avewright/proofwiki-math huggingface dataset:
Theorem: Space with Open Point is Non-Meager

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$ be an open point.
Then $T$ is a non-meager space.

Proof:
{{Recall|Meager Space|meager}}
{{:Definition:Meager Space}}
{{Recall|Non-Meager Space|non-meager}}
{{:Definition:Non-Meager Space}}
Let $x \in S$ be an open point of $T$.
That is:
:$\set x \in \tau$
{{AimForCont}} that $T$ is meager.
Let:
:$\ds T = \bigcup \SS$
where $\SS$ is a countable set of subsets of $S$ which are nowhere dense in $S$.
Then:
:$\exists H \in \SS: x \in H$
and so:
:$\set x \subseteq H$
{{Recall|Nowhere Dense|index = 2|nowhere dense}}
{{:Definition:Nowhere Dense/Definition 2}}
We have that $H$ is nowhere dense in $T$.
From Set is Subset of its Topological Closure we have that:
:$H \subseteq H^-$
So by Subset Relation is Transitive:
:$\set x \subseteq H^-$
But $\set x$ is an open set of $T$.
So $H$ is not nowhere dense.
Therefore $T$ cannot be a countable union of subsets of $S$ which are nowhere dense in $S$.
That is, $T$ is not meager.
Hence the result by definition of non-meager.
{{qed}}
Theorem: Trivial Topological Space is Non-Meager

Let $T = \struct {S, \tau}$ be a trivial topological space.
Then $T$ is non-meager.

Proof:
As $T$ is a trivial topological space, by definition $S$ is a singleton: $S = \set s$, say.
Then $\set s$ is an open set.
That is, $s$ is an open point.
The result follows from Space with Open Point is Non-Meager.
{{qed}}
Theorem: Equivalence of Definitions of T2 Space

{{TFAE|def = T2 Space|view = $T_2$ (Hausdorff) space}}
Let $T = \struct {S, \tau}$ be a topological space.

Proof:
=== Definition 1 implies Definition 2 ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
Let us take any arbitrary $x, y \in S: x \ne y$.
Let $\CC_x$ be the set of all closed neighborhoods of $x$:
:$\CC_x = \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$
where $\relcomp S H$ is the complement of $H$ in $S$.
We need to demonstrate that the only element in the intersection of $\CC_x$ is $x$:
:$\bigcap \CC_x = \set x$
and to do that we show that if $y \ne x$ then $y \notin \bigcap \CC_x$.
Let $C = \bigcap C_x$.
Clearly $x \in C$ and so $\set x \subseteq C_x$.
We have that $\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ by hypothesis.
As $x \in U$ it follows that $x \notin V$ and so $x \in \relcomp S V$.
Thus $x \in U \subseteq \relcomp S V$.
That is, $\relcomp S V$ is a closed neighborhood of $x$ and so $\relcomp S V \in \CC_x$.
As $y \in V$ it follows that $y \notin \relcomp S V$.
So $\relcomp S V$ is a closed neighborhood of $x$ which does not contain $y$.
So $y \notin \bigcap C_x$.
As $y$ is arbitrary:
:$\forall y \in S, y \ne x: \exists H: \relcomp S H \in \tau: y \notin H$
and so $C_x \subseteq \set x$.
That is:
:$\ds \forall x \in S: \set x = \bigcap \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$
or, each point is the intersection of all its closed neighborhoods.
{{qed|lemma}}
Theorem: Equivalence of Definitions of T3 Space

{{TFAE|def = T3 Space|view = $T_3$ space}}
Let $T = \struct {S, \tau}$ be a topological space.

Proof:
=== Definition by Open Sets implies Definition by Closed Neighborhoods ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
Let $U \in \tau$, and let $x \in U$.
Then by Relative Complement of Relative Complement:
:$\relcomp S {\relcomp S U} \in \tau$
So by definition of closed:
:$\relcomp S U$ is closed
and:
:$x \notin \relcomp S U$
We have {{hypothesis}}:
:$\exists A, B \in \tau: \relcomp S U \subseteq A, x \in B: A \cap B = \O$ 
It follows that:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = B
      | c = {{hypothesis}}
}}
{{eqn | o = \subseteq
      | r = \relcomp S A
      | c = Empty Intersection iff Subset of Complement: from $A \cap B$
}}
{{eqn | o = \subseteq
      | r = \relcomp S {\relcomp S U}
      | c = Set Complement inverts Subsets: from $\relcomp S U \subseteq A$
}}
{{eqn | r = U
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
That is:
:$x \in \relcomp S A \subseteq U$
So we have demonstrated that there exists a closed neighborhood $\relcomp S A$ of $x$ contained in $U$.
As $U, x$ are arbitrary:
:$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$
{{qed|lemma}}
Theorem: Big Implies Saturated

Let $\MM$ be an $\LL$-structure.
Let $\kappa$ be a cardinal.
If $\MM$ is $\kappa$-big, then it is $\kappa$-saturated.

Proof:
The idea of the proof is:
:to go to some elementary equivalent structure where a type is realized
and:
:to interpret a new relation symbol as the singleton containing some realization of the type.
This lets us write sentences about the extension saying that:
:there must exist an element satisfying the relation
and:
:every element satisfying the relation satisfies all of the formulas in the type.
Then using bigness, these sentences carry over into $\MM$.
Let $p$ be a complete $n$-type over $A$, where $A$ is a subset of the universe of $\MM$ with cardinality strictly less than $\kappa$.
We are to show that $p$ is realized in $\MM$.
Let $\LL_A$ be the language obtained from $\LL$ by adding constant symbols for each element of $A$.
Viewing $\MM$ as an $\LL_A$-structure in the natural way, let $\map {\operatorname {Th}_A} \MM$ be the collection of $\LL_A$-sentences satisfied by $\MM$.
By definition of type, $p \cup \map {\operatorname {Th}_A} \MM$ is satisfiable by some $\LL_A$-structure $\NN$.
Since $\NN$ satisfies $p$, some $n$-ordered tuple $\bar b$ in it realizes $p$.
Now, let $\LL_A^*$ be the language obtained from $\LL_A$ by adding a new relation symbol $R$.
We can extend $\NN$ to be an $\LL_A^*$-structure by interpreting the symbol $R$ as the singleton $R^\NN = \set {\bar b}$.
Since $\NN \models \map {\operatorname {Th}_A} \MM$, we have that $\MM$ and $\NN$ are elementary equivalent as $\LL_A$-structures.
We have that $\MM$ is $\kappa$-big by assumption.
Hence there is some relation $R^\MM$ on the universe of $\MM$ such that:
:$\struct {\MM, R^\MM}$ is elementary equivalent to $\struct {\NN, R^\NN}$ as $\LL_A^*$-structures.
But $\struct {\NN, R^\NN}$ satisfies the $\LL_A^*$-sentences:
:$\exists \bar x \map R {\bar x}$
and
:$\forall \bar x \paren {\map R {\bar x} \to \map \phi {\bar x} }$
for each $\map \phi {\bar x}$ in $p$.
Hence $\struct {\MM, R^\MM}$ also satisfies these sentences, and so there must be some $\bar d \in R^\MM$ realizing $p$.
So $\MM$ realizes $p$.
{{qed}}
Category:Model Theory for Predicate Logic
f2fi7fjzqxw5d74fpyufri34t8r01ga
Theorem: Limit Points in T1 Space

Let $T = \struct {S, \tau}$ be a $T_1$ space.
Let $H \subset S$ be any subset of $S$.
Let $x \in H$.
Then $x$ is a limit point of $H$ {{iff}} every neighborhood of $x$ contains infinitely many points of $H$.

Proof:
=== Necessary Condition ===
Suppose every neighborhood of $x$ contains infinitely many points of $H$.
Let $N$ be an open neighborhood of $x$.
Then $H \cap N$ is infinite, and so is $H \cap \paren {N \setminus \set x}$.
Since $N$ is arbitrary, $x$ is a limit point of $H$ by definition.
{{qed|lemma}}
Theorem: Singleton Point is Isolated

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then $x$ is an isolated point of the singleton set $\set x$, but not necessarily an isolated point of $T$.

Proof:
Let $U \in \tau$ be an open set of $T$ such that $x \in T$.
The fact that such a $U$ exists follows from the fact that:
:from {{Open-set-axiom|3}}, $S$ is open in $T$
:$x \in S$.
Hence:
:$\set x \subseteq S$
and so from Intersection with Subset is Subset:
$\set x \cap U = \set x$
So by definition, $x$ is an isolated point of $\set x$.
From Topological Space is Discrete iff All Points are Isolated, unless $T$ is the discrete space on $S$, not all elements of $T$ are isolated points of $T$.
{{qed}}
Category:Isolated Points
Category:Singletons
d6d92dtdi96y0yog7sd1uniihoyf3cs
Theorem: Singleton Set is not Dense-in-itself

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then the singleton set $\set x$ is not dense-in-itself.

Proof:
{{Recall|Dense-in-itself|dense-in-itself}}
{{:Definition:Dense-in-itself}}
From Singleton Point is Isolated, $x$ is isolated in $\set x$.
So by definition $\set x$ is not dense-in-itself.
{{qed}}
Category:Dense-in-itself
Category:Singletons
70m0ywinp90b6v7zyolicpff9l7xa3b
Theorem: Tietze Extension Theorem

Let $T = \struct {S, \tau}$ be a topological space which is normal.
Let $A \subseteq S$ be a closed set in $T$.
Let $f: A \to \R$ be a continuous mapping from $A \subseteq S$ to the real number line under the usual (Euclidean) topology.
Then there exists a continuous extension $g: S \to \R$, that is, such that:
:$\forall s \in A: \map f s = \map g s$

Proof:
=== Lemma ===
{{:Tietze Extension Theorem/Lemma}}{{qed|lemma}}
First suppose that for any continuous mapping on a closed set there is a continuous extension.
Let $C$ and $D$ be disjoint sets which are closed in $S$.
Define $f: C \cup D \to \R$ by:
:<nowiki>$\map f x = \begin {cases}
0 & : x \in C \\
1 & : x \in D \end {cases}$</nowiki>
Now $f$ is continuous and we can extend it to a continuous mapping $g: X \to \R$.
By Urysohn's Lemma, $S$ is normal because $g$ is a continuous mapping such that $\map g x = 0$ for $x \in C$ and $\map g x = 1$ for $x \in D$.
Conversely, let $S$ be normal and $A$ be closed in $S$.
By the {{Lemma|Tietze Extension Theorem}}, there exists a continuous mapping $g_0: S \to \R$ such that:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \size {\map {g_0} x}
      | o = \le
      | r = \dfrac 1 3
}}
{{eqn | q = \forall x \in A
      | l = \size {\map f x − \map {g_0} x}
      | o = \le
      | r = \dfrac 2 3
}}
{{end-eqn}}
Since $\paren {f − g_0}: A \to \R$ is continuous, the lemma tells us there is a continuous mapping $g_1: X \to \R$ such that:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \size {\map {g_1} x}
      | o = \le
      | r = \dfrac 1 3 \times \dfrac 2 3
}}
{{eqn | q = \forall x \in A
      | l = \size {\map f x − \map {g_0} x - \map {g_1} x}
      | o = \le
      | r = \dfrac 2 3 \times \dfrac 2 3
}}
{{end-eqn}}
By repeated application of the {{Lemma|Tietze Extension Theorem}} we can construct a sequence of continuous mappings $g_0, g_1, g_2, \ldots$ such that:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \size {\map {g_n} x}
      | o = \le
      | r = \dfrac 1 3 \times \paren {\dfrac 2 3}^n
}}
{{eqn | q = \forall x \in A
      | l = \size {\map f x − \map {g_0} x − \map {g_1} x - \map {g_2} x - \cdots - \map {g_n} x}
      | o = \le
      | r = \dfrac 2 3 \times \paren {\dfrac 2 3}^n
}}
{{end-eqn}}
Define:
:$\ds \map g x = \sum_{n \mathop = 0}^\infty \map {g_n} x$
We have that:
:$\size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$
and:
:$\ds \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n$
converges as a geometric series.
Thus $\ds \sum_{n \mathop = 0}^\infty \map {g_n} x$ converges absolutely and uniformly.
So $g$ is a continuous mapping defined everywhere.
Also, $\ds \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n = 1$ implies that $\size {\map g x} \le 1$.
Now for $x \in A$, we have that:
:$\ds \size {\map f x − \sum_{n \mathop = 0}^k \map {g_n} x} \le \paren {\dfrac 2 3}^{k + 1}$
As $k \to \infty$, the {{RHS}} $\to 0$ and so the sum goes to $\map g x$.
Thus:
:$\size {\map f x − \map g x} = 0$
Therefore $g$ extends $f$. 
{{qed}}
{{Namedfor|Heinrich Franz Friedrich Tietze|cat = Tietze}}
Theorem: Hermitian Operator has Orthogonal Eigenvectors

The eigenvectors of a Hermitian operator are orthogonal.

Proof:
This proof uses Dirac notation.
Note that the inner product is linear in the second argument.
Let $V$ be an inner product space over $\C$.
Let $\hat H$ be a Hermitian operator on $V$.
Let $\lambda_i$ and $\lambda_j$ be distinct eigenvalues of $\hat H$.
Let $\ket {x_i}$ and $\ket {x_j}$ be the eigenvectors associated with $\lambda_i$ and $\lambda_j$, respectively.
We have:
{{begin-eqn}}
{{eqn | l = \bra {x_i} \hat H \ket {x_j}
      | r = \braket {x_i} {\hat H x_j}
      | c = {{Defof|Dirac Notation/Inner Product}}
}}
{{eqn | r = \braket {x_i} {\lambda_j x_j}
      | c = 
}}
{{eqn | r = \lambda_j \braket {x_i} {x_j}
      | c = Inner Product is Sesquilinear
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \bra {x_i} \hat H \ket {x_j}
      | r = \braket {\hat H x_i} {x_j}
      | c = {{Defof|Hermitian Operator}} and {{Defof|Dirac Notation/Inner Product}}
}}
{{eqn | r = \braket {\lambda_i x_i} {x_j}
      | c = 
}}
{{eqn | r = \bar \lambda_i \braket {x_i} {x_j}
      | c = Inner Product is Sesquilinear
}}
{{eqn | r = \lambda_i \braket {x_i} {x_j}
      | c = Eigenvalues of Hermitian Operator are Real
}}
{{end-eqn}}
This now gives us the equations:
{{begin-eqn}}
{{eqn | n = 1
      | l = \bra {x_i} \hat H \ket {x_j}
      | r = \lambda_i \braket {x_i} {x_j}
}}
{{eqn | n = 2
      | l = \bra {x_i} \hat H \ket {x_j}
      | r = \lambda_j \braket {x_i} {x_j}
}}
{{end-eqn}}
Subtracting $(2)$ from $(1)$ gives:
:$\paren {\lambda_i - \lambda_j} \braket {x_i} {x_j} = 0$
Since $\lambda_i - \lambda_j \ne 0$, we can divide both sides by it.
Therefore:
:$\braket {x_i} {x_j} = 0$
Therefore the eigenvectors of $\hat H$ are orthogonal.
{{qed}}
Theorem: Hermitian Operator has Orthogonal Eigenvectors

The eigenvectors of a Hermitian operator are orthogonal.

Proof:
Follows from Hermitian Operator is Normal and Eigenvalues of Normal Operator have Orthogonal Eigenspaces.
{{qed}}
Theorem: Hausdorff Space iff Diagonal Set on Product is Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $\Delta_S$ be the diagonal set on $S$:
:$\Delta_S = \set {\tuple {x, x} \in S \times S: x \in S}$
where $S \times S$ is the Cartesian product of $S$ with itself.
Let $T^2 = \struct {S \times S, \TT}$ be the product space with product topology $\TT$.
Then $T$ is a $T_2$ (Hausdorff) space {{iff}} $\Delta_S$ is closed in $T^2$.

Proof:
Suppose $T$ is a $T_2$ (Hausdorff) space.
Let $\Delta: S \to S \times S$ be the diagonal mapping on $S$.
We have that $\map \Delta S$ consists of all the elements of $S \times S$ whose coordinates are equal.
:$\map \Delta S = \set {\tuple {x, x} \in S \times S: x \in S} = \set {\tuple {x, y} \in S \times S: x = y}$
By definition, $\Delta_S$ is the same set as $\map \Delta S$.
Now consider $H := \relcomp {S \times S} {\map \Delta S}$, its complement in $S \times S$.
We have that $H$ consists of all the elements of $S \times S$ whose coordinates are different:
:$H = \set {\tuple {x, y} \in S \times S: x \ne y}$
Let $\tuple{x, y} \in H$.
Because $T$ is a $T_2$ space:
:$\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
Because $U \cap V = \O$ it follows that $U \times V \cap \map \Delta S = \O$.
That is:
:$U \times V \subseteq H$
Now from Natural Basis of Product Topology of Finite Product, we have that $U \times V$ is an element of the natural basis of $T^2 = \struct {S \times S, \TT}$.
Hence $H$ is open in $T^2$.
Therefore, by definition, $\Delta_S$ is closed in $T^2 = \struct {S \times S, \TT}$.
{{qed|lemma}}
Now suppose $\Delta_S$ is closed in $T^2 = \struct {S \times S, \TT}$.
Then $H = \set {\tuple {x, y} \in S \times S: x \ne y}$ is open in $T^2$.
Let $\tuple {x, y} \in H$.
We have that $x \ne y$ by the nature of $H$.
Then there exists an element $U \times V$ in the natural basis of $T^2$, and therefore open in $T^2$.
As $U \times V \subseteq H$ it follows that $U \cap V = \O$.
From Natural Basis of Product Topology of Finite Product, as $U \times V$ is open in $T^2$, then $U$ and $V$ are both open in $T$.
So, for any $\tuple {x, y}$ we have $U, V \subseteq S$ such that $x \in U, y \in V, U \cap V = \O$.
So $T$ is a $T_2$ (Hausdorff) space.
{{qed}}
Theorem: T0 Property is Hereditary

:The property of being a $T_0$ space is hereditary.

Proof:
Let $T$ be a a $T_0$ space.
{{Recall|T0 Space|$T_0$ space}}
{{:Definition:T0 Space/Definition 1}}
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U \in \tau: x \in U, y \notin U$
or:
:$\exists U \in \tau: y \in U, x \notin U$
Then either:
:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
or:
:$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$
and so the $T_0$ axiom is satisfied.
{{qed}}
Theorem: T1 Property is Hereditary

:The property of being a $T_1$ space is hereditary.

Proof:
Let $T$ be a $T_1$ space.
{{Recall|T1 Space|$T_1$ space}}
{{:Definition:T1 Space/Definition 1}}
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x \in H$ such that $x \ne y$.
Then as $x \in S$:
:$\forall y \in S: \exists U \in \tau: x \in U, y \notin U$
Then:
:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
and so the $T_1$ axiom is satisfied.
{{qed}}
Theorem: T2 Property is Hereditary

:The property of being a $T_2$ (Hausdorff) space is hereditary.

Proof:
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
{{Recall|T2 Space|$T_2$ (Hausdorff) space}}
{{:Definition:T2 Space/Definition 1}}
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U, V \in \tau: x \in U, y \in V, U \cap V = \O$
As $x, y \in H$ we have that:
:$x \in U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_2$ axiom is satisfied in $H$.
{{qed}}
Theorem: T2.5 Property is Hereditary

:The property of being a $T_{2 \frac 1 2}$ space is hereditary.

Proof:
Let $T = \struct {S, \tau}$ be a $T_{2 \frac 1 2}$ space.
{{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space}}
{{:Definition:T2.5 Space/Definition 1}}
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U, V \in \tau: x \in U, y \in V, U^- \cap V^- = \O$
As $x, y \in H$ we have that:
:$x \in U \cap H, y \in V \cap H: \paren {U^- \cap H} \cap \paren {V^- \cap H} = \O$
From {{Corollary|Closure of Subset in Subspace|1}} we have that:
{{begin-eqn}}
{{eqn | l = \paren {U \cap H}^-
      | o = \subseteq
      | r = U^- \cap H
}}
{{eqn | l = \paren {V \cap H}^-
      | o = \subseteq
      | r = V^- \cap H
}}
{{end-eqn}}
where $\paren {U \cap H}^-$ and $\paren {V \cap H}^-$ are the closures in $H$ of the respective sets.
So:
:$x \in U \cap H, y \in V \cap H: \paren {U \cap H}^- \cap \paren {V \cap H}^- = \O$
and so the $T_{2 \frac 1 2}$ axiom is satisfied in $H$.
{{qed}}
Theorem: T3 Property is Hereditary

:The property of being a $T_3$ space is hereditary.

Proof:
Let $T = \struct {S, \tau}$ be a $T_3$ space.
{{Recall|T3 Space|$T_3$ space}}
{{:Definition:T3 Space/Definition 1}}
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $F \subseteq H$ such that $F$ is closed in $H$.
Let $y \in H$ such that $y \notin F$.
From Closed Set in Topological Subspace $F$ is also closed in $T$.
Because $T$ is a $T_3$ space, we have that:
:$\exists U, V \in \tau: F \subseteq U, y \in V, U \cap V = \O$
As $F \subseteq H$ and $y \in H, y \notin F$ we have that:
:$F \subseteq U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_3$ axiom is satisfied in $H$.
{{qed}}
Category:T3 Spaces
Category:Separation Properties Preserved in Subspace
Category:Examples of Hereditary Properties
c1gbj7lt1byi17h0oqovqiq6s8fdo1i
Theorem: T3.5 Property is Hereditary

:The property of being a $T_{3 \frac 1 2}$ space is hereditary.

Proof:
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}}
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $F \subseteq H$ such that $F$ is closed in $H$.
Let $y \in H$ such that $y \notin F$.
From Closed Set in Topological Subspace $F$ is also closed in $T$.
Because $T$ is a $T_{3 \frac 1 2}$ space, we have that there exists an Urysohn function for $F$ and $\set y$:
That is, there exists a continuous mapping $f: S \to \closedint 0 1$, where $\closedint 0 1$ is the closed unit interval, such that:
:$f {\restriction_F} = 0, f {\restriction_{\set y} } = 1$
where $f {\restriction_F}$ denotes the restriction of $f$ to $F$.
That is:
{{begin-eqn}}
{{eqn | q = \forall a \in F
      | l = \map f a
      | r = 0
}}
{{eqn | q = \forall b \in \set y
      | l = \map f b
      | r = 1
}}
{{end-eqn}}
From Continuity of Composite with Inclusion, as $f$ is continuous on $S$, then $f {\restriction_H}$ is continuous on $H$.
Thus $f {\restriction_H}$ is an Urysohn function for $F$ and $\set y$ in $H$.
So the $T_{3 \frac 1 2}$ axiom is satisfied in $H$.
{{qed}}
Category:T3.5 Spaces
Category:Separation Properties Preserved in Subspace
Category:Examples of Hereditary Properties
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Theorem: T5 Property is Hereditary

:The property of being a $T_5$ space is hereditary.

Proof:
Let $T = \struct {S, \tau}$ be a $T_5$ space.
{{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $A, B \subseteq H$ such that $\map {\cl_H} A \cap B = A \cap \map {\cl_H} B = \O$.
That is, $A$ and $B$ are separated in $H$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\cl_S} A \cap B
      | r = \map {\cl_S} A \cap \paren {H \cap B}
      | c = as $B \subseteq H$
}}
{{eqn | r = \paren {\map {\cl_S} A \cap H} \cap B
      | c = Intersection is Associative
}}
{{eqn | r = \map {\cl_H} A \cap B
      | c = Closure of Subset in Subspace
}}
{{eqn | r = \O
      | c = Assumption
}}
{{end-eqn}}
Similarly:
:$A \cap \map {\cl_S} B = \O$
So $A$ and $B$ are separated in $S$.
Because $T$ is a $T_5$ space, we have that:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
It follows that:
:$\exists U \cap H, V \cap H \in \tau_H : A \subseteq U \cap H, B \subseteq V \cap H, \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_5$ axiom is satisfied in $H$.
{{qed}}
Category:T5 Spaces
Category:Separation Properties Preserved in Subspace
Category:Examples of Hereditary Properties
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Theorem: Product Space is T1 iff Factor Spaces are T1

:$T$ is a $T_1$ space {{iff}} each of $\struct{S_\alpha, \tau_\alpha}$ is a $T_1$ space.

Proof:
=== Sufficient Condition ===
{{:Product Space is T1 iff Factor Spaces are T1/Sufficient Condition/Proof 1}}{{qed|lemma}}
Theorem: Product Space is T1 iff Factor Spaces are T1

:$T$ is a $T_1$ space {{iff}} each of $\struct{S_\alpha, \tau_\alpha}$ is a $T_1$ space.

Proof:
Let $T$ be a $T_1$ space.
As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
From Subspace of Product Space is Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_1$ property is hereditary, $T_\alpha$ is a $T_1$ space.
From $T_1$ Property is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is a $T_1$ space.
As $\alpha \in I$ was arbitrary, it follows that each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_1$ space.
Theorem: Gödel's Incompleteness Theorems/First

Let $T$ be the set of theorems of some recursive set of sentences in the language of arithmetic such that $T$ contains minimal arithmetic.
$T$ cannot be both consistent and complete.

Proof:
{{AimForCont}} that such a $T$ is consistent and complete.
By the Undecidability Theorem, since $T$ is consistent and contains $Q$, it is not recursive.
But, by Complete Recursively Axiomatized Theories are Recursive, since $T$ is complete and is the set of theorems of a recursive set, it is recursive.
The result follows by Proof by Contradiction.
{{qed}}