as... | Let $a, b \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {a + b}
| r = \eqclass {a + b} m
| c = Definition of $\phi$
}}
{{eqn | r = \eqclass a m +_m \eqclass b m
| c = {{Defof|Modulo Addition}}
}}
{{eqn | r = \map \phi a +_m \map \phi b
| c = Definition of $\phi$
}}
{{end-eqn}}
{{begin-e... | Ring Epimorphism from Integers to Integers Modulo m | https://proofwiki.org/wiki/Ring_Epimorphism_from_Integers_to_Integers_Modulo_m | https://proofwiki.org/wiki/Ring_Epimorphism_from_Integers_to_Integers_Modulo_m | [
"Ring Epimorphisms",
"Modulo Arithmetic",
"Integers"
] | [
"Definition:Ring of Integers",
"Definition:Ring of Integers Modulo m",
"Definition:Mapping",
"Definition:Residue Class",
"Definition:Ring Epimorphism",
"Definition:Ring Monomorphism",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Kernel of Ring Homomorphism",
"Definition:Set of Intege... | [
"Definition:Ring Homomorphism",
"Definition:Residue Class",
"Definition:Surjection",
"Definition:Injection",
"Definition:Ring Epimorphism",
"Definition:Ring Monomorphism",
"Definition:Kernel of Ring Homomorphism"
] |
proofwiki-3305 | Monotone Additive Function is Linear | Let $f: \R \to \R$ be a monotone real function which is additive, that is:
:$\forall x, y \in \R: \map f {x + y} = \map f x + \map f y$
Then:
:$\exists a \in \R: \forall x \in \R: \map f x = a x$
That is, $f$ is a linear function. | We use a Proof by Contraposition.
To that end, suppose $f$ is not linear.
We know that Graph of Nonlinear Additive Function is Dense in the Plane.
Therefore $f$ is not bounded on any nonempty open interval.
But then $f$ is certainly not monotone.
Hence, by Rule of Transposition, if $f$ is monotone, then it is linear. | Let $f: \R \to \R$ be a [[Definition:Monotone Real Function|monotone real function]] which is [[Definition:Additive Function (Algebra)|additive]], that is:
:$\forall x, y \in \R: \map f {x + y} = \map f x + \map f y$
Then:
:$\exists a \in \R: \forall x \in \R: \map f x = a x$
That is, $f$ is a [[Definition:Linear R... | We use a [[Proof by Contraposition]].
To that end, suppose $f$ is not linear.
We know that [[Graph of Nonlinear Additive Function is Dense in the Plane]].
Therefore $f$ is not bounded on any nonempty open interval.
But then $f$ is certainly not monotone.
Hence, by [[Rule of Transposition]], if $f$ is monotone, the... | Monotone Additive Function is Linear/Proof 2 | https://proofwiki.org/wiki/Monotone_Additive_Function_is_Linear | https://proofwiki.org/wiki/Monotone_Additive_Function_is_Linear/Proof_2 | [
"Monotone Additive Function is Linear",
"Monotone Real Functions",
"Additive Functions",
"Linear Real Functions"
] | [
"Definition:Monotone (Order Theory)/Real Function",
"Definition:Additive Function (Algebra)",
"Definition:Linear Real Function"
] | [
"Proof by Contraposition",
"Graph of Nonlinear Additive Function is Dense in the Plane",
"Rule of Transposition"
] |
proofwiki-3306 | Units of Ring of Polynomial Forms over Field | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $F \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $F$.
Then the units of $F \sqbrk X$ are all the elements of $F \sqbrk X$ whose degree is $0$. | An element of $F \sqbrk X$ whose degree is $0$ is merely an element of $F$.
But note that $0_F$, considered as an element of $F \sqbrk X$, has a degree which is not defined, so the null polynomial is seen to be excluded.
Any element $a$ of $F$ has an inverse $1_F / a$.
So all the elements of $F \sqbrk X$ whose degree i... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in an [[Definition:Indeterminate (Polynomial Theo... | An element of $F \sqbrk X$ whose [[Definition:Degree (Polynomial)|degree]] is $0$ is merely an element of $F$.
But note that $0_F$, considered as an element of $F \sqbrk X$, has a [[Definition:Degree (Polynomial)|degree]] which is not defined, so the [[Definition:Null Polynomial over Ring|null polynomial]] is seen to ... | Units of Ring of Polynomial Forms over Field | https://proofwiki.org/wiki/Units_of_Ring_of_Polynomial_Forms_over_Field | https://proofwiki.org/wiki/Units_of_Ring_of_Polynomial_Forms_over_Field | [
"Polynomial Theory",
"Units of Rings"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Ring of Polynomial Forms",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Unit of Ring",
"Definition:Degree of Polynomial"
] | [
"Definition:Degree of Polynomial",
"Definition:Degree of Polynomial",
"Definition:Null Polynomial/Ring",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Degree of Polynomial",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Null Polynomial/Ring",
"Properties of Degree"... |
proofwiki-3307 | Units of Quadratic Integers over 2 | Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Let $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ be the integral domain where $+$ and $\times$ are conventio... | For $a + b \sqrt 2$ to be a unit of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$, we require that:
:$\exists c, d \in \Z: \paren {a + b \sqrt 2} \paren {c + d \sqrt 2} = 1$
In Quadratic Integers over 2 are Not a Field it is shown that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} + \dfrac {b ... | Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]] of [[Definition:Quadratic Integer|quadratic integers]] over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
Let $\struct {\Z \sqbrk {\sqrt... | For $a + b \sqrt 2$ to be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$, we require that:
:$\exists c, d \in \Z: \paren {a + b \sqrt 2} \paren {c + d \sqrt 2} = 1$
In [[Quadratic Integers over 2 are Not a Field]] it is shown that the [[Definition:Product Inverse|product inverse]] of ... | Units of Quadratic Integers over 2 | https://proofwiki.org/wiki/Units_of_Quadratic_Integers_over_2 | https://proofwiki.org/wiki/Units_of_Quadratic_Integers_over_2 | [
"Examples of Integral Domains",
"Units of Rings",
"Quadratic Integers"
] | [
"Definition:Set",
"Definition:Algebraic Integer/Quadratic",
"Definition:Integer",
"Quadratic Integers over 2 form Subdomain of Reals",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Real Number",
"Definition:Unit of Ring"
] | [
"Definition:Unit of Ring",
"Quadratic Integers over 2 are Not a Field",
"Definition:Product Inverse",
"Definition:Integer"
] |
proofwiki-3308 | Units of Gaussian Integers | Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.
The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | By definition of the imaginary unit $i$:
{{begin-eqn}}
{{eqn | l = i^2
| r = -1
}}
{{eqn | l = i^3
| r = -i
}}
{{eqn | l = i^4
| r = 1
}}
{{end-eqn}}
thus demonstrating that $U_\C$ is generated by $i$.
Thus $\struct {U_\C, \times}$ is by definition a cyclic group of order $4$.
{{qed}} | Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
The [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | By definition of the [[Definition:Imaginary Unit|imaginary unit]] $i$:
{{begin-eqn}}
{{eqn | l = i^2
| r = -1
}}
{{eqn | l = i^3
| r = -i
}}
{{eqn | l = i^4
| r = 1
}}
{{end-eqn}}
thus demonstrating that $U_\C$ is [[Definition:Generator of Cyclic Group|generated]] by $i$.
Thus $\struct {U_\C, \time... | Units of Gaussian Integers form Group/Proof 1 | https://proofwiki.org/wiki/Units_of_Gaussian_Integers | https://proofwiki.org/wiki/Units_of_Gaussian_Integers_form_Group/Proof_1 | [
"Gaussian Integers",
"Units of Gaussian Integers"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Set",
"Definition:Unit of Ring"
] | [
"Definition:Complex Number/Imaginary Unit",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group",
"Definition:Order of Structure"
] |
proofwiki-3309 | Units of Gaussian Integers | Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.
The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | From Gaussian Integer Units are 4th Roots of Unity:
: $\left\{{1, i, -1, -i}\right\}$ constitutes the set of the $4$th roots of unity.
The result follows from Roots of Unity under Multiplication form Cyclic Group.
{{qed}} | Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
The [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | From [[Gaussian Integer Units are 4th Roots of Unity]]:
: $\left\{{1, i, -1, -i}\right\}$ constitutes the [[Definition:Set|set]] of the [[Definition:Complex Roots of Unity|$4$th roots of unity]].
The result follows from [[Roots of Unity under Multiplication form Cyclic Group]].
{{qed}} | Units of Gaussian Integers form Group/Proof 2 | https://proofwiki.org/wiki/Units_of_Gaussian_Integers | https://proofwiki.org/wiki/Units_of_Gaussian_Integers_form_Group/Proof_2 | [
"Gaussian Integers",
"Units of Gaussian Integers"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Set",
"Definition:Unit of Ring"
] | [
"Gaussian Integer Units are 4th Roots of Unity",
"Definition:Set",
"Definition:Root of Unity/Complex",
"Roots of Unity under Multiplication form Cyclic Group"
] |
proofwiki-3310 | Units of Gaussian Integers | Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.
The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | From Units of Gaussian Integers, $U_\C$ is the set of units of the ring of Gaussian integers.
From Group of Units is Group, $\left({U_\C, \times}\right)$ forms a group.
It remains to note that:
{{begin-eqn}}
{{eqn | l = i^2
| r = -1
}}
{{eqn | l = i^3
| r = -i
}}
{{eqn | l = i^4
| r = 1
}}
{{end-eqn}}... | Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
The [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | From [[Units of Gaussian Integers]], $U_\C$ is the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
From [[Group of Units is Group]], $\left({U_\C, \times}\right)$ forms a [[Definition:Group|group]].
It remains to note that:
{{begi... | Units of Gaussian Integers form Group/Proof 3 | https://proofwiki.org/wiki/Units_of_Gaussian_Integers | https://proofwiki.org/wiki/Units_of_Gaussian_Integers_form_Group/Proof_3 | [
"Gaussian Integers",
"Units of Gaussian Integers"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Set",
"Definition:Unit of Ring"
] | [
"Units of Gaussian Integers",
"Definition:Set",
"Definition:Unit of Ring",
"Definition:Ring of Gaussian Integers",
"Group of Units is Group",
"Definition:Group",
"Definition:Cyclic Group"
] |
proofwiki-3311 | Units of Gaussian Integers | Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.
The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | Let $a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$.
Then $a$ and $b$ are not both $0$ as then $a + b i$ would be the zero of $\struct {\Z \sqbrk i, +, \times}$.
Then:
:$\exists c, d \in \Z: \paren {a + b i} \paren {c + d i} = 1 + 0 i$
This leads (after algebra) to:
:$c = \dfrac a {a^2 + b^2}, d = \dfrac {-b}... | Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
The [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | Let $a + b i$ be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk i, +, \times}$.
Then $a$ and $b$ are not both $0$ as then $a + b i$ would be the [[Definition:Ring Zero|zero]] of $\struct {\Z \sqbrk i, +, \times}$.
Then:
:$\exists c, d \in \Z: \paren {a + b i} \paren {c + d i} = 1 + 0 i$
This leads (after ... | Units of Gaussian Integers/Proof 1 | https://proofwiki.org/wiki/Units_of_Gaussian_Integers | https://proofwiki.org/wiki/Units_of_Gaussian_Integers/Proof_1 | [
"Gaussian Integers",
"Units of Gaussian Integers"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Set",
"Definition:Unit of Ring"
] | [
"Definition:Unit of Ring",
"Definition:Ring Zero",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-3312 | Units of Gaussian Integers | Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.
The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | Let $\alpha = a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$.
Then by definition of unit:
:$\exists\beta = c + d i \in \Z \sqbrk i: \alpha \beta = 1$
Let $\cmod \alpha$ denote the modulus of $\alpha$.
Then:
{{begin-eqn}}
{{eqn | l = \cmod \alpha^2 \cdot \cmod \beta^2
| r = \cmod {\alpha \beta}^2
|... | Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
The [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$. | Let $\alpha = a + b i$ be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk i, +, \times}$.
Then by definition of [[Definition:Unit of Ring|unit]]:
:$\exists\beta = c + d i \in \Z \sqbrk i: \alpha \beta = 1$
Let $\cmod \alpha$ denote the [[Definition:Complex Modulus|modulus]] of $\alpha$.
Then:
{{begin-eqn}... | Units of Gaussian Integers/Proof 2 | https://proofwiki.org/wiki/Units_of_Gaussian_Integers | https://proofwiki.org/wiki/Units_of_Gaussian_Integers/Proof_2 | [
"Gaussian Integers",
"Units of Gaussian Integers"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Set",
"Definition:Unit of Ring"
] | [
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Complex Modulus",
"Complex Modulus of Product of Complex Numbers",
"Divisors of One",
"Definition:Positive/Integer",
"Definition:Set",
"Definition:Unit of Ring"
] |
proofwiki-3313 | Integers are Euclidean Domain | The integers $\Z$ with the mapping $\nu: \Z \to \Z$ defined as:
:$\forall x \in \Z: \map \nu x = \size x$
form a Euclidean domain. | From Integers form Ordered Integral Domain we have that $\struct {\Z, +, \times, \le}$ forms an ordered integral domain.
For all $a \in \Z$, the absolute value of $a$ is defined as:
:$\size a = \begin{cases} a & : 0 \le a \\ -a & : a < 0 \end{cases}$
By Product of Absolute Values on Ordered Integral Domain we have:
:$\... | The [[Definition:Integer|integers]] $\Z$ with the [[Definition:Mapping|mapping]] $\nu: \Z \to \Z$ defined as:
:$\forall x \in \Z: \map \nu x = \size x$
form a [[Definition:Euclidean Domain|Euclidean domain]]. | From [[Integers form Ordered Integral Domain]] we have that $\struct {\Z, +, \times, \le}$ forms an [[Definition:Ordered Integral Domain|ordered integral domain]].
For all $a \in \Z$, the [[Definition:Absolute Value on Ordered Integral Domain|absolute value]] of $a$ is defined as:
:$\size a = \begin{cases} a & : 0 \... | Integers are Euclidean Domain | https://proofwiki.org/wiki/Integers_are_Euclidean_Domain | https://proofwiki.org/wiki/Integers_are_Euclidean_Domain | [
"Integers",
"Examples of Euclidean Domains"
] | [
"Definition:Integer",
"Definition:Mapping",
"Definition:Euclidean Domain"
] | [
"Integers form Ordered Integral Domain",
"Definition:Ordered Integral Domain",
"Definition:Absolute Value/Ordered Integral Domain",
"Product of Absolute Values on Ordered Integral Domain",
"Relation Induced by Strict Positivity Property is Compatible with Multiplication",
"Division Theorem"
] |
proofwiki-3314 | Gaussian Integers form Euclidean Domain | Let $\struct {\Z \sqbrk i, +, \times}$ be the integral domain of Gaussian Integers.
Let $\nu: \Z \sqbrk i \to \R$ be the real-valued function defined as:
:$\forall a \in \Z \sqbrk i: \map \nu a = \cmod a^2$
where $\cmod a$ is the (complex) modulus of $a$.
Then $\nu$ is a Euclidean valuation on $\Z \sqbrk i$.
Hence $\... | We have by definition that $\Z \sqbrk i \subseteq \C$.
Let $a, b \in \Z \sqbrk i$.
We have from Modulus of Product that $\cmod a \cdot \cmod b = \cmod {a b}$.
From Complex Modulus is Non-Negative:
:$\forall a \in \C: \cmod a \ge 0$
and:
:$\cmod a = 0 \iff a = 0$
Let $a = x + i y$.
Suppose $a \in \Z \sqbrk i \ne 0$.
The... | Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Gaussian Integers form Integral Domain|integral domain of Gaussian Integers]].
Let $\nu: \Z \sqbrk i \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
:$\forall a \in \Z \sqbrk i: \map \nu a = \cmod a^2$
where $\cmod a$ is the [[Defini... | We have by definition that $\Z \sqbrk i \subseteq \C$.
Let $a, b \in \Z \sqbrk i$.
We have from [[Modulus of Product]] that $\cmod a \cdot \cmod b = \cmod {a b}$.
From [[Complex Modulus is Non-Negative]]:
:$\forall a \in \C: \cmod a \ge 0$
and:
:$\cmod a = 0 \iff a = 0$
Let $a = x + i y$.
Suppose $a \in \Z \sqbr... | Gaussian Integers form Euclidean Domain/Proof 1 | https://proofwiki.org/wiki/Gaussian_Integers_form_Euclidean_Domain | https://proofwiki.org/wiki/Gaussian_Integers_form_Euclidean_Domain/Proof_1 | [
"Gaussian Integers form Euclidean Domain",
"Examples of Euclidean Domains",
"Gaussian Integers"
] | [
"Gaussian Integers form Integral Domain",
"Definition:Real-Valued Function",
"Definition:Complex Modulus",
"Definition:Euclidean Domain/Valuation",
"Definition:Euclidean Domain"
] | [
"Complex Modulus of Product of Complex Numbers",
"Complex Modulus is Non-Negative",
"Definition:Euclidean Domain/Valuation",
"Definition:Complex Number",
"Definition:Gaussian Integer",
"Definition:Extension of Mapping",
"Definition:Complex Number",
"Definition:Complex Number/Complex Plane",
"Definit... |
proofwiki-3315 | Gaussian Integers form Euclidean Domain | Let $\struct {\Z \sqbrk i, +, \times}$ be the integral domain of Gaussian Integers.
Let $\nu: \Z \sqbrk i \to \R$ be the real-valued function defined as:
:$\forall a \in \Z \sqbrk i: \map \nu a = \cmod a^2$
where $\cmod a$ is the (complex) modulus of $a$.
Then $\nu$ is a Euclidean valuation on $\Z \sqbrk i$.
Hence $\... | We have by definition that $\Z \sqbrk i \subseteq \C$.
Let $a, b \in \Z \sqbrk i$.
We have from Modulus of Product that:
:$\cmod a \cdot \cmod b = \cmod {a b}$
From Complex Modulus is Norm we have that:
:$\forall a \in \C: \cmod a \ge 0$
:$\cmod a = 0 \iff a = 0$
Suppose $a = x + i y \in \Z \sqbrk i$ and $a \ne 0$.
The... | Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Gaussian Integers form Integral Domain|integral domain of Gaussian Integers]].
Let $\nu: \Z \sqbrk i \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
:$\forall a \in \Z \sqbrk i: \map \nu a = \cmod a^2$
where $\cmod a$ is the [[Defini... | We have by definition that $\Z \sqbrk i \subseteq \C$.
Let $a, b \in \Z \sqbrk i$.
We have from [[Modulus of Product]] that:
:$\cmod a \cdot \cmod b = \cmod {a b}$
From [[Complex Modulus is Norm]] we have that:
:$\forall a \in \C: \cmod a \ge 0$
:$\cmod a = 0 \iff a = 0$
Suppose $a = x + i y \in \Z \sqbrk i$ an... | Gaussian Integers form Euclidean Domain/Proof 2 | https://proofwiki.org/wiki/Gaussian_Integers_form_Euclidean_Domain | https://proofwiki.org/wiki/Gaussian_Integers_form_Euclidean_Domain/Proof_2 | [
"Gaussian Integers form Euclidean Domain",
"Examples of Euclidean Domains",
"Gaussian Integers"
] | [
"Gaussian Integers form Integral Domain",
"Definition:Real-Valued Function",
"Definition:Complex Modulus",
"Definition:Euclidean Domain/Valuation",
"Definition:Euclidean Domain"
] | [
"Complex Modulus of Product of Complex Numbers",
"Complex Modulus is Norm",
"Definition:Complex Number",
"Definition:Gaussian Integer",
"Definition:Complex Number",
"Definition:Extension of Mapping",
"Definition:Complex Number",
"Definition:Gaussian Integer",
"Definition:Minimal/Element",
"Definit... |
proofwiki-3316 | Degree of Product of Polynomials over Ring | :$\forall f, g \in R \sqbrk X: \map \deg {f g} \le \map \deg f + \map \deg g$ | Let the leading coefficient of:
:$\map f X$ be $a_n$
:$\map g X$ be $b_n$.
Then:
{{begin-eqn}}
{{eqn | l = \map f X
| r = a_n X^n + \cdots + a_0
}}
{{eqn | l = \map g X
| r = b_n X^n + \cdots + b_0
}}
{{end-eqn}}
Consider the leading coefficient of the product $\map f X \map g X$: call it $c$.
From the defi... | :$\forall f, g \in R \sqbrk X: \map \deg {f g} \le \map \deg f + \map \deg g$ | Let the [[Definition:Leading Coefficient (Polynomial)|leading coefficient]] of:
:$\map f X$ be $a_n$
:$\map g X$ be $b_n$.
Then:
{{begin-eqn}}
{{eqn | l = \map f X
| r = a_n X^n + \cdots + a_0
}}
{{eqn | l = \map g X
| r = b_n X^n + \cdots + b_0
}}
{{end-eqn}}
Consider the [[Definition:Leading Coefficien... | Degree of Product of Polynomials over Ring | https://proofwiki.org/wiki/Degree_of_Product_of_Polynomials_over_Ring | https://proofwiki.org/wiki/Degree_of_Product_of_Polynomials_over_Ring | [
"Polynomial Theory",
"Degree of Product of Polynomials over Ring"
] | [] | [
"Definition:Leading Coefficient of Polynomial",
"Definition:Leading Coefficient of Polynomial",
"Definition:Polynomial Addition",
"Definition:Multiplication of Polynomials",
"Definition:Ring with Unity",
"Definition:Proper Zero Divisor",
"Definition:Ring with Unity"
] |
proofwiki-3317 | Degree of Sum of Polynomials | :$\forall f, g \in R \sqbrk X: \map \deg {f + g} \le \max \set {\map \deg f, \map \deg g}$ | First we associate to $f = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$ a formal vector $x_f = \tuple {a_0, a_1, \ldots, a_n, 0_R, \ldots} \in R^\infty$.
Let $x_f^i \in R$ denote the element at the $i$th position.
Then:
:$\map \deg f = \sup \set {i \in \N : x_f^i \ne 0_R}$
The sum $+$ in the polynomial ring $R \sqbr... | :$\forall f, g \in R \sqbrk X: \map \deg {f + g} \le \max \set {\map \deg f, \map \deg g}$ | First we associate to $f = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$ a formal [[Definition:Vector Space|vector]] $x_f = \tuple {a_0, a_1, \ldots, a_n, 0_R, \ldots} \in R^\infty$.
Let $x_f^i \in R$ denote the element at the $i$th position.
Then:
:$\map \deg f = \sup \set {i \in \N : x_f^i \ne 0_R}$
The sum $+$ ... | Degree of Sum of Polynomials | https://proofwiki.org/wiki/Degree_of_Sum_of_Polynomials | https://proofwiki.org/wiki/Degree_of_Sum_of_Polynomials | [
"Polynomial Theory"
] | [] | [
"Definition:Vector Space",
"Definition:Polynomial Ring"
] |
proofwiki-3318 | Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors | :$\forall f, g \in R \sqbrk X: \map \deg {f g} \ge \map \deg f$ | From Degree of Product of Polynomials over Integral Domain, we have:
:$\map \deg {f g} = \map \deg f + \map \deg g$
But $\map \deg g \ge 0$ by definition of degree, as $g$ is not null.
Hence the result.
{{qed}} | :$\forall f, g \in R \sqbrk X: \map \deg {f g} \ge \map \deg f$ | From [[Degree of Product of Polynomials over Integral Domain]], we have:
:$\map \deg {f g} = \map \deg f + \map \deg g$
But $\map \deg g \ge 0$ by definition of [[Definition:Degree of Polynomial over Integral Domain|degree]], as $g$ is not [[Definition:Null Polynomial over Ring|null]].
Hence the result.
{{qed}} | Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors | https://proofwiki.org/wiki/Degree_of_Product_of_Polynomials_over_Integral_Domain_not_Less_than_Degree_of_Factors | https://proofwiki.org/wiki/Degree_of_Product_of_Polynomials_over_Integral_Domain_not_Less_than_Degree_of_Factors | [
"Polynomial Theory"
] | [] | [
"Degree of Product of Polynomials over Ring/Corollary 2",
"Definition:Degree of Polynomial/Integral Domain",
"Definition:Null Polynomial/Ring"
] |
proofwiki-3319 | Polynomial Forms over Field is Euclidean Domain | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental in $F$.
Let $F \sqbrk X$ be the ring of polynomial forms in $X$ over $F$.
Then $F \sqbrk X$ is a Euclidean domain. | From Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors we have that:
:$\forall a, b \in F \sqbrk X, a \ne 0_F, b \ne 0_F: \map \deg {a b} \ge \map \deg a$
where $\map \deg a$ denotes the degree of $a$.
From Division Theorem for Polynomial Forms over Field:
:$\forall a, b \in F \sqbrk... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental]] in $F$.
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|... | From [[Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors]] we have that:
:$\forall a, b \in F \sqbrk X, a \ne 0_F, b \ne 0_F: \map \deg {a b} \ge \map \deg a$
where $\map \deg a$ denotes the [[Definition:Degree (Polynomial)|degree]] of $a$.
From [[Division Theorem for Polynomial Fo... | Polynomial Forms over Field is Euclidean Domain | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_is_Euclidean_Domain | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_is_Euclidean_Domain | [
"Field Theory",
"Examples of Euclidean Domains",
"Polynomial Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomial Forms",
"Definition:Euclidean Domain"
] | [
"Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors",
"Definition:Degree of Polynomial",
"Division Theorem for Polynomial Forms over Field",
"Definition:Euclidean Domain/Valuation"
] |
proofwiki-3320 | Euclidean Domain is Principal Ideal Domain | A Euclidean domain is a principal ideal domain. | Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$ and whose Euclidean valuation is $\nu$.
We need to show that every ideal of $\struct {D, +, \times}$ is a principal ideal.
Let $U$ be an ideal of $\struct {D, +, \times}$ such that $U \ne \set 0$.
Let $d \in U$ such that $d \ne 0$ and $\map \nu d$ is ... | A [[Definition:Euclidean Domain|Euclidean domain]] is a [[Definition:Principal Ideal Domain|principal ideal domain]]. | Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$ and whose [[Definition:Euclidean Valuation|Euclidean valuation]] is $\nu$.
We need to show that every [[Definition:Ideal of Ring|ideal]] of $\struct {D, +, \times}$ is a [[Definition:Principal ... | Euclidean Domain is Principal Ideal Domain | https://proofwiki.org/wiki/Euclidean_Domain_is_Principal_Ideal_Domain | https://proofwiki.org/wiki/Euclidean_Domain_is_Principal_Ideal_Domain | [
"Euclidean Domains",
"Principal Ideal Domains"
] | [
"Definition:Euclidean Domain",
"Definition:Principal Ideal Domain"
] | [
"Definition:Euclidean Domain",
"Definition:Ring Zero",
"Definition:Euclidean Domain/Valuation",
"Definition:Ideal of Ring",
"Definition:Principal Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Natural Numbers",
"Well-Ordering Principle",
"Defini... |
proofwiki-3321 | Polynomial Forms is PID Implies Coefficient Ring is Field | Let $D$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Let $D \sqbrk X$ be a principal ideal domain;
Then $D$ is a field. | Let $y \in D$ be non-zero.
Then, using the principal ideal property, for some $f \in D \sqbrk X$ we have:
:$\gen {y, X} = \gen f \subseteq D \sqbrk X$
Therefore:
:$\exists p, q \in D \sqbrk X: y = f p, X = f q$
By Properties of Degree we conclude that $f = a$ and $q = b + c X$ for some $a, b, c \in D$.
Substituting int... | Let $D$ be an [[Definition:Integral Domain|integral domain]].
Let $D \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $D$.
Let $D \sqbrk X$ be a [[Definition:Principal Ideal Domain|principal ideal domain]];
Then $D$ is a [[Definition:Field (Abstract Algebra)|field]]. | Let $y \in D$ be non-zero.
Then, using the [[Definition:Principal Ideal Domain|principal ideal property]], for some $f \in D \sqbrk X$ we have:
:$\gen {y, X} = \gen f \subseteq D \sqbrk X$
Therefore:
:$\exists p, q \in D \sqbrk X: y = f p, X = f q$
By [[Properties of Degree]] we conclude that $f = a$ and $q = b + c... | Polynomial Forms is PID Implies Coefficient Ring is Field | https://proofwiki.org/wiki/Polynomial_Forms_is_PID_Implies_Coefficient_Ring_is_Field | https://proofwiki.org/wiki/Polynomial_Forms_is_PID_Implies_Coefficient_Ring_is_Field | [
"Polynomial Theory",
"Principal Ideal Domains"
] | [
"Definition:Integral Domain",
"Definition:Ring of Polynomial Forms",
"Definition:Principal Ideal Domain",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Principal Ideal Domain",
"Properties of Degree",
"Definition:Group of Units/Ring",
"Definition:Field (Abstract Algebra)"
] |
proofwiki-3322 | Euclidean Domain is GCD Domain | Let $\struct {D, +, \times}$ be a Euclidean domain.
Then any two elements $a, b \in D$ have a greatest common divisor $d$ such that:
:$d \divides a \land d \divides b$
:$x \divides a \land x \divides b \implies x \divides d$
and $d$ is written $\gcd \set {a, b}$.
For any $a, b \in D$:
:$\exists s, t \in D: s a + t b = ... | Let $a, b \in D$.
Let $U \subseteq D$ be the set of all elements $h a + k b$ of $D$ where $h, k \in D$.
Then $U$ is an ideal of $D$.
Note that $U = \ideal a + \ideal b$ where $\ideal a$ and $\ideal b$ are Principal Ideal.
By Sum of Ideals is Ideal, $U$ is an ideal.
By Euclidean Domain is Principal Ideal Domain, $U$ is ... | Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]].
Then any two elements $a, b \in D$ have a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] $d$ such that:
:$d \divides a \land d \divides b$
:$x \divides a \land x \divides b \implies x \divides d$
and $d... | Let $a, b \in D$.
Let $U \subseteq D$ be the set of all elements $h a + k b$ of $D$ where $h, k \in D$.
Then $U$ is an [[Definition:Ideal of Ring|ideal]] of $D$.
Note that $U = \ideal a + \ideal b$ where $\ideal a$ and $\ideal b$ are [[Definition:Ideal of Ring|Principal Ideal]].
By [[Sum of Ideals is Ideal]], $U$ i... | Euclidean Domain is GCD Domain | https://proofwiki.org/wiki/Euclidean_Domain_is_GCD_Domain | https://proofwiki.org/wiki/Euclidean_Domain_is_GCD_Domain | [
"Euclidean Domains",
"GCD Domains"
] | [
"Definition:Euclidean Domain",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Greatest Common Divisor/Integral Domain",
"Definition:Associate/Integral Domain"
] | [
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Sum of Ideals is Ideal",
"Euclidean Domain is Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Definition:Divisor (Algebra)/Ring with Unity",
"Common Divisor in Integral Domain Divides Linear Combination",
"Definition:Greatest Com... |
proofwiki-3323 | Associates in Ring of Polynomial Forms over Field | Let $F \sqbrk X$ be the ring of polynomial forms over the field $F$.
Let $\map d X$ and $\map {d'} X$ be polynomial forms in $F \sqbrk X$.
Then $\map d X$ is an associate of $\map {d'} X$ {{iff}} $\map d X = c \cdot \map {d'} X$ for some $c \in F, c \ne 0_F$.
Hence any two polynomials in $F \sqbrk X$ have a unique moni... | From the definition of associate, there exist $\map e X$ and $\map {e'} X$ \in $F \sqbrk X$ such that:
:$\map d X = \map e X \cdot \map {d'} X$
:$\map {d'} X = \map {e'} X \cdot \map d X$
From Field is Integral Domain, $F$ is an integral domain.
From Degree of Product of Polynomials over Integral Domain it follows that... | Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field (Abstract Algebra)|field]] $F$.
Let $\map d X$ and $\map {d'} X$ be [[Definition:Polynomial Form|polynomial forms]] in $F \sqbrk X$.
Then $\map d X$ is an [[Definition:Associate in Integral Domain|ass... | From the definition of [[Definition:Associate in Integral Domain|associate]], there exist $\map e X$ and $\map {e'} X$ \in $F \sqbrk X$ such that:
:$\map d X = \map e X \cdot \map {d'} X$
:$\map {d'} X = \map {e'} X \cdot \map d X$
From [[Field is Integral Domain]], $F$ is an [[Definition:Integral Domain|integral dom... | Associates in Ring of Polynomial Forms over Field | https://proofwiki.org/wiki/Associates_in_Ring_of_Polynomial_Forms_over_Field | https://proofwiki.org/wiki/Associates_in_Ring_of_Polynomial_Forms_over_Field | [
"Factorization",
"Polynomial Theory",
"Euclidean Domains",
"Associates"
] | [
"Definition:Ring of Polynomial Forms",
"Definition:Field (Abstract Algebra)",
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Associate/Integral Domain",
"Definition:Polynomial over Ring as Function on Free Monoid on Set",
"Definition:Monic Polynomial",
"Definition:Great... | [
"Definition:Associate/Integral Domain",
"Field is Integral Domain",
"Definition:Integral Domain",
"Degree of Product of Polynomials over Ring/Corollary 2",
"Definition:Proper Zero Divisor",
"Definition:Multiplicative Identity"
] |
proofwiki-3324 | Zero Element Generates Null Ideal | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
For $r \in R$, let $\ideal r$ denote the ideal generated by $r$.
Then $\ideal {0_R}$ is the null ideal. | By definition:
:$\ideal {0_R} = \set {r \circ 0_R: r \in R}$
but for each $r \in R$ we have by Ring Product with Zero that $r \circ 0_R = 0_R$ for all $r \in R$.
Therefore $\ideal {0_R}$ is the null ideal.
{{qed}}
Category:Ideal Theory
g0bxzwsul4zh97e4us5incfu0yaqszb | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
For $r \in R$, let $\ideal r$ denote the [[Definition:Ideal of Ring|ideal]] [[Definition:Generator of Ideal|generated]] by $r$.
Then $\ideal {0_R}$ is the [[Definition:Null Ideal|null ideal]]. | By definition:
:$\ideal {0_R} = \set {r \circ 0_R: r \in R}$
but for each $r \in R$ we have by [[Ring Product with Zero]] that $r \circ 0_R = 0_R$ for all $r \in R$.
Therefore $\ideal {0_R}$ is the [[Definition:Null Ideal|null ideal]].
{{qed}}
[[Category:Ideal Theory]]
g0bxzwsul4zh97e4us5incfu0yaqszb | Zero Element Generates Null Ideal | https://proofwiki.org/wiki/Zero_Element_Generates_Null_Ideal | https://proofwiki.org/wiki/Zero_Element_Generates_Null_Ideal | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Ideal of Ring",
"Definition:Generator of Ideal of Ring",
"Definition:Null Ideal"
] | [
"Ring Product with Zero",
"Definition:Null Ideal",
"Category:Ideal Theory"
] |
proofwiki-3325 | Irreducible Elements of Ring of Integers | Let $\struct {\Z, +, \times}$ be the ring of integers.
The irreducible elements of $\struct {\Z, +, \times}$ are the prime numbers and their negatives. | We have that Integers form Integral Domain.
Therefore the concept of irreducible is defined.
Let $p$ be a prime number.
By definition, the only divisors of $p$ are $1, -1, p, -p$.
From Units of Ring of Integers, $1$ and $-1$ are (the only) units of $\Z$.
From Associates are Unit Multiples, $p$ and $-p$ are (the only) a... | Let $\struct {\Z, +, \times}$ be the [[Integers form Totally Ordered Ring|ring of integers]].
The [[Definition:Irreducible Element of Ring|irreducible elements]] of $\struct {\Z, +, \times}$ are the [[Definition:Prime Number|prime numbers]] and their [[Definition:Negative|negatives]]. | We have that [[Integers form Integral Domain]].
Therefore the concept of [[Definition:Irreducible Element of Ring|irreducible]] is defined.
Let $p$ be a [[Definition:Prime Number|prime number]].
By [[Definition:Prime Number|definition]], the only [[Definition:Divisor of Integer|divisors]] of $p$ are $1, -1, p, -p$.
... | Irreducible Elements of Ring of Integers | https://proofwiki.org/wiki/Irreducible_Elements_of_Ring_of_Integers | https://proofwiki.org/wiki/Irreducible_Elements_of_Ring_of_Integers | [
"Integers"
] | [
"Integers form Totally Ordered Ring",
"Definition:Irreducible Element of Ring",
"Definition:Prime Number",
"Definition:Negative"
] | [
"Integers form Integral Domain",
"Definition:Irreducible Element of Ring",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Invertible Integers under Multiplication",
"Definition:Unit of Ring",
"Equivalence of Definitions of Associate in Integral Domain/De... |
proofwiki-3326 | Euclid's Lemma for Euclidean Domains | Let $\struct {D, +, \times}$ be a Euclidean domain whose unity is $1$.
Let $a, b, c \in D$.
Let $a \divides b \times c$, where $\divides$ denotes divisibility.
Let $a \perp b$, where $\perp$ denotes relative primeness.
Then $a \divides c$. | {{begin-eqn}}
{{eqn | l = a
| o = \perp
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| r = 1
| c = {{Defof|Coprime Elements of Euclidean Domain}}
}}
{{eqn | ll= \leadsto
| q = \exists x, y \in D
| l = a \times x + b \times y
| r = 1
| c = Bézo... | Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Unity of Ring|unity]] is $1$.
Let $a, b, c \in D$.
Let $a \divides b \times c$, where $\divides$ denotes [[Definition:Divisor of Ring Element|divisibility]].
Let $a \perp b$, where $\perp$ denotes [[Definition:Copr... | {{begin-eqn}}
{{eqn | l = a
| o = \perp
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {a, b}
| r = 1
| c = {{Defof|Coprime Elements of Euclidean Domain}}
}}
{{eqn | ll= \leadsto
| q = \exists x, y \in D
| l = a \times x + b \times y
| r = 1
| c = [[Bé... | Euclid's Lemma for Euclidean Domains | https://proofwiki.org/wiki/Euclid's_Lemma_for_Euclidean_Domains | https://proofwiki.org/wiki/Euclid's_Lemma_for_Euclidean_Domains | [
"Euclidean Domains",
"Euclid's Lemma"
] | [
"Definition:Euclidean Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Coprime/Euclidean Domain"
] | [
"Bézout's Identity/Euclidean Domain",
"Bézout's Identity/Euclidean Domain"
] |
proofwiki-3327 | Rational Polynomial is Content Times Primitive Polynomial | Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
where:
:$\cont f$ is the content of $\map f X$
:$\map {f^*} X$ is a primitive polynomial.
For a given polynomial $\map f X$, both $\c... | === Proof of Existence ===
{{:Rational Polynomial is Content Times Primitive Polynomial/Existence}}{{qed|lemma}} | Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
w... | === [[Rational Polynomial is Content Times Primitive Polynomial/Existence|Proof of Existence]] ===
{{:Rational Polynomial is Content Times Primitive Polynomial/Existence}}{{qed|lemma}} | Rational Polynomial is Content Times Primitive Polynomial | https://proofwiki.org/wiki/Rational_Polynomial_is_Content_Times_Primitive_Polynomial | https://proofwiki.org/wiki/Rational_Polynomial_is_Content_Times_Primitive_Polynomial | [
"Polynomial Theory",
"Rational Polynomial is Content Times Primitive Polynomial"
] | [
"Definition:Ring of Polynomial Forms",
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Content of Polynomial/Rational",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Unique"
] | [
"Rational Polynomial is Content Times Primitive Polynomial/Existence"
] |
proofwiki-3328 | Product of Rational Polynomials | Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$.
Using Rational Polynomial is Content Times Primitive Polynomial, let these be expressed as:
:$\map f X = \cont f \cdot \map {f^*} X$
:$\map g X = \cont g \cdot \map {g... | From Rational Polynomial is Content Times Primitive Polynomial:
:$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \map {g^*} X$
and this expression is unique.
By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \map {g^*} X$ is primitive.
From Content of Rational Polynomial is ... | Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$.
Using [[Rational Polynomial is Cont... | From [[Rational Polynomial is Content Times Primitive Polynomial]]:
:$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \map {g^*} X$
and this expression is [[Definition:Unique|unique]].
By [[Gauss's Lemma on Primitive Rational Polynomials]] we have that $\map {f^*} X \map {g^*} X$ is [[Definition:Pri... | Product of Rational Polynomials | https://proofwiki.org/wiki/Product_of_Rational_Polynomials | https://proofwiki.org/wiki/Product_of_Rational_Polynomials | [
"Polynomial Theory"
] | [
"Definition:Ring of Polynomial Forms",
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring/Indeterminate",
"Rational Polynomial is Content Times Primitive Polynomial",
"Definition:Content of Polynomial/Rational",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Multiplication... | [
"Rational Polynomial is Content Times Primitive Polynomial",
"Definition:Unique",
"Gauss's Lemma on Primitive Rational Polynomials",
"Definition:Primitive Polynomial (Ring Theory)",
"Content of Rational Polynomial is Multiplicative"
] |
proofwiki-3329 | Equivalence of Definitions of Perfect Set | {{TFAE|def = Perfect Set}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset of $S$. | Let $T = \struct {S, \tau}$ be a topological space and let $H \subseteq S$. | {{TFAE|def = Perfect Set}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$. | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] and let $H \subseteq S$. | Equivalence of Definitions of Perfect Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Perfect_Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Perfect_Set | [
"Perfect Sets"
] | [
"Definition:Topological Space",
"Definition:Subset"
] | [
"Definition:Topological Space"
] |
proofwiki-3330 | Complement of Interior equals Closure of Complement | Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.
Let $\map \complement H$ be the complement of $H$ in $T$:
:$\map \complement H = T \setminus H$
Then:
:$\map \complement {H^\circ} = \paren {\map \complement H}^-$
This can alternatively be... | Let $\tau$ be the topology on $T$.
Let $\mathbb K = \set {K \in \tau: K \subseteq H}$.
Then:
{{begin-eqn}}
{{eqn | l = T \setminus H^\circ
| r = T \setminus \bigcup_{K \mathop \in \mathbb K} K
| c = {{Defof|Interior (Topology)}} of $H$
}}
{{eqn | r = \bigcap_{K \mathop \in \mathbb K} \paren {T \setminus K}
... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ and $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of $H$.
Let $\map \complement H$ be the [[Definition:Relative Complement|complement of $H$ i... | Let $\tau$ be the [[Definition:Topology|topology]] on $T$.
Let $\mathbb K = \set {K \in \tau: K \subseteq H}$.
Then:
{{begin-eqn}}
{{eqn | l = T \setminus H^\circ
| r = T \setminus \bigcup_{K \mathop \in \mathbb K} K
| c = {{Defof|Interior (Topology)}} of $H$
}}
{{eqn | r = \bigcap_{K \mathop \in \mathb... | Complement of Interior equals Closure of Complement | https://proofwiki.org/wiki/Complement_of_Interior_equals_Closure_of_Complement | https://proofwiki.org/wiki/Complement_of_Interior_equals_Closure_of_Complement | [
"Set Closures",
"Set Interiors"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)",
"Definition:Interior (Topology)",
"Definition:Relative Complement"
] | [
"Definition:Topology",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Set Complement inverts Subsets"
] |
proofwiki-3331 | Interior of Finite Intersection equals Intersection of Interiors | Let $T$ be a topological space.
Let $n \in \N$.
Let:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
Then:
:$\ds \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ = \bigcap_{i \mathop = 1}^n H_i^\circ$
where $H_i^\circ$ denotes the interior of $H_i$. | In the following, $H_i^-$ denotes the closure of the set $H_i$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ
| r = T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^-
| c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {\paren {\... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $n \in \N$.
Let:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
Then:
:$\ds \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ = \bigcap_{i \mathop = 1}^n H_i^\circ$
where $H_i^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of ... | In the following, $H_i^-$ denotes the [[Definition:Closure (Topology)|closure]] of the set $H_i$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ
| r = T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^-
| c = [[Complement of Interior equals Closure of Complement]]
}}
{{... | Interior of Finite Intersection equals Intersection of Interiors | https://proofwiki.org/wiki/Interior_of_Finite_Intersection_equals_Intersection_of_Interiors | https://proofwiki.org/wiki/Interior_of_Finite_Intersection_equals_Intersection_of_Interiors | [
"Set Interiors",
"Set Intersection"
] | [
"Definition:Topological Space",
"Definition:Interior (Topology)"
] | [
"Definition:Closure (Topology)",
"Complement of Interior equals Closure of Complement",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Closure of Finite Union equals Union of Closures",
"Complement of Interior equals Closure of Complement",
"De Morgan's Laws (Se... |
proofwiki-3332 | Finite Intersection of Regular Open Sets is Regular Open | Let $T$ be a topological space.
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are regular open in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{- \circ}$
where $H_i^{- \circ}$ denotes the interior of the closure of $H_i$.
Then $\ds \bigcap_... | {{begin-eqn}}
{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}
| r = \paren {T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^\circ}^\circ
| c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \paren {T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setmi... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are [[Definition:Regular Open Set|regular open]] in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{- \circ}$
where $H_i^{- \ci... | {{begin-eqn}}
{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}
| r = \paren {T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^\circ}^\circ
| c = [[Complement of Interior equals Closure of Complement]]
}}
{{eqn | r = \paren {T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \s... | Finite Intersection of Regular Open Sets is Regular Open | https://proofwiki.org/wiki/Finite_Intersection_of_Regular_Open_Sets_is_Regular_Open | https://proofwiki.org/wiki/Finite_Intersection_of_Regular_Open_Sets_is_Regular_Open | [
"Regular Open Sets",
"Set Intersection"
] | [
"Definition:Topological Space",
"Definition:Regular Open Set",
"Definition:Interior (Topology)",
"Definition:Closure (Topology)",
"Definition:Regular Open Set"
] | [
"Complement of Interior equals Closure of Complement",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Complement of Interior equals Closure of Complement",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union",
"Relative Complement of R... |
proofwiki-3333 | Finite Union of Regular Closed Sets is Regular Closed | Let $T$ be a topological space.
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are regular closed in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{\circ -}$
where $H_i^{\circ -}$ denotes the closure of the interior of $H_i$
Then $\ds \bigcup... | {{begin-eqn}}
{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}
| r = \paren {T \setminus \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^-}^-
| c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i}... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are [[Definition:Regular Closed Set|regular closed]] in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{\circ -}$
where $H_i^{\... | {{begin-eqn}}
{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}
| r = \paren {T \setminus \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^-}^-
| c = [[Complement of Interior equals Closure of Complement]]
}}
{{eqn | r = \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus ... | Finite Union of Regular Closed Sets is Regular Closed | https://proofwiki.org/wiki/Finite_Union_of_Regular_Closed_Sets_is_Regular_Closed | https://proofwiki.org/wiki/Finite_Union_of_Regular_Closed_Sets_is_Regular_Closed | [
"Regular Closed Sets",
"Set Union"
] | [
"Definition:Topological Space",
"Definition:Regular Closed Set",
"Definition:Closure (Topology)",
"Definition:Interior (Topology)",
"Definition:Regular Closed Set"
] | [
"Complement of Interior equals Closure of Complement",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union",
"Complement of Interior equals Closure of Complement",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Relative Complement of R... |
proofwiki-3334 | Boundary is Intersection of Closure with Closure of Complement | Let $T = \struct {S, \tau}$ be a topological space.
Let $X \subseteq S$.
Let $\partial X$ denote the boundary of $X$ in $T$, defined as:
:$\partial X = X^- \setminus X^\circ$
Let $\overline X = S \setminus X$ denote the complement of $X$ in $S$.
Let $X^-$ denote the closure of $X$ in $T$.
Then:
:$\partial X = X^- \cap ... | {{begin-eqn}}
{{eqn | l = \partial X
| r = X^- \setminus X^\circ
| c = {{Defof|Boundary (Topology)|index = 1}}: $X^\circ$ is the interior of $X$
}}
{{eqn | r = X^- \cap \overline {\paren {X^\circ} }
| c = Set Difference as Intersection with Relative Complement
}}
{{eqn | r = X^- \cap \paren {\overline... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $X \subseteq S$.
Let $\partial X$ denote the [[Definition:Boundary (Topology)|boundary]] of $X$ in $T$, defined as:
:$\partial X = X^- \setminus X^\circ$
Let $\overline X = S \setminus X$ denote the [[Definition:Relative Complem... | {{begin-eqn}}
{{eqn | l = \partial X
| r = X^- \setminus X^\circ
| c = {{Defof|Boundary (Topology)|index = 1}}: $X^\circ$ is the [[Definition:Interior (Topology)|interior]] of $X$
}}
{{eqn | r = X^- \cap \overline {\paren {X^\circ} }
| c = [[Set Difference as Intersection with Relative Complement]]
}}... | Boundary is Intersection of Closure with Closure of Complement | https://proofwiki.org/wiki/Boundary_is_Intersection_of_Closure_with_Closure_of_Complement | https://proofwiki.org/wiki/Boundary_is_Intersection_of_Closure_with_Closure_of_Complement | [
"Set Closures",
"Set Boundaries",
"Set Intersection"
] | [
"Definition:Topological Space",
"Definition:Boundary (Topology)",
"Definition:Relative Complement",
"Definition:Closure (Topology)"
] | [
"Definition:Interior (Topology)",
"Set Difference as Intersection with Relative Complement",
"Complement of Interior equals Closure of Complement"
] |
proofwiki-3335 | Set is Closed iff it Contains its Boundary | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is closed in $T$ {{iff}}:
:$\partial H \subseteq H$
where $\partial H$ is the boundary of $H$. | From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the closure of $H$ in $T$.
Hence from Intersection is Subset we have that:
:$\partial H \subseteq H^-$
Then from Closed Set Equals its Closure, $H$ is closed in $T$ {{iff}} $H = H^-$.
He... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Then $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}}:
:$\partial H \subseteq H$
where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$. | From [[Boundary is Intersection of Closure with Closure of Complement]]:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
Hence from [[Intersection is Subset]] we have that:
:$\partial H \subseteq H^-$
Then from [[Closed Set Equals its Closu... | Set is Closed iff it Contains its Boundary | https://proofwiki.org/wiki/Set_is_Closed_iff_it_Contains_its_Boundary | https://proofwiki.org/wiki/Set_is_Closed_iff_it_Contains_its_Boundary | [
"Closed Sets",
"Set Boundaries"
] | [
"Definition:Topological Space",
"Definition:Closed Set/Topology",
"Definition:Boundary (Topology)"
] | [
"Boundary is Intersection of Closure with Closure of Complement",
"Definition:Closure (Topology)",
"Intersection is Subset",
"Set is Closed iff Equals Topological Closure",
"Definition:Closed Set/Topology"
] |
proofwiki-3336 | Set is Open iff Disjoint from Boundary | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is open in $T$ {{iff}}:
:$\partial H \cap H = \O$
where $\partial H$ is the boundary of $H$ in $T$. | From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the closure of $H$.
Hence from Intersection is Subset we have that:
:$\partial H \subseteq \paren {S \setminus H}^-$
But from Closed Set Equals its Closure, $\paren {S \setminus H}^- = S... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Then $H$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}}:
:$\partial H \cap H = \O$
where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$ in $T$. | From [[Boundary is Intersection of Closure with Closure of Complement]]:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$.
Hence from [[Intersection is Subset]] we have that:
:$\partial H \subseteq \paren {S \setminus H}^-$
But from [[Closed Set Equ... | Set is Open iff Disjoint from Boundary | https://proofwiki.org/wiki/Set_is_Open_iff_Disjoint_from_Boundary | https://proofwiki.org/wiki/Set_is_Open_iff_Disjoint_from_Boundary | [
"Open Sets",
"Set Boundaries"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:Boundary (Topology)"
] | [
"Boundary is Intersection of Closure with Closure of Complement",
"Definition:Closure (Topology)",
"Intersection is Subset",
"Set is Closed iff Equals Topological Closure",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Intersection with Complement i... |
proofwiki-3337 | Set is Clopen iff Boundary is Empty | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is both closed and open in $T$ {{iff}}:
:$\partial H = \O$
where $\partial H$ is the boundary of $H$. | From Set is Open iff Disjoint from Boundary we have that:
:$H$ is open in $T$ {{iff}} $\partial H \cap H = \O$
From Set is Closed iff it Contains its Boundary we have that:
:$H$ is closed in $T$ {{iff}} $\partial H \subseteq H$
From Intersection with Subset is Subset:
:$\partial H \subseteq H \iff \partial H \cap H = \... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Then $H$ is [[Definition:Clopen Set|both closed and open]] in $T$ {{iff}}:
:$\partial H = \O$
where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$. | From [[Set is Open iff Disjoint from Boundary]] we have that:
:$H$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}} $\partial H \cap H = \O$
From [[Set is Closed iff it Contains its Boundary]] we have that:
:$H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}} $\partial H \subseteq H$
From [[... | Set is Clopen iff Boundary is Empty | https://proofwiki.org/wiki/Set_is_Clopen_iff_Boundary_is_Empty | https://proofwiki.org/wiki/Set_is_Clopen_iff_Boundary_is_Empty | [
"Clopen Sets",
"Set Boundaries"
] | [
"Definition:Topological Space",
"Definition:Clopen Set",
"Definition:Boundary (Topology)"
] | [
"Set is Open iff Disjoint from Boundary",
"Definition:Open Set/Topology",
"Set is Closed iff it Contains its Boundary",
"Definition:Closed Set/Topology",
"Intersection with Subset is Subset",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology"
] |
proofwiki-3338 | Boundary of Set is Closed | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\partial H$ denote the boundary of $H$ in $T$.
Then $\partial H$ is closed in $T$. | From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the closure of $H$
From Topological Closure is Closed, both $H^-$ and $\paren {S \setminus H}^-$ are closed in $T$.
From Topology Defined by Closed Sets, the intersection of arbitrarily ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $\partial H$ denote the [[Definition:Boundary (Topology)|boundary]] of $H$ in $T$.
Then $\partial H$ is [[Definition:Closed Set (Topology)|closed]] in $T$. | From [[Boundary is Intersection of Closure with Closure of Complement]]:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$
From [[Topological Closure is Closed]], both $H^-$ and $\paren {S \setminus H}^-$ are [[Definition:Closed Set (Topology)|closed]... | Boundary of Set is Closed | https://proofwiki.org/wiki/Boundary_of_Set_is_Closed | https://proofwiki.org/wiki/Boundary_of_Set_is_Closed | [
"Closed Sets",
"Set Boundaries"
] | [
"Definition:Topological Space",
"Definition:Boundary (Topology)",
"Definition:Closed Set/Topology"
] | [
"Boundary is Intersection of Closure with Closure of Complement",
"Definition:Closure (Topology)",
"Topological Closure is Closed",
"Definition:Closed Set/Topology",
"Topology Defined by Closed Sets",
"Definition:Set Intersection",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
... |
proofwiki-3339 | Boundary of Boundary is Contained in Boundary | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq T$.
Then:
:$\map \partial {\partial H} \subseteq \partial H$
where $\partial H$ is the boundary of $H$ in $T$.
That is, the boundary of the boundary of $H$ is contained in the boundary of $H$. | Let $B = \partial H$.
From Boundary of Set is Closed we have that $B$ is closed in $T$.
Let $B^-$ denote the closure of $B$.
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial B = B^- \cap \paren {S \setminus B}^-$
and so from Intersection is Subset:
:$\partial B \subseteq B^-$
But from Clo... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Then:
:$\map \partial {\partial H} \subseteq \partial H$
where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$ in $T$.
That is, the [[Definition:Boundary (Topology)|boundary]] of the [[D... | Let $B = \partial H$.
From [[Boundary of Set is Closed]] we have that $B$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
Let $B^-$ denote the [[Definition:Closure (Topology)|closure]] of $B$.
From [[Boundary is Intersection of Closure with Closure of Complement]]:
:$\partial B = B^- \cap \paren {S \setminus ... | Boundary of Boundary is Contained in Boundary | https://proofwiki.org/wiki/Boundary_of_Boundary_is_Contained_in_Boundary | https://proofwiki.org/wiki/Boundary_of_Boundary_is_Contained_in_Boundary | [
"Set Boundaries"
] | [
"Definition:Topological Space",
"Definition:Boundary (Topology)",
"Definition:Boundary (Topology)",
"Definition:Boundary (Topology)",
"Definition:Boundary (Topology)"
] | [
"Boundary of Set is Closed",
"Definition:Closed Set/Topology",
"Definition:Closure (Topology)",
"Boundary is Intersection of Closure with Closure of Complement",
"Intersection is Subset",
"Set is Closed iff Equals Topological Closure"
] |
proofwiki-3340 | Equivalence of Definitions of Exterior | {{TFAE|def = Exterior (Topology)|view = exterior|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$. | Let $H^e$ be defined as:
:$H^e$ is the complement of the closure of $H$ in $T$.
Then:
{{begin-eqn}}
{{eqn | l = H^e
| r = S \setminus H^-
| c =
}}
{{eqn | r = \paren {S \setminus H}^\circ
| c = Complement of Interior equals Closure of Complement
}}
{{end-eqn}}
Thus:
:$H^e$ is the interior of the comp... | {{TFAE|def = Exterior (Topology)|view = exterior|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$. | Let $H^e$ be defined as:
:$H^e$ is the [[Definition:Set Complement|complement]] of the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
Then:
{{begin-eqn}}
{{eqn | l = H^e
| r = S \setminus H^-
| c =
}}
{{eqn | r = \paren {S \setminus H}^\circ
| c = [[Complement of Interior equals Closure of... | Equivalence of Definitions of Exterior | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Exterior | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Exterior | [
"Set Exteriors"
] | [
"Definition:Topological Space"
] | [
"Definition:Set Complement",
"Definition:Closure (Topology)",
"Complement of Interior equals Closure of Complement",
"Definition:Interior (Topology)",
"Definition:Set Complement"
] |
proofwiki-3341 | Interior is Subset of Exterior of Exterior | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $H^e$ denote the exterior of $H$ in $T$, and let $H^\circ$ denote the interior of $H$ in $T$.
Then:
:$H^\circ \subseteq \paren {H^e}^e$ | {{begin-eqn}}
{{eqn | l = \paren {H^e}^e
| r = \paren {S \setminus H^e}^\circ
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {S \setminus \paren {S \setminus H^-} }^\circ
| c = Equivalence of Definitions of Exterior
}}
{{eqn | r = \paren {H^-}^\circ
| c = Relative Complement ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $H^e$ denote the [[Definition:Exterior (Topology)|exterior]] of $H$ in $T$, and let $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of $H$ in $T$.
Then:
:$H^\circ \subseteq \paren {H^e}^e$ | {{begin-eqn}}
{{eqn | l = \paren {H^e}^e
| r = \paren {S \setminus H^e}^\circ
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {S \setminus \paren {S \setminus H^-} }^\circ
| c = [[Equivalence of Definitions of Exterior]]
}}
{{eqn | r = \paren {H^-}^\circ
| c = [[Relative Compl... | Interior is Subset of Exterior of Exterior | https://proofwiki.org/wiki/Interior_is_Subset_of_Exterior_of_Exterior | https://proofwiki.org/wiki/Interior_is_Subset_of_Exterior_of_Exterior | [
"Set Interiors",
"Set Exteriors"
] | [
"Definition:Topological Space",
"Definition:Exterior (Topology)",
"Definition:Interior (Topology)"
] | [
"Equivalence of Definitions of Exterior",
"Relative Complement of Relative Complement",
"Interior is Subset of Interior of Closure"
] |
proofwiki-3342 | Interior is Subset of Interior of Closure | Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^\circ$ denote the interior of $H$.
Let $H^-$ denote the closure of $H$.
Then:
:$H^\circ \subseteq \left({H^-}\right)^\circ$ | From Set is Subset of its Topological Closure, we have $H \subseteq H^-$.
The result follows directly from Interior of Subset.
{{qed}}
Category:Set Closures
Category:Set Interiors
hpi3i0ga2cmk8ylx6ricx5jzzr61qz3 | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Let $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of $H$.
Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$.
Then:
:$H^\circ \subseteq \left({H^-}\right)^\circ$ | From [[Set is Subset of its Topological Closure]], we have $H \subseteq H^-$.
The result follows directly from [[Interior of Subset]].
{{qed}}
[[Category:Set Closures]]
[[Category:Set Interiors]]
hpi3i0ga2cmk8ylx6ricx5jzzr61qz3 | Interior is Subset of Interior of Closure | https://proofwiki.org/wiki/Interior_is_Subset_of_Interior_of_Closure | https://proofwiki.org/wiki/Interior_is_Subset_of_Interior_of_Closure | [
"Set Closures",
"Set Interiors"
] | [
"Definition:Topological Space",
"Definition:Interior (Topology)",
"Definition:Closure (Topology)"
] | [
"Set is Subset of its Topological Closure",
"Interior of Subset",
"Category:Set Closures",
"Category:Set Interiors"
] |
proofwiki-3343 | Exterior of Finite Union equals Intersection of Exteriors | Let $T = \struct {S, \tau}$ be a topological space.
Let $n \in \N$.
Let $\forall i \in \closedint 1 n: H_i \subseteq S$.
Then:
:$\ds \paren {\bigcup_{i \mathop = 1}^n H_i}^e = \bigcap_{i \mathop = 1}^n H_i^e$
where $H_i^e$ denotes the exterior of $H_i$. | In the following, $H_i^\circ$ denotes the interior of the set $H_i$ in $T$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^e
| r = \paren {S \setminus \bigcup_{i \mathop = 1}^n H_i}^\circ
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {\bigcap_{i \mathop = 1}^n \paren {... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $n \in \N$.
Let $\forall i \in \closedint 1 n: H_i \subseteq S$.
Then:
:$\ds \paren {\bigcup_{i \mathop = 1}^n H_i}^e = \bigcap_{i \mathop = 1}^n H_i^e$
where $H_i^e$ denotes the [[Definition:Exterior (Topology)|exterior]] of ... | In the following, $H_i^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of the set $H_i$ in $T$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^e
| r = \paren {S \setminus \bigcup_{i \mathop = 1}^n H_i}^\circ
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \pare... | Exterior of Finite Union equals Intersection of Exteriors | https://proofwiki.org/wiki/Exterior_of_Finite_Union_equals_Intersection_of_Exteriors | https://proofwiki.org/wiki/Exterior_of_Finite_Union_equals_Intersection_of_Exteriors | [
"Set Exteriors"
] | [
"Definition:Topological Space",
"Definition:Exterior (Topology)"
] | [
"Definition:Interior (Topology)",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union",
"Interior of Finite Intersection equals Intersection of Interiors"
] |
proofwiki-3344 | Closure of Union contains Union of Closures | Let $T = \struct {S, \tau}$ be a topological space.
Let $\mathbb H$ be a set of subsets of $S$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ denotes the power set of $S$.
Then the union of the closures of the elements of $\mathbb H$ is a subset of the closure of the union of $\mathbb H$:
:$\ds \bi... | Let $\ds K = \bigcup_{H \mathop \in \mathbb H} \map \cl H$ and $\ds L = \bigcup_{H \mathop \in \mathbb H} H$.
We have:
:$\forall H \in \mathbb H: H \subseteq L$
so from Topological Closure of Subset is Subset of Topological Closure:
:$\map \cl H \subseteq \map \cl L$
It follows from Union is Smallest Superset: General ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
Then the [[Definition:S... | Let $\ds K = \bigcup_{H \mathop \in \mathbb H} \map \cl H$ and $\ds L = \bigcup_{H \mathop \in \mathbb H} H$.
We have:
:$\forall H \in \mathbb H: H \subseteq L$
so from [[Topological Closure of Subset is Subset of Topological Closure]]:
:$\map \cl H \subseteq \map \cl L$
It follows from [[Union is Smallest Superset/... | Closure of Union contains Union of Closures | https://proofwiki.org/wiki/Closure_of_Union_contains_Union_of_Closures | https://proofwiki.org/wiki/Closure_of_Union_contains_Union_of_Closures | [
"Set Union",
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Subset",
"Definition:Power Set",
"Definition:Set Union",
"Definition:Closure (Topology)",
"Definition:Element",
"Definition:Subset",
"Definition:Closure (Topology)",
"Definition:Set Union"
] | [
"Topological Closure of Subset is Subset of Topological Closure",
"Union is Smallest Superset/General Result"
] |
proofwiki-3345 | Set is Subset of its Topological Closure | Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ be the closure of $H$ in $T$.
Then:
:$H \subseteq H^-$ | From the definition of closure, we have:
:$H^-$ is the union of $H$ and its limit points.
From Subset of Union it follows directly that:
:$H \subseteq H^-$
{{qed}}
Category:Set Closures
4d7jedgi469fn642g55pjs0t2diepw3 | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Let $H^-$ be the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
Then:
:$H \subseteq H^-$ | From the [[Definition:Closure (Topology)|definition of closure]], we have:
:$H^-$ is the [[Definition:Set Union|union]] of $H$ and its [[Definition:Limit Point of Set|limit points]].
From [[Subset of Union]] it follows directly that:
:$H \subseteq H^-$
{{qed}}
[[Category:Set Closures]]
4d7jedgi469fn642g55pjs0t2diepw... | Set is Subset of its Topological Closure | https://proofwiki.org/wiki/Set_is_Subset_of_its_Topological_Closure | https://proofwiki.org/wiki/Set_is_Subset_of_its_Topological_Closure | [
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)"
] | [
"Definition:Closure (Topology)",
"Definition:Set Union",
"Definition:Limit Point/Topology/Set",
"Set is Subset of Union",
"Category:Set Closures"
] |
proofwiki-3346 | Set Closure as Intersection of Closed Sets | Let $T$ be a topological space.
Let $H \subseteq T$.
Let the closure of $H$ (in $T$) be defined as:
:$H^- := H \cup H'$
where $H'$ is the derived set of $H$.
Let $\mathbb K$ be defined as:
:$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$
That is, let $\mathbb K$ be the set of all closed sets of $T... | What needs to be proved here is:
:$\ds H^- = \bigcap \mathbb K$
where:
:$H^- = H \cup H'$
:$H'$ denotes the set of all limit points of $H$
:$\ds \bigcap \mathbb K$ is the intersection of all closed sets of $T$ which contain $H$.
Let $K \in \mathbb K$.
From Topological Closure of Subset is Subset of Topological Closure,... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Let the [[Definition:Closure (Topology)/Definition 1|closure of $H$ (in $T$)]] be defined as:
:$H^- := H \cup H'$
where $H'$ is the [[Definition:Derived Set|derived set]] of $H$.
Let $\mathbb K$ be defined as:
:$\mathbb K := \lef... | What needs to be proved here is:
:$\ds H^- = \bigcap \mathbb K$
where:
:$H^- = H \cup H'$
:$H'$ denotes the [[Definition:Set|set]] of all [[Definition:Limit Point of Set|limit points]] of $H$
:$\ds \bigcap \mathbb K$ is the [[Definition:Intersection of Set of Sets|intersection]] of all [[Definition:Closed Set (Topolog... | Set Closure as Intersection of Closed Sets | https://proofwiki.org/wiki/Set_Closure_as_Intersection_of_Closed_Sets | https://proofwiki.org/wiki/Set_Closure_as_Intersection_of_Closed_Sets | [
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)/Definition 1",
"Definition:Derived Set",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closure (Topology)/Definition 2",
"Definition:Set Intersection/Set of Sets",
"Definition:Closed Set/Topology"
] | [
"Definition:Set",
"Definition:Limit Point/Topology/Set",
"Definition:Set Intersection/Set of Sets",
"Definition:Closed Set/Topology",
"Topological Closure of Subset is Subset of Topological Closure",
"Set is Closed iff Equals Topological Closure",
"Intersection is Largest Subset",
"Topological Closure... |
proofwiki-3347 | Set Closure is Smallest Closed Set/Topology | Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ in $T$.
Then $H^-$ is the smallest superset of $H$ that is closed in $T$. | Define:
:$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$
That is, let $\mathbb K$ be the set of all supersets of $H$ that are closed in $T$.
The claim is that $H^-$ is the smallest set of $\mathbb K$.
From Set is Subset of its Topological Closure:
:$H \subseteq H^-$
From Topological Closure is Clo... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
Then $H^-$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Superset|superset]] of $H$ that is [[Definition:Closed Set (Topolog... | Define:
:$\mathbb K := \leftset {K \supseteq H: K}$ is [[Definition:Closed Set (Topology)|closed]] in $\rightset T$
That is, let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Superset|supersets]] of $H$ that are [[Definition:Closed Set (Topology)|closed]] in $T$.
The claim is that $H^-$ is the [[Defini... | Set Closure is Smallest Closed Set/Topology | https://proofwiki.org/wiki/Set_Closure_is_Smallest_Closed_Set/Topology | https://proofwiki.org/wiki/Set_Closure_is_Smallest_Closed_Set/Topology | [
"Set Closure is Smallest Closed Set",
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)",
"Definition:Smallest Set by Set Inclusion",
"Definition:Subset/Superset",
"Definition:Closed Set/Topology"
] | [
"Definition:Closed Set/Topology",
"Definition:Set",
"Definition:Subset/Superset",
"Definition:Closed Set/Topology",
"Definition:Smallest Set by Set Inclusion",
"Set is Subset of its Topological Closure",
"Topological Closure is Closed",
"Definition:Closed Set/Topology",
"Set Closure as Intersection ... |
proofwiki-3348 | Equivalence of Definitions of Interior (Topology) | {{TFAE|def = Interior (Topology)|view = interior|context = Topology (Mathematical Branch)|contextview = topology}}
Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$.
=== Definition 1 ===
{{:Definition:Interior (Topology)/Definition 1}}
=== Definition 2 ===
{{:Definition:Interior (Topology)/Definition ... | === Definition $1$ is equivalent to Definition $2$ ===
Let $\mathbb K$ be defined as:
:$\mathbb K := \set {K \in \tau: K \subseteq H}$
That is, let $\mathbb K$ be the set of all open sets of $T$ contained in $H$.
Then from definition 1 of the interior of $H$, we have:
:$\ds H^\circ = \bigcup_{K \mathop \in \mathbb K} K... | {{TFAE|def = Interior (Topology)|view = interior|context = Topology (Mathematical Branch)|contextview = topology}}
Let $\struct {T, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
=== [[Definition:Interior (Topology)/Definition 1|Definition 1]] ===
{{:Definition:Interior (Topolo... | === Definition $1$ is equivalent to Definition $2$ ===
Let $\mathbb K$ be defined as:
:$\mathbb K := \set {K \in \tau: K \subseteq H}$
That is, let $\mathbb K$ be the set of all [[Definition:Open Set (Topology)|open sets]] of $T$ contained in $H$.
Then from [[Definition:Interior (Topology)/Definition 1|definition 1]... | Equivalence of Definitions of Interior (Topology) | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Interior_(Topology) | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Interior_(Topology) | [
"Equivalence of Definitions of Interior (Topology)",
"Set Interiors"
] | [
"Definition:Topological Space",
"Definition:Interior (Topology)/Definition 1",
"Definition:Interior (Topology)/Definition 2",
"Definition:Interior (Topology)/Definition 3"
] | [
"Definition:Open Set/Topology",
"Definition:Interior (Topology)/Definition 1",
"Definition:Interior (Topology)",
"Definition:Set Union",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Set is Subset of Union",
"Definition:Open Set/Topology",
"Definition:Interior (Topology)/Definitio... |
proofwiki-3349 | Intersection of Interiors contains Interior of Intersection | Let $T$ be a topological space.
Let $\mathbb H$ be a set of subsets of $T$.
That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then the interior of the intersection of $\mathbb H$ is a subset of the intersection of the interiors of the elements of $\mathbb H$.
:$\ds \paren {\big... | In the following, $H^-$ denotes the closure of the set $H$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ
| r = T \setminus \paren {T \setminus \bigcap_{H \mathop \in \mathbb H} H}^-
| c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {\... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $T$.
That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the [[Definition:Power Set|power set]] of $T$.
Then the [[Definition:Interior (Topology)|interi... | In the following, $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of the set $H$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ
| r = T \setminus \paren {T \setminus \bigcap_{H \mathop \in \mathbb H} H}^-
| c = [[Complement of Interior equals Closure of Complement... | Intersection of Interiors contains Interior of Intersection | https://proofwiki.org/wiki/Intersection_of_Interiors_contains_Interior_of_Intersection | https://proofwiki.org/wiki/Intersection_of_Interiors_contains_Interior_of_Intersection | [
"Set Interiors",
"Set Intersection"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Subset",
"Definition:Power Set",
"Definition:Interior (Topology)",
"Definition:Set Intersection",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Interior (Topology)",
"Definition:Element"
] | [
"Definition:Closure (Topology)",
"Complement of Interior equals Closure of Complement",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Closure of Union contains Union of Closures",
"Set Complement inverts Subsets",
"Complement of Interior equals Closure of Compl... |
proofwiki-3350 | Intersection of Exteriors contains Exterior of Union | Let $T = \struct {S, \tau}$ be a topological space.
Let $\mathbb H$ be a set of subsets of $S$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ is the power set of $S$.
Then:
:$\ds \paren {\bigcup_{H \mathop \in \mathbb H} H}^e \subseteq \bigcap_{H \mathop \in \mathbb H} H^e$
where $H^e$ denotes the ... | In the following, $H^\circ$ denotes the interior of the set $H$ in $T$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e
| r = \paren {S \setminus \bigcup_{H \mathop \in \mathbb H} H}^\circ
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {\bigcap_{H \mathop \in \ma... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ is the [[Definition:Power Set|power set]] of $S$.
Then:
:$\ds \paren {\bigcup_... | In the following, $H^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of the set $H$ in $T$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e
| r = \paren {S \setminus \bigcup_{H \mathop \in \mathbb H} H}^\circ
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r ... | Intersection of Exteriors contains Exterior of Union | https://proofwiki.org/wiki/Intersection_of_Exteriors_contains_Exterior_of_Union | https://proofwiki.org/wiki/Intersection_of_Exteriors_contains_Exterior_of_Union | [
"Set Exteriors"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Subset",
"Definition:Power Set",
"Definition:Exterior (Topology)"
] | [
"Definition:Interior (Topology)",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union",
"Intersection of Interiors contains Interior of Intersection"
] |
proofwiki-3351 | Exterior of Intersection contains Union of Exteriors | Let $T = \struct {S, \tau}$ be a topological space.
Let $\mathbb H$ be a set of subsets of $TS$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ is the power set of $S$.
Then:
:$\ds \bigcup_{H \mathop \in \mathbb H} H^e \subseteq \paren {\bigcap_{H \mathop \in \mathbb H} H}^e$
where $H^e$ denotes the... | We have:
{{begin-eqn}}
{{eqn | l = \bigcup_{H \mathop \in \mathbb H} H^e
| r = \bigcup_{H \mathop \in \mathbb H} \paren {S \setminus H^-})
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = S \setminus \bigcap_{H \mathop \in \mathbb H} H^-
| c = De Morgan's Laws: Difference with Intersection
... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $TS$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ is the [[Definition:Power Set|power set]] of $S$.
Then:
:$\ds \bigcup_{H \mat... | We have:
{{begin-eqn}}
{{eqn | l = \bigcup_{H \mathop \in \mathbb H} H^e
| r = \bigcup_{H \mathop \in \mathbb H} \paren {S \setminus H^-})
| c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = S \setminus \bigcap_{H \mathop \in \mathbb H} H^-
| c = [[De Morgan's Laws (Set Theory)/Set Difference... | Exterior of Intersection contains Union of Exteriors | https://proofwiki.org/wiki/Exterior_of_Intersection_contains_Union_of_Exteriors | https://proofwiki.org/wiki/Exterior_of_Intersection_contains_Union_of_Exteriors | [
"Set Exteriors"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Subset",
"Definition:Power Set",
"Definition:Exterior (Topology)"
] | [
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Closure of Intersection is Subset of Intersection of Closures",
"Set Complement inverts Subsets",
"Definition:Exterior (Topology)"
] |
proofwiki-3352 | Separated Sets are Disjoint | Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B \subseteq S$ such that $A$ and $B$ are separated in $T$.
Then $A$ and $B$ are disjoint:
:$A \cap B = \O$ | Let $A$ and $B$ be separated in $T$.
Then:
{{begin-eqn}}
{{eqn | l = A^- \cap B
| r = \O
| c = {{Defof|Separated Sets}}: $A^-$ is the closure of $A$
}}
{{eqn | ll= \leadsto
| l = \paren {A \cup A'} \cap B
| r = \O
| c = {{Defof|Closure (Topology)|Set Closure}}: $A'$ is the derived set of $... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A, B \subseteq S$ such that $A$ and $B$ are [[Definition:Separated Sets|separated in $T$]].
Then $A$ and $B$ are [[Definition:Disjoint Sets|disjoint]]:
:$A \cap B = \O$ | Let $A$ and $B$ be [[Definition:Separated Sets|separated in $T$]].
Then:
{{begin-eqn}}
{{eqn | l = A^- \cap B
| r = \O
| c = {{Defof|Separated Sets}}: $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$
}}
{{eqn | ll= \leadsto
| l = \paren {A \cup A'} \cap B
| r = \O
| c = {{D... | Separated Sets are Disjoint | https://proofwiki.org/wiki/Separated_Sets_are_Disjoint | https://proofwiki.org/wiki/Separated_Sets_are_Disjoint | [
"Separated Sets"
] | [
"Definition:Topological Space",
"Definition:Separated Sets",
"Definition:Disjoint Sets"
] | [
"Definition:Separated Sets",
"Definition:Closure (Topology)",
"Definition:Derived Set",
"Intersection Distributes over Union",
"Union is Empty iff Sets are Empty"
] |
proofwiki-3353 | Second-Countable Space is First-Countable | Let $T = \struct {S, \tau}$ be a topological space which is a second-countable space.
Then $T$ is also a first-countable space. | {{Recall|First-Countable Space|first-countable space}}
{{:Definition:First-Countable Space}}
{{Recall|Second-Countable Space|second-countable space}}
{{:Definition:Second-Countable Space}}
Consider the underlying set $S$ as an open set.
From Set is Open iff Neighborhood of all its Points, $S$ has that property.
As $T$ ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:Second-Countable Space|second-countable space]].
Then $T$ is also a [[Definition:First-Countable Space|first-countable space]]. | {{Recall|First-Countable Space|first-countable space}}
{{:Definition:First-Countable Space}}
{{Recall|Second-Countable Space|second-countable space}}
{{:Definition:Second-Countable Space}}
Consider the [[Definition:Underlying Set of Topological Space|underlying set]] $S$ as an [[Definition:Open Set (Topology)|open se... | Second-Countable Space is First-Countable | https://proofwiki.org/wiki/Second-Countable_Space_is_First-Countable | https://proofwiki.org/wiki/Second-Countable_Space_is_First-Countable | [
"Second-Countable Spaces",
"First-Countable Spaces"
] | [
"Definition:Topological Space",
"Definition:Second-Countable Space",
"Definition:First-Countable Space"
] | [
"Definition:Underlying Set/Topological Space",
"Definition:Open Set/Topology",
"Set is Open iff Neighborhood of all its Points",
"Definition:Property",
"Definition:Countable Basis",
"Definition:Element",
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:Second-Countable Space",
"De... |
proofwiki-3354 | Second-Countability is Hereditary | Let $T = \struct {S, \tau}$ be a topological space which is second-countable.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is second-countable.
That is, the property of second-countability is hereditary. | From the definition of second-countable, $\struct {S, \tau}$ has a countable basis.
That is, $\exists \BB \subseteq \tau$ such that:
:for all $U \in \tau$, $U$ is a union of sets from $\BB$
:$\BB$ is countable.
As $H \subseteq S$ it follows that a $H$ itself is a union of sets from $\BB$.
The result follows from Basis ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Second-Countable Space|second-countable]].
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
Then $T_H$ is [[Definition:Second-Countable... | From the definition of [[Definition:Second-Countable Space|second-countable]], $\struct {S, \tau}$ has a [[Definition:Countable Basis|countable basis]].
That is, $\exists \BB \subseteq \tau$ such that:
:for all $U \in \tau$, $U$ is a [[Definition:Set Union|union]] of sets from $\BB$
:$\BB$ is [[Definition:Countable Se... | Second-Countability is Hereditary | https://proofwiki.org/wiki/Second-Countability_is_Hereditary | https://proofwiki.org/wiki/Second-Countability_is_Hereditary | [
"Second-Countable Spaces",
"Examples of Hereditary Properties"
] | [
"Definition:Topological Space",
"Definition:Second-Countable Space",
"Definition:Topological Subspace",
"Definition:Second-Countable Space",
"Definition:Second-Countable Space",
"Definition:Hereditary Property (Topology)"
] | [
"Definition:Second-Countable Space",
"Definition:Countable Basis",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Set Union",
"Basis for Topological Subspace"
] |
proofwiki-3355 | First-Countability is Hereditary | Let $T = \struct {S, \tau}$ be a topological space which is first-countable.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is first-countable.
That is, the property of first-countability is hereditary. | From the definition of first-countable, every point in $S$ has a countable local basis in $T$.
The intersection of $H$ with the countable local basis of $S$ provides a countable local basis for $H$.
As every point in $H$ is also a point in $S$, the result follows from Basis for Topological Subspace.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:First-Countable Space|first-countable]].
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
Then $T_H$ is [[Definition:First-Countable Sp... | From the definition of [[Definition:First-Countable Space|first-countable]], every [[Definition:Point of Set|point]] in $S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] in $T$.
The [[Definition:Set Intersection|intersection]] of $H$ with the [[Definition:Countable Set|countable]]... | First-Countability is Hereditary | https://proofwiki.org/wiki/First-Countability_is_Hereditary | https://proofwiki.org/wiki/First-Countability_is_Hereditary | [
"First-Countable Spaces",
"Examples of Hereditary Properties"
] | [
"Definition:Topological Space",
"Definition:First-Countable Space",
"Definition:Topological Subspace",
"Definition:First-Countable Space",
"Definition:First-Countable Space",
"Definition:Hereditary Property (Topology)"
] | [
"Definition:First-Countable Space",
"Definition:Element",
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:Set Intersection",
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:Element",
"Definition:Element",
... |
proofwiki-3356 | Basis for Topological Subspace | Let $T = \struct {A, \tau}$ be a topological space.
Let $\O \subseteq H \subseteq A$ and so let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.
Let $\BB$ be a (synthetic) basis for $T$.
Let $\BB_H$ be defined as:
:$\BB_H = \set {U \cap H: U \in \BB}$
Then $\BB_H$ is a (synthetic) basis for $H$. | $\BB \subseteq \powerset A$ is a synthetic basis for $T$ {{iff}}:
:$(\text B 1): \quad$ $A$ is a union of sets from $\BB$
:$(\text B 2): \quad$ If $B_1, B_2 \in B$, then $B_1 \cap B_2$ is a union of sets from $\BB$.
Let $A = \mathbb S$ be a union of sets from $\BB$.
Then:
:$\ds A = \bigcup_{S \mathop \in \mathbb S} S$
... | Let $T = \struct {A, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\O \subseteq H \subseteq A$ and so let $T_H = \struct {H, \tau_H}$ be a [[Definition:Topological Subspace|subspace]] of $T$.
Let $\BB$ be a [[Definition:Synthetic Basis|(synthetic) basis]] for $T$.
Let $\BB_H$ be defined as:
:$... | $\BB \subseteq \powerset A$ is a [[Definition:Synthetic Basis|synthetic basis]] for $T$ {{iff}}:
:$(\text B 1): \quad$ $A$ is a [[Definition:Set Union|union]] of sets from $\BB$
:$(\text B 2): \quad$ If $B_1, B_2 \in B$, then $B_1 \cap B_2$ is a [[Definition:Set Union|union]] of sets from $\BB$.
Let $A = \mathbb S$ ... | Basis for Topological Subspace | https://proofwiki.org/wiki/Basis_for_Topological_Subspace | https://proofwiki.org/wiki/Basis_for_Topological_Subspace | [
"Topological Subspaces",
"Topological Bases"
] | [
"Definition:Topological Space",
"Definition:Topological Subspace",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Basis (Topology)/Synthetic Basis"
] | [
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Set Union",
"Definition:Set Union",
"Definition:Set Union",
"Intersection with Subset is Subset",
"Intersection Distributes over Union",
"Definition:Set Union",
"Definition:Basis (Topology)/Synthetic Basis",
"Intersection Distributes over Un... |
proofwiki-3357 | Characterisation of Local Rings | Let $R$ be a ring.
Let $J \lhd R$ be a maximal ideal.
:$(1): \quad$ If the set $R \setminus J$ is precisely the group of units $R^\times$ of $R$, then $\tuple {R, J}$ is a local ring.
:$(2): \quad$ If $1 + x$ is a unit in $R$ for all $x \in J$ then $\tuple {R, J}$ is local.
{{explain|The specific interpretation of the ... | $(1): \quad$ Suppose that $J$ is the set of non-units of $R$.
Then by Ideal of Unit is Whole Ring, every ideal not equal to $R$ is contained in $J$.
Therefore $J$ is the unique maximal ideal of $R$, so $\tuple {R, J}$ is local.
$(2): \quad$ Let $x \in R \setminus J$.
Then:
:$J \subsetneq \map I {J \cup \set x} \subsete... | Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $J \lhd R$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
:$(1): \quad$ If the set $R \setminus J$ is precisely the [[Definition:Group of Units of Ring|group of units]] $R^\times$ of $R$, then $\tuple {R, J}$ is a [[Definition:Local Ring|local ri... | $(1): \quad$ Suppose that $J$ is the set of non-units of $R$.
Then by [[Ideal of Unit is Whole Ring]], every [[Definition:Ideal of Ring|ideal]] not equal to $R$ is contained in $J$.
Therefore $J$ is the unique [[Definition:Maximal Ideal of Ring|maximal ideal]] of $R$, so $\tuple {R, J}$ is [[Definition:Local Ring|loc... | Characterisation of Local Rings | https://proofwiki.org/wiki/Characterisation_of_Local_Rings | https://proofwiki.org/wiki/Characterisation_of_Local_Rings | [
"Local Rings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Maximal Ideal of Ring",
"Definition:Group of Units/Ring",
"Definition:Local Ring",
"Definition:Unit of Ring",
"Definition:Local Ring"
] | [
"Ideal of Unit is Whole Ring",
"Definition:Ideal of Ring",
"Definition:Maximal Ideal of Ring",
"Definition:Local Ring",
"Definition:Generator of Ideal of Ring",
"Definition:Maximal Ideal of Ring",
"Definition:Generator of Ideal of Ring",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Categ... |
proofwiki-3358 | Continuity Defined by Closure | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous {{iff}}:
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
where $H^-$ denotes the closure of $H$ in $T_1$.
That is, {{iff}} the image of the clos... | === Necessary Condition ===
{{:Continuity Defined by Closure/Necessary Condition/Proof 1}}{{qed|lemma}} | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Mapping|mapping]].
Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] {{iff}}:
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subs... | === [[Continuity Defined by Closure/Necessary Condition|Necessary Condition]] ===
{{:Continuity Defined by Closure/Necessary Condition/Proof 1}}{{qed|lemma}} | Continuity Defined by Closure | https://proofwiki.org/wiki/Continuity_Defined_by_Closure | https://proofwiki.org/wiki/Continuity_Defined_by_Closure | [
"Continuity Defined by Closure",
"Continuous Mappings",
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Closure (Topology)",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Closure (Topology)",
"Definition:Subset",
"Definition:Closure (Topology)",
"Definition:Image (Set ... | [
"Continuity Defined by Closure/Necessary Condition"
] |
proofwiki-3359 | Continuity Defined by Closure | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous {{iff}}:
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
where $H^-$ denotes the closure of $H$ in $T_1$.
That is, {{iff}} the image of the clos... | Let $f$ be continuous.
Let $y \in f \sqbrk {\map \cl H}$.
Then:
:$\exists x \in \map \cl H: y = \map f x$
Let $U$ be an open set of $T_2$ such that $y \in U$.
Then by definition of continuous mapping:
:$f^{-1} \sqbrk U$ is an open set of $T_1$ such that:
::$x \in f^{-1} \sqbrk U$
Hence:
:$f^{-1} \sqbrk U \cap H \ne \O$... | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Mapping|mapping]].
Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] {{iff}}:
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subs... | Let $f$ be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
Let $y \in f \sqbrk {\map \cl H}$.
Then:
:$\exists x \in \map \cl H: y = \map f x$
Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T_2$ such that $y \in U$.
Then by definition of [[Definition:Everywhere Continuous Mapping ... | Continuity Defined by Closure/Necessary Condition/Proof 1 | https://proofwiki.org/wiki/Continuity_Defined_by_Closure | https://proofwiki.org/wiki/Continuity_Defined_by_Closure/Necessary_Condition/Proof_1 | [
"Continuity Defined by Closure",
"Continuous Mappings",
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Closure (Topology)",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Closure (Topology)",
"Definition:Subset",
"Definition:Closure (Topology)",
"Definition:Image (Set ... | [
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Set/Topology"
] |
proofwiki-3360 | Valuation Ring is Local | Let $R$ be a valuation ring.
Then $R$ is a local ring. | By definition of local ring:
:$R$ is a '''local ring''' {{iff}}:
::$R$ is non-trivial
::the sum of two arbitrary non-units is a non-unit.
First we recall that as a valuation ring is an integral domain, then {{afortiori}}:
:$R$ is non-trivial
:$R$ has no (proper) zero divisors
:$R$ is a commutative and unitary ring.
Let... | Let $R$ be a [[Definition:Valuation Ring|valuation ring]].
Then $R$ is a [[Definition:Local Ring|local ring]]. | By definition of [[Definition:Local Ring|local ring]]:
:$R$ is a '''[[Definition:Local Ring|local ring]]''' {{iff}}:
::$R$ is [[Definition:Non-Trivial Ring|non-trivial]]
::the [[Definition:Ring Addition|sum]] of two arbitrary [[Definition:Unit of Ring|non-units]] is a [[Definition:Unit of Ring|non-unit]].
First we re... | Valuation Ring is Local | https://proofwiki.org/wiki/Valuation_Ring_is_Local | https://proofwiki.org/wiki/Valuation_Ring_is_Local | [
"Valuation Rings",
"Local Rings"
] | [
"Definition:Valuation Ring",
"Definition:Local Ring"
] | [
"Definition:Local Ring",
"Definition:Local Ring",
"Definition:Non-Trivial Ring",
"Definition:Ring (Abstract Algebra)/Addition",
"Definition:Unit of Ring",
"Definition:Unit of Ring",
"Definition:Valuation Ring",
"Definition:Integral Domain",
"Definition:Non-Trivial Ring",
"Definition:Proper Zero Di... |
proofwiki-3361 | Bijection is Open iff Inverse is Continuous | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a bijection.
Then $f$ is open {{iff}} $f^{-1}$ is continuous. | Let $f$ be a bijection.
Let $g := f^{-1}$.
By Bijection iff Inverse is Bijection we have that $g$ is a bijection and that $g^{-1} = f$.
Let $f$ be open.
Then by definition of open mapping:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
taking $H \in \tau_1$ by definition of open in $T_1$.
But $f = g^{-1}$ and so:
:$\fo... | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Bijection|bijection]].
Then $f$ is [[Definition:Open Mapping|open]] {{iff}} $f^{-1}$ is [[Definition:Continuous Mapping (Topology)|continuous]]. | Let $f$ be a [[Definition:Bijection|bijection]].
Let $g := f^{-1}$.
By [[Bijection iff Inverse is Bijection]] we have that $g$ is a [[Definition:Bijection|bijection]] and that $g^{-1} = f$.
Let $f$ be [[Definition:Open Mapping|open]].
Then by definition of [[Definition:Open Mapping|open mapping]]:
:$\forall H \in ... | Bijection is Open iff Inverse is Continuous | https://proofwiki.org/wiki/Bijection_is_Open_iff_Inverse_is_Continuous | https://proofwiki.org/wiki/Bijection_is_Open_iff_Inverse_is_Continuous | [
"Open Mappings",
"Continuous Mappings"
] | [
"Definition:Topological Space",
"Definition:Bijection",
"Definition:Open Mapping",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Definition:Open Mapping",
"Definition:Open Mapping",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)... |
proofwiki-3362 | Bijection is Open iff Closed | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a bijection.
Then $f$ is open {{iff}} $f$ is closed. | Let $f$ be a bijection.
Suppose $f$ is an open mapping.
From the definition of open mapping:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
As $f$ is a bijection:
:$f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \setminus f \sqbrk H = S_2 \setminus f \sqbrk H$
By definition of closed set:
:$S_1 \setminus H$ is closed in $... | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Bijection|bijection]].
Then $f$ is [[Definition:Open Mapping|open]] {{iff}} $f$ is [[Definition:Closed Mapping|closed]]. | Let $f$ be a [[Definition:Bijection|bijection]].
Suppose $f$ is an [[Definition:Open Mapping|open mapping]].
From the definition of [[Definition:Open Mapping|open mapping]]:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
As $f$ is a [[Definition:Bijection|bijection]]:
:$f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \s... | Bijection is Open iff Closed | https://proofwiki.org/wiki/Bijection_is_Open_iff_Closed | https://proofwiki.org/wiki/Bijection_is_Open_iff_Closed | [
"Open Mappings",
"Closed Mappings",
"Bijections"
] | [
"Definition:Topological Space",
"Definition:Bijection",
"Definition:Open Mapping",
"Definition:Closed Mapping"
] | [
"Definition:Bijection",
"Definition:Open Mapping",
"Definition:Open Mapping",
"Definition:Bijection",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Mapping",
"Definition:Closed Set/Topology",
"Definition:Closed Mapping",
"Definition:Closed Mapping",
"Defini... |
proofwiki-3363 | T2 Space is T1 | Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then $T$ is also a $T_1$ space. | {{Recall|T1 Space|$T_1$ space}}
{{:Definition:T1 Space/Definition 1}}
Let $T$ be a $T_2$ space.
{{Recall|T2 Space|$T_2$ space}}
{{:Definition:T2 Space/Definition 1}}
As $U \cap V = \O$ it follows from the definition of disjoint sets that:
{{begin-itemize}}
{{item||$x \in U \implies x \notin V$}}
{{item||$y \in V \impli... | Let $T = \struct {S, \tau}$ be a [[Definition:T2 Space|$T_2$ (Hausdorff) space]].
Then $T$ is also a [[Definition:T1 Space|$T_1$ space]]. | {{Recall|T1 Space|$T_1$ space}}
{{:Definition:T1 Space/Definition 1}}
Let $T$ be a [[Definition:T2 Space|$T_2$ space]].
{{Recall|T2 Space|$T_2$ space}}
{{:Definition:T2 Space/Definition 1}}
As $U \cap V = \O$ it follows from the definition of [[Definition:Disjoint Sets|disjoint sets]] that:
{{begin-itemize}}
{{item|... | T2 Space is T1 | https://proofwiki.org/wiki/T2_Space_is_T1 | https://proofwiki.org/wiki/T2_Space_is_T1 | [
"T1 Spaces",
"Hausdorff Spaces"
] | [
"Definition:T2 Space",
"Definition:T1 Space"
] | [
"Definition:T2 Space",
"Definition:Disjoint Sets",
"Definition:Arbitrary",
"Definition:T1 Space"
] |
proofwiki-3364 | T1 Space is T0 | Let $\struct {S, \tau}$ be a $T_1$ space.
Then $\struct {S, \tau}$ is also a $T_0$ space. | {{Recall|T0 Space|$T_0$ space|index = 1}}
{{:Definition:T0 Space/Definition 1}}
Let $\struct {S, \tau}$ be a $T_1$ space.
Let $x, y \in S: x \ne y$.
{{Recall|T1 Space|$T_1$ space|index = 1}}
{{:Definition:T1 Space/Definition 1}}
From the Rule of Addition:
:'''Either'''
::$\exists U \in \tau: x \in U, y \notin U$
:'''or... | Let $\struct {S, \tau}$ be a [[Definition:T1 Space|$T_1$ space]].
Then $\struct {S, \tau}$ is also a [[Definition:T0 Space|$T_0$ space]]. | {{Recall|T0 Space|$T_0$ space|index = 1}}
{{:Definition:T0 Space/Definition 1}}
Let $\struct {S, \tau}$ be a [[Definition:T1 Space |$T_1$ space]].
Let $x, y \in S: x \ne y$.
{{Recall|T1 Space|$T_1$ space|index = 1}}
{{:Definition:T1 Space/Definition 1}}
From the [[Rule of Addition]]:
:'''Either'''
::$\exists U \... | T1 Space is T0 | https://proofwiki.org/wiki/T1_Space_is_T0 | https://proofwiki.org/wiki/T1_Space_is_T0 | [
"T0 Spaces",
"T1 Spaces"
] | [
"Definition:T1 Space",
"Definition:T0 Space"
] | [
"Definition:T1 Space ",
"Rule of Addition",
"Definition:T0 Space"
] |
proofwiki-3365 | T5 Space is T4 | Let $\struct {S, \tau}$ be a $T_5$ space.
Then $\struct {S, \tau}$ is also a $T_4$ space. | Let $\struct {S, \tau}$ be a $T_5$ space.
{{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
Let $C, D \subseteq S$ be disjoint sets which are closed in $T$.
Thus $C, D \in \map \complement \tau$ from the definition of closed set.
From Topological Closure is Closed:
:$C^- = C, D^- = D$
and so from $C... | Let $\struct {S, \tau}$ be a [[Definition:T5 Space|$T_5$ space]].
Then $\struct {S, \tau}$ is also a [[Definition:T4 Space|$T_4$ space]]. | Let $\struct {S, \tau}$ be a [[Definition:T5 Space|$T_5$ space]].
{{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
Let $C, D \subseteq S$ be [[Definition:Disjoint Sets|disjoint sets]] which are [[Definition:Closed Set (Topology)|closed]] in $T$.
Thus $C, D \in \map \complement \tau$ from the de... | T5 Space is T4 | https://proofwiki.org/wiki/T5_Space_is_T4 | https://proofwiki.org/wiki/T5_Space_is_T4 | [
"T4 Spaces",
"T5 Spaces"
] | [
"Definition:T5 Space",
"Definition:T4 Space"
] | [
"Definition:T5 Space",
"Definition:Disjoint Sets",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Topological Closure is Closed",
"Definition:T5 Space",
"Definition:T4 Space"
] |
proofwiki-3366 | Completely Normal Space is Normal | Let $\struct {S, \tau}$ be a completely normal space.
Then $\struct {S, \tau}$ is also a normal space. | Let $\struct {S, \tau}$ be a completely normal space.
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
We have that a $T_5$ space is $T_4$.
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
Hence the result.
... | Let $\struct {S, \tau}$ be a [[Definition:Completely Normal Space|completely normal space]].
Then $\struct {S, \tau}$ is also a [[Definition:Normal Space|normal space]]. | Let $\struct {S, \tau}$ be a [[Definition:Completely Normal Space|completely normal space]].
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
We have that a [[T5 Space is T4|$T_5$ space is $T_4$]].
{{Recall|Normal Space|normal space|index = 1... | Completely Normal Space is Normal | https://proofwiki.org/wiki/Completely_Normal_Space_is_Normal | https://proofwiki.org/wiki/Completely_Normal_Space_is_Normal | [
"Completely Normal Spaces",
"Normal Spaces"
] | [
"Definition:Completely Normal Space",
"Definition:Normal Space"
] | [
"Definition:Completely Normal Space",
"T5 Space is T4"
] |
proofwiki-3367 | Space which is T3 and T0 is also T2 | Let $\struct {S, \tau}$ be a topological space which is both a $T_3$ space and a $T_0$ space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space. | {{Recall|T2 Space|$T_2$ Space|index = 1}}
{{:Definition:T2 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be both a $T_3$ space and a $T_0$ space.
{{Recall|T0 Space|$T_0$ Space|index = 1}}
{{:Definition:T0 Space/Definition 1}}
Let $x, y \in S$.
{{WLOG}}, suppose that $\exists V \in \tau: y \in V, x \notin V$.
Then by... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is both a [[Definition:T3 Space|$T_3$ space]] and a [[Definition:T0 Space|$T_0$ space]].
Then $\struct {S, \tau}$ is also a [[Definition:T2 Space|$T_2$ (Hausdorff) space]]. | {{Recall|T2 Space|$T_2$ Space|index = 1}}
{{:Definition:T2 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be both a [[Definition:T3 Space|$T_3$ space]] and a [[Definition:T0 Space|$T_0$ space]].
{{Recall|T0 Space|$T_0$ Space|index = 1}}
{{:Definition:T0 Space/Definition 1}}
Let $x, y \in S$.
{{WLOG}}, suppose th... | Space which is T3 and T0 is also T2 | https://proofwiki.org/wiki/Space_which_is_T3_and_T0_is_also_T2 | https://proofwiki.org/wiki/Space_which_is_T3_and_T0_is_also_T2 | [
"T3 Spaces",
"T0 Spaces",
"Hausdorff Spaces"
] | [
"Definition:Topological Space",
"Definition:T3 Space",
"Definition:T0 Space",
"Definition:T2 Space"
] | [
"Definition:T3 Space",
"Definition:T0 Space",
"Definition:Relative Complement",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:T2 Space"
] |
proofwiki-3368 | Normal Space is T3 | Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a $T_3$ space. | {{Recall|T3 Space|$T_3$ space|index = 1}}
{{:Definition:T3 Space/Definition 1}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
{{Recall|T1 Space|$T_1$ space|index = 3}}
{{:Definition:T1 Space/Definition 3}}
Let $F$ be an arbitrary closed set in $T$.
Let $y \in \relcomp S F$, th... | Let $\struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
Then $\struct {S, \tau}$ is also a [[Definition:T3 Space|$T_3$ space]]. | {{Recall|T3 Space|$T_3$ space|index = 1}}
{{:Definition:T3 Space/Definition 1}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
{{Recall|T1 Space|$T_1$ space|index = 3}}
{{:Definition:T1 Space/Definition 3}}
Let $F$ be an [[Definition:Arbitrary|arbitrary]] [[Definition:Closed... | Normal Space is T3 | https://proofwiki.org/wiki/Normal_Space_is_T3 | https://proofwiki.org/wiki/Normal_Space_is_T3 | [
"T3 Spaces",
"Normal Spaces"
] | [
"Definition:Normal Space",
"Definition:T3 Space"
] | [
"Definition:Arbitrary",
"Definition:Closed Set/Topology",
"Definition:T1 Space",
"Definition:Closed Set/Topology",
"Definition:Disjoint Sets",
"Definition:Closed Set/Topology",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Definition:Arbitrary",
"Definition:T3 Space"
] |
proofwiki-3369 | T2.5 Space is T2 | Let $\struct {S, \tau}$ be a $T_{2 \frac 1 2}$ space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space. | {{Recall|T2 Space|$T_2$ space|index = 1}}
{{:Definition:T2 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a $T_{2 \frac 1 2}$ space.
{{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space|index = 1}}
{{:Definition:T2.5 Space/Definition 1}}
We have that Set is Subset of its Topological Closure and so $U \subseteq U^-$ and $V ... | Let $\struct {S, \tau}$ be a [[Definition:T2.5 Space|$T_{2 \frac 1 2}$ space]].
Then $\struct {S, \tau}$ is also a [[Definition:T2 Space|$T_2$ (Hausdorff) space]]. | {{Recall|T2 Space|$T_2$ space|index = 1}}
{{:Definition:T2 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a [[Definition:T2.5 Space|$T_{2 \frac 1 2}$ space]].
{{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space|index = 1}}
{{:Definition:T2.5 Space/Definition 1}}
We have that [[Set is Subset of its Topological Closure]]... | T2.5 Space is T2 | https://proofwiki.org/wiki/T2.5_Space_is_T2 | https://proofwiki.org/wiki/T2.5_Space_is_T2 | [
"T2.5 Spaces",
"Hausdorff Spaces"
] | [
"Definition:T2.5 Space",
"Definition:T2 Space"
] | [
"Definition:T2.5 Space",
"Set is Subset of its Topological Closure",
"Definition:T2 Space"
] |
proofwiki-3370 | Regular Space is T2.5 | Let $\struct {S, \tau}$ be a regular space.
Then $\struct {S, \tau}$ is also a $T_{2 \frac 1 2}$ space. | {{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space|index = 2}}
{{:Definition:T2.5 Space/Definition 2}}
Let $T = \struct {S, \tau}$ be a regular space.
{{Recall|Regular Space|index = 3}}
{{:Definition:Regular Space/Definition 3}}
Let $x, y \in S$ with $x \ne y$.
{{Recall|T2 Space|$T_2$ space|index = 1}}
{{:Definition:T2 Space/... | Let $\struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
Then $\struct {S, \tau}$ is also a [[Definition:T2.5 Space|$T_{2 \frac 1 2}$ space]]. | {{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space|index = 2}}
{{:Definition:T2.5 Space/Definition 2}}
Let $T = \struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
{{Recall|Regular Space|index = 3}}
{{:Definition:Regular Space/Definition 3}}
Let $x, y \in S$ with $x \ne y$.
{{Recall|T2 Space|$T_2$ space|in... | Regular Space is T2.5 | https://proofwiki.org/wiki/Regular_Space_is_T2.5 | https://proofwiki.org/wiki/Regular_Space_is_T2.5 | [
"Regular Spaces",
"T2.5 Spaces"
] | [
"Definition:Regular Space",
"Definition:T2.5 Space"
] | [
"Definition:Regular Space",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Subsets of Disjoint Sets are Disjoint",
"Definition:Closed Neighborhood"
] |
proofwiki-3371 | Equivalence of Definitions of T0 Space | {{TFAE|def = T0 Space|view = $T_0$ (Kolmogorov) space}}
Let $T = \struct {S, \tau}$ be a topological space. | === Definition by Open Sets implies Definition by Limit Points ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall x, y \in S$, either:
::$(1): \quad \exists U \in \tau: x \in U, y \notin U$
::$(2): \quad \exists U \in \tau: y \in U, x \notin U$
Let $x, y \in S$ such that $x \ne y$.
By the defin... | {{TFAE|def = T0 Space|view = $T_0$ (Kolmogorov) space}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]]. | === Definition by Open Sets implies Definition by Limit Points ===
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
:$\forall x, y \in S$, either:
::$(1): \quad \exists U \in \tau: x \in U, y \notin U$
::$(2): \quad \exists U \in \tau: y \in U, x \notin U$
Let $x, y \in... | Equivalence of Definitions of T0 Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_T0_Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_T0_Space | [
"T0 Spaces"
] | [
"Definition:Topological Space"
] | [
"Definition:Topological Space",
"Definition:Limit Point/Topology/Point",
"Definition:Limit Point/Topology/Point",
"Definition:Open Set/Topology",
"Definition:Limit Point/Topology/Point",
"Definition:Open Set/Topology",
"Definition:Topological Space",
"Definition:Limit Point/Topology/Point",
"Definit... |
proofwiki-3372 | Equivalence of Definitions of T1 Space | {{TFAE|def = T1 Space|view = $T_1$ space}}
Let $T = \struct {S, \tau}$ be a topological space. | === Definition by Open Sets implies Definition by Closed Points ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall x \in S: \forall y \in S, x \ne y: \exists U \in \tau: x \in U, y \notin U$
Let $x, y \in S$.
By the definition of limit point of a point, the above condition means:
:$x$ is a limi... | {{TFAE|def = T1 Space|view = $T_1$ space}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]]. | === Definition by Open Sets implies Definition by Closed Points ===
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
:$\forall x \in S: \forall y \in S, x \ne y: \exists U \in \tau: x \in U, y \notin U$
Let $x, y \in S$.
By the definition of [[Definition:Limit Point of... | Equivalence of Definitions of T1 Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_T1_Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_T1_Space | [
"T1 Spaces"
] | [
"Definition:Topological Space"
] | [
"Definition:Topological Space",
"Definition:Limit Point/Topology/Point",
"Definition:Limit Point/Topology/Point",
"Definition:Open Set/Topology",
"Definition:Limit Point/Topology/Point",
"Definition:Limit Point/Topology/Point",
"Definition:Vacuous Truth",
"Definition:Limit Point/Topology/Point",
"Eq... |
proofwiki-3373 | T3.5 Space is T3 | Let $T$ be a $T_{3 \frac 1 2}$ space.
Then $T$ is also a $T_3$ space. | Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space.
{{Recall|T3.5 Space|$T_{3 \frac 1 2}$ space}}
{{:Definition:T3.5 Space/Definition 1}}
{{Recall|Urysohn Function|Urysohn function}}
{{:Definition:Urysohn Function}}
Let $F \subseteq S$ be a closed set in $T$.
Let $y \in \relcomp S F$.
Let:
{{begin-eqn}}
{{eqn | l... | Let $T$ be a [[Definition:T3.5 Space|$T_{3 \frac 1 2}$ space]].
Then $T$ is also a [[Definition:T3 Space|$T_3$ space]]. | Let $T = \struct {S, \tau}$ be a [[Definition:T3.5 Space|$T_{3 \frac 1 2}$ space]].
{{Recall|T3.5 Space|$T_{3 \frac 1 2}$ space}}
{{:Definition:T3.5 Space/Definition 1}}
{{Recall|Urysohn Function|Urysohn function}}
{{:Definition:Urysohn Function}}
Let $F \subseteq S$ be a [[Definition:Closed Set (Topology)|closed se... | T3.5 Space is T3/Proof 1 | https://proofwiki.org/wiki/T3.5_Space_is_T3 | https://proofwiki.org/wiki/T3.5_Space_is_T3/Proof_1 | [
"T3.5 Space is T3",
"T3.5 Spaces",
"T3 Spaces"
] | [
"Definition:T3.5 Space",
"Definition:T3 Space"
] | [
"Definition:T3.5 Space",
"Definition:Closed Set/Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Open Set/Topology",
"Definition:Contradiction",
"Definition:Many-to-One Relation",
"Definition:Closed Set/Topology",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Defin... |
proofwiki-3374 | Completely Regular Space is Regular | Let $\struct {S, \tau}$ be a completely regular space.
Then $\struct {S, \tau}$ is also a regular space. | Let $T = \struct {S, \tau}$ be a completely regular space.
{{Recall|Completely Regular Space|completely regular space|index = 1}}
{{:Definition:Completely Regular Space/Definition 1}}
We have:
:$T_{3 \frac 1 2}$ Space is $T_3$.
{{Recall|Regular Space|regular space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
... | Let $\struct {S, \tau}$ be a [[Definition:Completely Regular Space|completely regular space]].
Then $\struct {S, \tau}$ is also a [[Definition:Regular Space|regular space]]. | Let $T = \struct {S, \tau}$ be a [[Definition:Completely Regular Space|completely regular space]].
{{Recall|Completely Regular Space|completely regular space|index = 1}}
{{:Definition:Completely Regular Space/Definition 1}}
We have:
:[[T3.5 Space is T3|$T_{3 \frac 1 2}$ Space is $T_3$]].
{{Recall|Regular Space|regu... | Completely Regular Space is Regular | https://proofwiki.org/wiki/Completely_Regular_Space_is_Regular | https://proofwiki.org/wiki/Completely_Regular_Space_is_Regular | [
"Completely Regular Spaces",
"Regular Spaces"
] | [
"Definition:Completely Regular Space",
"Definition:Regular Space"
] | [
"Definition:Completely Regular Space",
"T3.5 Space is T3"
] |
proofwiki-3375 | Normal Space is Completely Regular | Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a completely regular space. | {{Recall|Completely Regular Space|completely regular space|index = 2}}
{{:Definition:Completely Regular Space/Definition 2}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
From Normal Space is $T_{3 \frac 1 2}$, we have that $T$ is a $T_{3 \frac 1 2}$ space.
So, by definition, ... | Let $\struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
Then $\struct {S, \tau}$ is also a [[Definition:Completely Regular Space|completely regular space]]. | {{Recall|Completely Regular Space|completely regular space|index = 2}}
{{:Definition:Completely Regular Space/Definition 2}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
From [[Normal Space is T3.5|Normal Space is $T_{3 \frac 1 2}$]], we have that $T$ is a [[Definition:T3.5... | Normal Space is Completely Regular | https://proofwiki.org/wiki/Normal_Space_is_Completely_Regular | https://proofwiki.org/wiki/Normal_Space_is_Completely_Regular | [
"Normal Spaces",
"Completely Regular Spaces"
] | [
"Definition:Normal Space",
"Definition:Completely Regular Space"
] | [
"Normal Space is T3.5",
"Definition:T3.5 Space",
"Definition:Completely Regular Space"
] |
proofwiki-3376 | T4 and T3 Space is T3.5 | Let a topological space $T = \struct {S, \tau}$ be:
:a $T_4$ space
and also:
:a $T_3$ space.
Then $T$ is also a $T_{3 \frac 1 2}$ space. | Let $T = \struct {S, \tau}$ be a $T_4$ space which is also a $T_3$ space.
{{Recall|T3 Space|$T_3$ Space|index = 1}}
{{:Definition:T3 Space/Definition 1}}
Consider this $U \in \tau$, which is disjoint from $\set y$.
Then $\relcomp S U$ is a closed set which is disjoint from $F$ but such that $\set y \subseteq \relcomp S... | Let a [[Definition:Topological Space|topological space]] $T = \struct {S, \tau}$ be:
:a [[Definition:T4 Space|$T_4$ space]]
and also:
:a [[Definition:T3 Space|$T_3$ space]].
Then $T$ is also a [[Definition:T3.5 Space|$T_{3 \frac 1 2}$ space]]. | Let $T = \struct {S, \tau}$ be a [[Definition:T4 Space|$T_4$ space]] which is also a [[Definition:T3 Space|$T_3$ space]].
{{Recall|T3 Space|$T_3$ Space|index = 1}}
{{:Definition:T3 Space/Definition 1}}
Consider this $U \in \tau$, which is [[Definition:Disjoint Sets|disjoint]] from $\set y$.
Then $\relcomp S U$ is a... | T4 and T3 Space is T3.5 | https://proofwiki.org/wiki/T4_and_T3_Space_is_T3.5 | https://proofwiki.org/wiki/T4_and_T3_Space_is_T3.5 | [
"T4 Spaces",
"T3 Spaces",
"T3.5 Spaces"
] | [
"Definition:Topological Space",
"Definition:T4 Space",
"Definition:T3 Space",
"Definition:T3.5 Space"
] | [
"Definition:T4 Space",
"Definition:T3 Space",
"Definition:Disjoint Sets",
"Definition:Closed Set/Topology",
"Definition:Disjoint Sets",
"Definition:T4 Space",
"Urysohn's Lemma",
"Definition:Urysohn Function",
"Definition:Urysohn Function",
"Definition:Closed Set/Topology",
"Definition:Element",
... |
proofwiki-3377 | T0 Property is Preserved under Homeomorphism | If $T_A$ is a $T_0$ space, then so is $T_B$. | By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_0$ Property is Preserved under Closed Bijection.
{{qed}} | If $T_A$ is a [[Definition:T0 Space|$T_0$ space]], then so is $T_B$. | By definition of [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
The result follows from [[T0 Property is Preserved under Closed Bijection|$T_0$ Prope... | T0 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T0_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T0_Property_is_Preserved_under_Homeomorphism | [
"T0 Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T0 Space"
] | [
"Definition:Homeomorphism/Topological Spaces",
"Definition:Closed Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Bijection",
"T0 Property is Preserved under Closed Bijection"
] |
proofwiki-3378 | T1 Property is Preserved under Homeomorphism | If $T_A$ is a $T_1$ space, then so is $T_B$. | By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_1$ Property is Preserved under Closed Bijection.
{{qed}} | If $T_A$ is a [[Definition:T1 Space|$T_1$ space]], then so is $T_B$. | By definition of [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
The result follows from [[T1 Property is Preserved under Closed Bijection|$T_1$ Prop... | T1 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T1_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T1_Property_is_Preserved_under_Homeomorphism | [
"T1 Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T1 Space"
] | [
"Definition:Homeomorphism/Topological Spaces",
"Definition:Closed Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Bijection",
"T1 Property is Preserved under Closed Bijection"
] |
proofwiki-3379 | T2 Property is Preserved under Homeomorphism | If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$. | By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_2$ Property is Preserved under Closed Bijection.
{{qed}} | If $T_A$ is a [[Definition:T2 Space|$T_2$ (Hausdorff) space]], then so is $T_B$. | By definition of [[Definition:Homeomorphism/Topological Spaces/Definition 4|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
The result follows from [[T2 Property is Preserved under Closed Bijection|... | T2 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T2_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T2_Property_is_Preserved_under_Homeomorphism | [
"Hausdorff Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T2 Space"
] | [
"Definition:Homeomorphism/Topological Spaces/Definition 4",
"Definition:Closed Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Bijection",
"T2 Property is Preserved under Closed Bijection"
] |
proofwiki-3380 | T2.5 Property is Preserved under Homeomorphism | If $T_A$ is a $T_{2 \frac 1 2}$ space, then so is $T_B$. | By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_{2 \frac 1 2}$ Property is Preserved under Closed Bijection.
{{qed}} | If $T_A$ is a [[Definition:T2.5 Space|$T_{2 \frac 1 2}$ space]], then so is $T_B$. | By definition of [[Definition:Homeomorphism/Topological Spaces/Definition 4|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
The result follows from [[T2.5 Property is Preserved under Closed Bijectio... | T2.5 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T2.5_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T2.5_Property_is_Preserved_under_Homeomorphism | [
"T2.5 Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T2.5 Space"
] | [
"Definition:Homeomorphism/Topological Spaces/Definition 4",
"Definition:Closed Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Bijection",
"T2.5 Property is Preserved under Closed Bijection"
] |
proofwiki-3381 | Regular Property is Preserved under Homeomorphism | If $T_A$ is a regular space, then so is $T_B$. | {{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
From $T_3$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T_3$ space, then so is $T_B$.
From $T_0$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T_0$ space, then so is $T_B$.
Hence the result.
{{qed}} | If $T_A$ is a [[Definition:Regular Space|regular space]], then so is $T_B$. | {{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
From [[T3 Property is Preserved under Homeomorphism|$T_3$ Property is Preserved under Homeomorphism]]:
:If $T_A$ is a [[Definition:T3 Space|$T_3$ space]], then so is $T_B$.
From [[T0 Property is Preserved under Homeomorphism|$T_0$ Property ... | Regular Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/Regular_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/Regular_Property_is_Preserved_under_Homeomorphism | [
"Regular Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:Regular Space"
] | [
"T3 Property is Preserved under Homeomorphism",
"Definition:T3 Space",
"T0 Property is Preserved under Homeomorphism",
"Definition:T0 Space"
] |
proofwiki-3382 | T3 Property is Preserved under Homeomorphism | If $T_A$ is a $T_3$ space, then so is $T_B$. | Suppose that $T_A$ is a $T_3$ space.
Let $F$ be closed in $T_B$.
Let $y \in S_B$ such that $y \notin F$.
From Preimage of Intersection under Mapping it follows that $\phi^{-1} \sqbrk F$ and $\map {\phi^{-1}} y$ are disjoint.
Also, as $\phi$ is a homeomorphism, it is a fortiori continuous.
Thus by Continuity Defined fro... | If $T_A$ is a [[Definition:T3 Space|$T_3$ space]], then so is $T_B$. | Suppose that $T_A$ is a [[Definition:T3 Space|$T_3$ space]].
Let $F$ be [[Definition:Closed Set (Topology)|closed]] in $T_B$.
Let $y \in S_B$ such that $y \notin F$.
From [[Preimage of Intersection under Mapping]] it follows that $\phi^{-1} \sqbrk F$ and $\map {\phi^{-1}} y$ are [[Definition:Disjoint Sets|disjoint... | T3 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T3_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T3_Property_is_Preserved_under_Homeomorphism | [
"T3 Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T3 Space"
] | [
"Definition:T3 Space",
"Definition:Closed Set/Topology",
"Preimage of Intersection under Mapping",
"Definition:Disjoint Sets",
"Definition:Homeomorphism/Topological Spaces",
"Definition:A Fortiori",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Continuity Defined from Closed Sets",
"Defini... |
proofwiki-3383 | T3.5 Property is Preserved under Homeomorphism | If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$. | Let $F \subseteq S_B$ be closed, and let $y \in S_B$ such that $y \notin F$.
Let $G = \phi^{-1} \sqbrk F$ and let $z = \map {\phi^{-1} } y$.
Since $T_A$ is a $T_{3 \frac 1 2}$ space, there exists an Urysohn function $f: X_A \to \closedint 0 1$ for $G$ and $\set z$.
Define $g: S_B \to \closedint 0 1$ by:
:$\map g x = \m... | If $T_A$ is a [[Definition:T3.5 Space|$T_{3 \frac 1 2}$ space]], then so is $T_B$. | Let $F \subseteq S_B$ be [[Definition:Closed Set (Topology)|closed]], and let $y \in S_B$ such that $y \notin F$.
Let $G = \phi^{-1} \sqbrk F$ and let $z = \map {\phi^{-1} } y$.
Since $T_A$ is a [[Definition:T3.5 Space|$T_{3 \frac 1 2}$ space]], there exists an [[Definition:Urysohn Function|Urysohn function]] $f: X_... | T3.5 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T3.5_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T3.5_Property_is_Preserved_under_Homeomorphism | [
"T3.5 Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T3.5 Space"
] | [
"Definition:Closed Set/Topology",
"Definition:T3.5 Space",
"Definition:Urysohn Function",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mapping",
"Definition:Urysohn Function",
"Definition:T3.5 Space"
] |
proofwiki-3384 | Completely Regular Property is Preserved under Homeomorphism | If $T_A$ is a completely regular space, then so is $T_B$. | {{Recall|Completely Regular Space|completely regular space|index = 1}}
{{:Definition:Completely Regular Space/Definition 1}}
From $T_{3 \frac 1 2}$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$.
From $T_0$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T... | If $T_A$ is a [[Definition:Completely Regular Space|completely regular space]], then so is $T_B$. | {{Recall|Completely Regular Space|completely regular space|index = 1}}
{{:Definition:Completely Regular Space/Definition 1}}
From [[T3.5 Property is Preserved under Homeomorphism|$T_{3 \frac 1 2}$ Property is Preserved under Homeomorphism]]:
:If $T_A$ is a [[Definition:T3.5 Space|$T_{3 \frac 1 2}$ space]], then so is ... | Completely Regular Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/Completely_Regular_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/Completely_Regular_Property_is_Preserved_under_Homeomorphism | [
"Completely Regular Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:Completely Regular Space"
] | [
"T3.5 Property is Preserved under Homeomorphism",
"Definition:T3.5 Space",
"T0 Property is Preserved under Homeomorphism",
"Definition:T0 Space"
] |
proofwiki-3385 | Normal Property is Preserved under Homeomorphism | If $T_A$ is a normal space, then so is $T_B$. | {{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
From $T_4$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T_4$ space, then so is $T_B$.
From $T_1$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T_1$ space, then so is $T_B$.
Hence the result.
{{qed}} | If $T_A$ is a [[Definition:Normal Space|normal space]], then so is $T_B$. | {{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
From [[T4 Property is Preserved under Homeomorphism|$T_4$ Property is Preserved under Homeomorphism]]:
:If $T_A$ is a [[Definition:T4 Space|$T_4$ space]], then so is $T_B$.
From [[T1 Property is Preserved under Homeomorphism|$T_1... | Normal Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/Normal_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/Normal_Property_is_Preserved_under_Homeomorphism | [
"Normal Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:Normal Space"
] | [
"T4 Property is Preserved under Homeomorphism",
"Definition:T4 Space",
"T1 Property is Preserved under Homeomorphism",
"Definition:T1 Space"
] |
proofwiki-3386 | T4 Property is Preserved under Homeomorphism | If $T_A$ is a $T_4$ space, then so is $T_B$. | {{Recall|T4 Space|$T_4$ space|index = 1}}
{{:Definition:T4 Space/Definition 1}}
Suppose that $T_A$ is a $T_4$ space.
Let $B_1$ and $B_2$ be closed in $T_B$ that are disjoint.
Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 := \phi^{-1} \sqbrk {B_2}$ the preimages of $B_1$ and $B_2$ under $\phi$, respectively.
From P... | If $T_A$ is a [[Definition:T4 Space|$T_4$ space]], then so is $T_B$. | {{Recall|T4 Space|$T_4$ space|index = 1}}
{{:Definition:T4 Space/Definition 1}}
Suppose that $T_A$ is a [[Definition:T4 Space|$T_4$ space]].
Let $B_1$ and $B_2$ be [[Definition:Closed Set (Topology)|closed]] in $T_B$ that are [[Definition:Disjoint Sets|disjoint]].
Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 ... | T4 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T4_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T4_Property_is_Preserved_under_Homeomorphism | [
"T4 Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T4 Space"
] | [
"Definition:T4 Space",
"Definition:Closed Set/Topology",
"Definition:Disjoint Sets",
"Definition:Preimage/Mapping/Subset",
"Preimage of Intersection under Mapping",
"Definition:Disjoint Sets",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Cont... |
proofwiki-3387 | Completely Normal Property is Preserved under Homeomorphism | If $T_A$ is a completely normal space, then so is $T_B$. | {{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
From $T_5$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T_5$ space, then so is $T_B$.
From $T_1$ Property is Preserved under Homeomorphism:
:If $T_A$ is a $T_1$ space, then so is $T_B$... | If $T_A$ is a [[Definition:Completely Normal Space|completely normal space]], then so is $T_B$. | {{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
From [[T5 Property is Preserved under Homeomorphism|$T_5$ Property is Preserved under Homeomorphism]]:
:If $T_A$ is a [[Definition:T5 Space|$T_5$ space]], then so is $T_B$.
From [[T1 Property is P... | Completely Normal Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/Completely_Normal_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/Completely_Normal_Property_is_Preserved_under_Homeomorphism | [
"Completely Normal Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:Completely Normal Space"
] | [
"T5 Property is Preserved under Homeomorphism",
"Definition:T5 Space",
"T1 Property is Preserved under Homeomorphism",
"Definition:T1 Space"
] |
proofwiki-3388 | T5 Property is Preserved under Homeomorphism | If $T_A$ is a $T_5$ space, then so is $T_B$. | {{Recall|T5 Space|$T_5$ space|index = 1}}
{{:Definition:T5 Space/Definition 1}}
Suppose that $T_A$ is a $T_5$ space.
Let $B_1$ and $B_2$ be separated in $T_B$.
Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 := \phi^{-1} \sqbrk {B_2}$ the preimages of $B_1$ and $B_2$ under $\phi$, respectively.
From Preimage of Inte... | If $T_A$ is a [[Definition:T5 Space|$T_5$ space]], then so is $T_B$. | {{Recall|T5 Space|$T_5$ space|index = 1}}
{{:Definition:T5 Space/Definition 1}}
Suppose that $T_A$ is a [[Definition:T5 Space|$T_5$ space]].
Let $B_1$ and $B_2$ be [[Definition:Separated Sets|separated]] in $T_B$.
Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 := \phi^{-1} \sqbrk {B_2}$ the [[Definition:Preimag... | T5 Property is Preserved under Homeomorphism | https://proofwiki.org/wiki/T5_Property_is_Preserved_under_Homeomorphism | https://proofwiki.org/wiki/T5_Property_is_Preserved_under_Homeomorphism | [
"T5 Spaces",
"Separation Axioms Preserved under Homeomorphism"
] | [
"Definition:T5 Space"
] | [
"Definition:T5 Space",
"Definition:Separated Sets",
"Definition:Preimage/Mapping/Subset",
"Preimage of Intersection under Mapping",
"Definition:Separated Sets",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Continuity Defined from Closed Sets",
... |
proofwiki-3389 | Inverse of Homeomorphism is Homeomorphism | Let $T, T'$ be topological spaces.
Let $f: T \to T'$ be a homeomorphism.
Then $f^{-1}: T' \to T$ is also a homeomorphism. | By definition, a homeomorphism is a bijection such that both $f$ and $f^{-1}$ are continuous.
As $f$ is a bijection then by Bijection iff Inverse is Bijection, so is $f^{-1}$.
So by definition $f^{-1}$ is a bijection such that both $f^{-1}$ and $\left({f^{-1}}\right)^{-1}$ are continuous.
The result follows from Invers... | Let $T, T'$ be [[Definition:Topological Space|topological spaces]].
Let $f: T \to T'$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
Then $f^{-1}: T' \to T$ is also a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]]. | By definition, a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] is a [[Definition:Bijection|bijection]] such that both $f$ and $f^{-1}$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
As $f$ is a [[Definition:Bijection|bijection]] then by [[Bijection iff Inverse is Bijection]],... | Inverse of Homeomorphism is Homeomorphism | https://proofwiki.org/wiki/Inverse_of_Homeomorphism_is_Homeomorphism | https://proofwiki.org/wiki/Inverse_of_Homeomorphism_is_Homeomorphism | [
"Homeomorphisms (Topological Spaces)"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Homeomorphism/Topological Spaces",
"Definition:Bijection",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Inverse of Inverse of Bijection",
... |
proofwiki-3390 | T0 Property is Preserved under Closed Bijection | If $T_A$ is a $T_0$ space, then so is $T_B$. | Let $T_A$ be a $T_0$ space.
By Bijection is Open iff Closed, $\phi$ is an open bijection.
By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous.
By definition:
:$\forall x, y \in S_A$, either:
::$\exists U \in \tau_A: x \in U, y \notin U$
:or:
::$\exists U \in \tau_A: y \in U, x \no... | If $T_A$ is a [[Definition:T0 Space|$T_0$ space]], then so is $T_B$. | Let $T_A$ be a [[Definition:T0 Space|$T_0$ space]].
By [[Bijection is Open iff Closed]], $\phi$ is an [[Definition:Open Mapping|open]] [[Definition:Bijection|bijection]].
By [[Bijection is Open iff Inverse is Continuous]], it follows that $\phi^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous... | T0 Property is Preserved under Closed Bijection | https://proofwiki.org/wiki/T0_Property_is_Preserved_under_Closed_Bijection | https://proofwiki.org/wiki/T0_Property_is_Preserved_under_Closed_Bijection | [
"T0 Spaces",
"Separation Axioms Preserved under Closed Bijection"
] | [
"Definition:T0 Space"
] | [
"Definition:T0 Space",
"Bijection is Open iff Closed",
"Definition:Open Mapping",
"Definition:Bijection",
"Bijection is Open iff Inverse is Continuous",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Preimage of Subset is Subset of Preimage",
"Definition:Continuous Mapping (Topology)/Everywhe... |
proofwiki-3391 | T2.5 Property is Preserved under Closed Bijection | If $T_A$ is a $T_{2 \frac 1 2}$ space, then so is $T_B$. | Let $T_A$ be a $T_{2 \frac 1 2}$ space.
Then:
:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A^- \cap V_A^- = \O$
where $U_A^-$ signifies the closure of $U_A$.
Suppose that $T_B$ is not a $T_{2 \frac 1 2}$ space.
Then:
:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: ... | If $T_A$ is a [[Definition:T2.5 Space|$T_{2 \frac 1 2}$ space]], then so is $T_B$. | Let $T_A$ be a [[Definition:T2.5 Space|$T_{2 \frac 1 2}$ space]].
Then:
:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A^- \cap V_A^- = \O$
where $U_A^-$ signifies the [[Definition:Closure (Topology)|closure]] of $U_A$.
Suppose that $T_B$ is not a [[Definition:T2.5 Space|$T_{2... | T2.5 Property is Preserved under Closed Bijection | https://proofwiki.org/wiki/T2.5_Property_is_Preserved_under_Closed_Bijection | https://proofwiki.org/wiki/T2.5_Property_is_Preserved_under_Closed_Bijection | [
"T2.5 Spaces",
"Separation Axioms Preserved under Closed Bijection"
] | [
"Definition:T2.5 Space"
] | [
"Definition:T2.5 Space",
"Definition:Closure (Topology)",
"Definition:T2.5 Space",
"Definition:Closure (Topology)",
"Definition:Open Set/Topology",
"Definition:Disjoint Sets",
"Topological Closure is Closed",
"Definition:Closed Set/Topology",
"Topology Defined by Closed Sets",
"Definition:Closed S... |
proofwiki-3392 | T1 Property is Preserved under Closed Bijection | If $T_A$ is a $T_1$ space, then so is $T_B$. | Let $T_A$ be a $T_1$ space.
By definition, all points in $T_A$ are closed.
Let $a \in S_A$.
Then $\set a$ is a closed set.
As $\phi$ is a closed mapping it follows directly that $\phi \sqbrk {\set a}$ is closed.
As $\phi$ is a bijection it follows that every point in $S_B$ is the image under $\phi$ of a single point in... | If $T_A$ is a [[Definition:T1 Space|$T_1$ space]], then so is $T_B$. | Let $T_A$ be a [[Definition:T1 Space|$T_1$ space]].
By definition, all points in $T_A$ are [[Definition:Closed Point|closed]].
Let $a \in S_A$.
Then $\set a$ is a [[Definition:Closed Set (Topology)|closed set]].
As $\phi$ is a [[Definition:Closed Mapping|closed mapping]] it follows directly that $\phi \sqbrk {\set... | T1 Property is Preserved under Closed Bijection | https://proofwiki.org/wiki/T1_Property_is_Preserved_under_Closed_Bijection | https://proofwiki.org/wiki/T1_Property_is_Preserved_under_Closed_Bijection | [
"T1 Spaces",
"Separation Axioms Preserved under Closed Bijection"
] | [
"Definition:T1 Space"
] | [
"Definition:T1 Space",
"Definition:Closed Point",
"Definition:Closed Set/Topology",
"Definition:Closed Mapping",
"Definition:Closed Set/Topology",
"Definition:Bijection",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Closed Point",
"Definition:T1 Space"
] |
proofwiki-3393 | T2 Property is Preserved under Closed Bijection | If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$. | Let $T_A$ be a $T_2$ (Hausdorff) space.
Then:
:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \O$
Suppose that $T_B$ is not $T_2$ (Hausdorff).
Then:
:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B \cap V_B \ne \O$
That is... | If $T_A$ is a [[Definition:T2 Space|$T_2$ (Hausdorff) space]], then so is $T_B$. | Let $T_A$ be a [[Definition:T2 Space|$T_2$ (Hausdorff) space]].
Then:
:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \O$
Suppose that $T_B$ is not [[Definition:T2 Space|$T_2$ (Hausdorff)]].
Then:
:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in ... | T2 Property is Preserved under Closed Bijection | https://proofwiki.org/wiki/T2_Property_is_Preserved_under_Closed_Bijection | https://proofwiki.org/wiki/T2_Property_is_Preserved_under_Closed_Bijection | [
"Hausdorff Spaces",
"Separation Axioms Preserved under Closed Bijection"
] | [
"Definition:T2 Space"
] | [
"Definition:T2 Space",
"Definition:T2 Space",
"Definition:Open Set/Topology",
"Definition:Disjoint Sets",
"Definition:Topology",
"Definition:Open Set/Topology",
"Preimage of Intersection under Mapping",
"Bijection is Open iff Closed",
"Definition:Open Mapping",
"Definition:Bijection",
"Bijection... |
proofwiki-3394 | Separation Properties Preserved under Expansion | Let $S$ be a set.
Let $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ be topological spaces such that $\tau_2$ is an expansion of $\tau_1$.
=== $T_0$ Property is Preserved under Expansion ===
{{:T0 Property is Preserved under Expansion}}
=== $T_1$ Property is Preserved under Expansion ===
{{:T1 Property is Preserved un... | Let $S$ be a set.
Let $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ be topological spaces such that $\tau_2$ is an expansion of $\tau_1$.
{{Recall|Expansion of Topology}}
{{:Definition:Expansion of Topology}}
Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the identity mapping from $\struct {S, \tau_1}$ to ... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ be [[Definition:Topological Space|topological spaces]] such that $\tau_2$ is an [[Definition:Expansion of Topology|expansion]] of $\tau_1$.
=== [[T0 Property is Preserved under Expansion|$T_0$ Property is Preserved under Expansi... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ be [[Definition:Topological Space|topological spaces]] such that $\tau_2$ is an [[Definition:Expansion of Topology|expansion]] of $\tau_1$.
{{Recall|Expansion of Topology}}
{{:Definition:Expansion of Topology}}
Let $I_S: \stru... | Separation Properties Preserved under Expansion | https://proofwiki.org/wiki/Separation_Properties_Preserved_under_Expansion | https://proofwiki.org/wiki/Separation_Properties_Preserved_under_Expansion | [
"Separation Properties Preserved under Expansion",
"Expansions of Topologies",
"Separation Axioms"
] | [
"Definition:Set",
"Definition:Topological Space",
"Definition:Expansion of Topology",
"T0 Property is Preserved under Expansion",
"T1 Property is Preserved under Expansion",
"T2 Property is Preserved under Expansion",
"T2.5 Property is Preserved under Expansion"
] | [
"Definition:Set",
"Definition:Topological Space",
"Definition:Expansion of Topology",
"Definition:Identity Mapping",
"Identity Mapping to Expansion is Closed",
"Definition:Closed Mapping",
"Identity Mapping is Bijection",
"T0 Property is Preserved under Closed Bijection",
"T1 Property is Preserved u... |
proofwiki-3395 | Identity Mapping to Expansion is Closed | Let $S$ be a set on which $\tau_1$ and $\tau_2$ are topologies such that:
:$\tau_1 \subseteq \tau_2$
that is, such that $\tau_2$ is an expansion of $\tau_1$.
Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the identity mapping from $\struct {S, \tau_1}$ to $\struct {S, \tau_2}$.
Then $I_S$ is closed. | $\tau_1 \subseteq \tau_2$ means that every open set of $\struct {S, \tau_1}$ is also an open set of $\struct {S, \tau_2}$.
Let $A \subseteq S$ be closed in $\struct {S, \tau_1}$
Then by definition $S \setminus A$ is open in $\struct {S, \tau_1}$.
Then $I_S \sqbrk {S \setminus A} = S \setminus A$ is open in $\struct {S,... | Let $S$ be a set on which $\tau_1$ and $\tau_2$ are [[Definition:Topology|topologies]] such that:
:$\tau_1 \subseteq \tau_2$
that is, such that $\tau_2$ is an [[Definition:Expansion of Topology|expansion]] of $\tau_1$.
Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the [[Definition:Identity Mapping|identity... | $\tau_1 \subseteq \tau_2$ means that every [[Definition:Open Set (Topology)|open set]] of $\struct {S, \tau_1}$ is also an [[Definition:Open Set (Topology)|open set]] of $\struct {S, \tau_2}$.
Let $A \subseteq S$ be [[Definition:Closed Set (Topology)|closed]] in $\struct {S, \tau_1}$
Then by definition $S \setminus A... | Identity Mapping to Expansion is Closed | https://proofwiki.org/wiki/Identity_Mapping_to_Expansion_is_Closed | https://proofwiki.org/wiki/Identity_Mapping_to_Expansion_is_Closed | [
"Identity Mappings",
"Closed Mappings",
"Expansions of Topologies"
] | [
"Definition:Topology",
"Definition:Expansion of Topology",
"Definition:Identity Mapping",
"Definition:Closed Mapping"
] | [
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Mapping"
] |
proofwiki-3396 | T4 Property is Weakly Hereditary | Let $T = \struct {S, \tau}$ be a topological space.
Let $T_K$ be a subspace of $T$ such that $K$ is closed in $T$.
If $T$ is a $T_4$ space then $T_K$ is also a $T_4$ space.
That is, the property of being a $T_4$ space is weakly hereditary. | Let $T = \struct {S, \tau}$ be a $T_4$ space.
{{Recall|T4 Space|$T_4$ space}}
{{:Definition:T4 Space/Definition 1}}
We have that the set $\tau_K$ is defined as:
:$\tau_K := \set {U \cap K: U \in \tau}$
where $K$ is closed in $T$.
Let $A, B \subseteq K$ be closed in $K$ such that:
:$A \cap B = \O$
From Intersection with... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $T_K$ be a [[Definition:Topological Subspace|subspace]] of $T$ such that $K$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
If $T$ is a [[Definition:T4 Space|$T_4$ space]] then $T_K$ is also a [[Definition:T4 Space|$T_4$... | Let $T = \struct {S, \tau}$ be a [[Definition:T4 Space|$T_4$ space]].
{{Recall|T4 Space|$T_4$ space}}
{{:Definition:T4 Space/Definition 1}}
We have that the set $\tau_K$ is defined as:
:$\tau_K := \set {U \cap K: U \in \tau}$
where $K$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
Let $A, B \subseteq K$ be... | T4 Property is Weakly Hereditary | https://proofwiki.org/wiki/T4_Property_is_Weakly_Hereditary | https://proofwiki.org/wiki/T4_Property_is_Weakly_Hereditary | [
"T4 Spaces",
"Separation Properties Preserved in Subspace",
"Examples of Weakly Hereditary Properties"
] | [
"Definition:Topological Space",
"Definition:Topological Subspace",
"Definition:Closed Set/Topology",
"Definition:T4 Space",
"Definition:T4 Space",
"Definition:T4 Space",
"Definition:Weakly Hereditary Property"
] | [
"Definition:T4 Space",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Intersection with Subset is Subset",
"Definition:Closed Set/Topology",
"Topology Defined by Closed Sets",
"Definition:T4 Space",
"Definition:Topological Subspace",
"Definition:Open Set/Topology",
"Definitio... |
proofwiki-3397 | Completely Normal iff Every Subspace is Normal | Let $T = \struct {S, \tau}$ be a topological space.
Then $T$ is a completely normal space {{iff}} every subspace of $T$ is normal. | From the definitions, we have that:
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
{{Recall|T5 Space|$T_5$ space|index = 3}}
{{:Definition:T5 Space/Definiti... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then $T$ is a [[Definition:Completely Normal Space|completely normal space]] {{iff}} every [[Definition:Topological Subspace|subspace]] of $T$ is [[Definition:Normal Space|normal]]. | From the definitions, we have that:
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
{{Recall|T5 Space|$T_5$ space|index = 3}}
{{:Definition:T5 Space/Defin... | Completely Normal iff Every Subspace is Normal | https://proofwiki.org/wiki/Completely_Normal_iff_Every_Subspace_is_Normal | https://proofwiki.org/wiki/Completely_Normal_iff_Every_Subspace_is_Normal | [
"Completely Normal Spaces",
"Normal Spaces",
"Topological Subspaces"
] | [
"Definition:Topological Space",
"Definition:Completely Normal Space",
"Definition:Topological Subspace",
"Definition:Normal Space"
] | [
"T1 Property is Hereditary",
"Definition:Topological Subspace",
"Definition:T1 Space",
"Definition:T1 Space"
] |
proofwiki-3398 | Little Picard Theorem | A non-constant entire function $f: \C \to \C$ omits at most one complex value $a \in \C$. | {{MissingLinks}}
For this proof, we use the fact that there is a holomorphic covering map:
:$\phi: \Bbb D \to \C \setminus \set {0, 1}$
where $\Bbb D = \set {z \in \C: \cmod z < 1}$.
This follows from the Riemann Uniformization Theorem, but is much easier to prove.
Indeed, such a covering is given by the elliptic modul... | A non-[[Definition:Constant Mapping|constant]] [[Definition:Entire Function|entire function]] $f: \C \to \C$ omits at most one [[Definition:Complex Number|complex]] [[Definition:Value|value]] $a \in \C$. | {{MissingLinks}}
For this proof, we use the fact that there is a [[Definition:Holomorphic Covering Map|holomorphic covering map]]:
:$\phi: \Bbb D \to \C \setminus \set {0, 1}$
where $\Bbb D = \set {z \in \C: \cmod z < 1}$.
This follows from the [[Riemann Uniformization Theorem]], but is much easier to prove.
Indeed,... | Little Picard Theorem | https://proofwiki.org/wiki/Little_Picard_Theorem | https://proofwiki.org/wiki/Little_Picard_Theorem | [
"Complex Analysis"
] | [
"Definition:Constant Mapping",
"Definition:Entire Function",
"Definition:Complex Number",
"Definition:Value"
] | [
"Definition:Holomorphic Covering Map",
"Riemann Uniformization Theorem",
"Definition:Elliptic Modular Function",
"Definition:Entire Function",
"Definition:Affine Transformation",
"Definition:Holomorphic Function",
"Liouville's Theorem (Complex Analysis)",
"Definition:Constant Mapping",
"Definition:C... |
proofwiki-3399 | Completely Regular Space is Urysohn | Let $\struct {S, \tau}$ be a completely regular space.
Then $\struct {S, \tau}$ is also an Urysohn space. | Let $T = \struct {S, \tau}$ be a completely regular space.
Let $x, y \in S: x \ne y$.
{{Recall|Completely Regular Space|completely regular space|index = 2}}
{{:Definition:Completely Regular Space/Definition 2}}
Clearly $\set x$ and $\set y$ are disjoint.
From the definition of $T_1$ space, we have that $\set x$ and $\s... | Let $\struct {S, \tau}$ be a [[Definition:Completely Regular Space|completely regular space]].
Then $\struct {S, \tau}$ is also an [[Definition:Urysohn Space|Urysohn space]]. | Let $T = \struct {S, \tau}$ be a [[Definition:Completely Regular Space|completely regular space]].
Let $x, y \in S: x \ne y$.
{{Recall|Completely Regular Space|completely regular space|index = 2}}
{{:Definition:Completely Regular Space/Definition 2}}
Clearly $\set x$ and $\set y$ are [[Definition:Disjoint Sets|disj... | Completely Regular Space is Urysohn | https://proofwiki.org/wiki/Completely_Regular_Space_is_Urysohn | https://proofwiki.org/wiki/Completely_Regular_Space_is_Urysohn | [
"Completely Regular Spaces",
"Urysohn Spaces"
] | [
"Definition:Completely Regular Space",
"Definition:Urysohn Space"
] | [
"Definition:Completely Regular Space",
"Definition:Disjoint Sets",
"Definition:T1 Space",
"Definition:Closed Set/Topology",
"Definition:T3.5 Space",
"Definition:Urysohn Function",
"Definition:Urysohn Space"
] |
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