avewright/proofwiki-math huggingface dataset:
Theorem: T1 Property is Preserved under Homeomorphism

If $T_A$ is a $T_1$ space, then so is $T_B$.

Proof:
By definition of homeomorphism, $\phi$ is a closed continuous  bijection.
The result follows from $T_1$ Property is Preserved under Closed Bijection.
{{qed}}
Theorem: T2 Property is Preserved under Homeomorphism

If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$.

Proof:
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_2$ Property is Preserved under Closed Bijection.
{{qed}}
Theorem: T2.5 Property is Preserved under Homeomorphism

If $T_A$ is a $T_{2 \frac 1 2}$ space, then so is $T_B$.

Proof:
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}}
Theorem: Regular Property is Preserved under Homeomorphism

If $T_A$ is a regular space, then so is $T_B$.

Proof:
{{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}}
Theorem: T3 Property is Preserved under Homeomorphism

If $T_A$ is a $T_3$ space, then so is $T_B$.

Proof:
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 from Closed Sets, $\phi^{-1} \sqbrk F$ is closed.
Now as $T_A$ is a $T_3$ space, we find disjoint open sets $U_1$ containing $\phi^{-1} \sqbrk F$ and $U_2$ containing $\map {\phi^{-1}} y$.
From Image of Subset is Subset of Image, we have $F = \phi \sqbrk {\phi^{-1} \sqbrk F} \subseteq \phi \sqbrk {U_1}$.
Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism.
Then {{mutatis}}, we deduce also $\set y \subseteq \phi \sqbrk {U_2}$.
From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint.
Since $\phi$ is a homeomorphism, they are also both open in $T_B$.
Therewith, we have construed two disjoint open sets in $T_B$, one containing $F$, and the other containing $y$.
Hence $T_B$ is shown to be a $T_3$ space as well.
{{qed}}
Theorem: T3.5 Property is Preserved under Homeomorphism

If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$.

Proof:
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:
:$\map g x = \map f {\map {\phi^{-1} } x}$
By Composite of Continuous Mappings is Continuous, $g$ is continuous.
Also, for all $x \in F$:
:$\map g x = 0$
as $\map {\phi^{-1} } x \in G$.
Similarly, because $\map {\phi^{-1} } y = z$:
:$\map g y = 1$
Hence $g$ is an Urysohn function for $F$ and $\set y$.
Since $F$ and $y$ were arbitrary, it follows that $T_B$ is a $T_{3 \frac 1 2}$ space.
{{qed}}
Theorem: Completely Regular Property is Preserved under Homeomorphism

If $T_A$ is a completely regular space, then so is $T_B$.

Proof:
{{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 $T_0$ space, then so is $T_B$.
Hence the result.
{{qed}}
Theorem: Normal Property is Preserved under Homeomorphism

If $T_A$ is a normal space, then so is $T_B$.

Proof:
{{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}}
Theorem: T4 Property is Preserved under Homeomorphism

If $T_A$ is a $T_4$ space, then so is $T_B$.

Proof:
{{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 Preimage of Intersection under Mapping it follows that $A_1$ and $A_2$ are disjoint.
Also, as $\phi$ is a homeomorphism, it is {{afortiori}} continuous.
Thus Continuity Defined from Closed Sets applies to yield that both $A_1$ and $A_2$ are closed.
Now as $T_A$ is a $T_4$ space, we find disjoint open sets $U_1$ containing $A_1$ and $U_2$ containing $A_2$.
From Image of Subset is Subset of Image, we have $B_1 = \phi \sqbrk {\phi^{-1} \sqbrk {B_1} } = \phi \sqbrk {A_1} \subseteq \phi \sqbrk {U_1}$.
Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism.
Then {{mutatis}}, we deduce also $B_2 \subseteq \phi \sqbrk {U_2}$.
From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint.
Since $\phi$ is a homeomorphism, they are also both open in $T_B$.
Therewith, we have construed two disjoint open sets in $T_B$, one containing $B_1$, and the other containing $B_2$.
Hence $T_B$ is shown to be a $T_4$ space as well.
{{qed}}
Theorem: Completely Normal Property is Preserved under Homeomorphism

If $T_A$ is a completely normal space, then so is $T_B$.

Proof:
{{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 $T_B$.
Hence the result.
{{qed}}
Theorem: T5 Property is Preserved under Homeomorphism

If $T_A$ is a $T_5$ space, then so is $T_B$.

Proof:
{{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 Intersection under Mapping:
:$f^{-1} \sqbrk { {B_1}^- \cap B_2} = f^{-1} \sqbrk { {B_1}^-} \cap f^{-1} \sqbrk {B_2} = \O$
:$f^{-1} \sqbrk {B_1 \cap {B_2}^-} = f^{-1} \sqbrk {B_1} \cap f^{-1} \sqbrk { {B_2}^-} = \O$
Hence it follows that both $A_1 \cap {A_2}^- = \O$ and ${A_1}^- \cap A_2 = \O$
That is, that $A_1$ and $A_2$ are separated in $T_A$.
Also, as $\phi$ is a homeomorphism, it is {{afortiori}} continuous.
Thus Continuity Defined from Closed Sets applies to yield that both ${A_1}^-$ and ${A_2}^-$ are closed.
Now as $T_A$ is a $T_5$ space, we find disjoint open sets $U_1$ containing $A_1$ and $U_2$ containing $A_2$.
From Image of Subset is Subset of Image, we have $B_1 = \phi \sqbrk {\phi^{-1} \sqbrk {B_1} } = \phi \sqbrk {A_1} \subseteq \phi \sqbrk {U_1}$.
Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism.
Then {{mutatis}}, we deduce also $B_2 \subseteq \phi \sqbrk {U_2}$.
From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint.
Since $\phi$ is a homeomorphism, they are also both open in $T_B$.
Therewith, we have construed two disjoint open sets in $T_B$, one containing $B_1$, and the other containing $B_2$.
Hence $T_B$ is shown to be a $T_5$ space as well.
{{qed}}
Theorem: 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.

Proof:
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 Inverse of Inverse of Bijection.
{{qed}}
Category:Homeomorphisms (Topological Spaces)
ks5ni1wl9vrbi9ryno80ek8o4d47e1y
Theorem: T0 Property is Preserved under Closed Bijection

If $T_A$ is a $T_0$ space, then so is $T_B$.

Proof:
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:
::$\exists U \in \tau_A: y \in U, x \notin U$
{{AimForCont}}:
:$\exists x, y \in S_B: \forall V \in \tau_B: x, y \in V \lor x, y \notin V$
That is:
:$\exists x, y \in S_B: \forall V \in \tau_B: \set {x, y} \subseteq V \lor \set {x, y} \cap V = \O$
From Preimage of Subset is Subset of Preimage:
:$\forall V \in \tau_B: \phi^{-1} \sqbrk {\set {x, y} } \subseteq \phi^{-1} \sqbrk V \lor \phi^{-1} \sqbrk {\set {x, y} } \cap \phi^{-1} \sqbrk V = \O$
that is:
:$\forall V \in \tau_B: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in \phi^{-1} \sqbrk V \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin \phi^{-1} \sqbrk V$
But by definition of continuous mapping, $U = \phi^{-1} \sqbrk V$ is open in $T_A$ {{iff}} $V$ is open in $T_B$.
Thus:
:$\forall U \in \tau_A: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in U \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin U$
This contradicts  the condition that $T_A$ is a $T_0$ space.
It follows by Proof by Contradiction that $T_B$ must also be a $T_0$ space.
{{qed}}
Theorem: T2.5 Property is Preserved under Closed Bijection

If $T_A$ is a $T_{2 \frac 1 2}$ space, then so is $T_B$.

Proof:
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:
:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B^- \cap V_B^- \ne \O$
That is, there exists at least one pair of points $a$ and $b$ for which the closures of all the open sets containing $a$ and $b$ are not disjoint.
From Topological Closure is Closed, both $U_B^-$ and $V_B^-$ are closed.
From Topology Defined by Closed Sets, it follows that $W_B = U_B^- \cap V_B^-$ is also a closed set.
From Preimage of Intersection under Mapping, we have:
:$\phi^{-1} \sqbrk {U_B^- \cap V_B^-} = \phi^{-1} \sqbrk {U_B^-} \cap \phi^{-1} \sqbrk {V_B^-}$
Let $H_A = \phi^{-1} \sqbrk {U_B^-}, K_A = \phi^{-1} \sqbrk {V_B^-}, W_A = \phi^{-1} \sqbrk {W_B}$.
Then we have:
:$H_A \cap K_A = W_A \ne \O$
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.
From Continuity Defined from Closed Sets, we have that $H_A$, $K_A$ and $W_A$ are all closed.
Also, let $U_A = \phi^{-1} \sqbrk {U_B}$ and $V_A = \phi^{-1} \sqbrk {V_B}$.
From Continuity Defined by Closure, we have that:
:$\phi^{-1} \sqbrk {U_B^-} \subseteq \paren {\phi^{-1} \sqbrk {U_B} }^-$
:$\phi^{-1} \sqbrk {V_B^-} \subseteq \paren {\phi^{-1} \sqbrk {V_B} }^-$
that is:
:$H_A \subseteq U_A^-$
:$K_A \subseteq V_A^-$
As $H_A \cap K_A \ne \O$ it follows that $U_A^- \cap V_A^- \ne \O$.
From the bijective nature of $\phi$, we have that:
:$a \in U_B \implies \map {\phi^{-1} } a \in U_A$
:$b \in V_B \implies \map {\phi^{-1} } b \in V_A$
Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.
Thus we have that:
:$\exists x, y \in X_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A^- \cap V_A^- \ne \O$
contradicting the fact that $T_A$ is a $T_{2 \frac 1 2}$ space.
Hence $T_B$ must after all be a $T_{2 \frac 1 2}$ space.
{{qed}}
Theorem: T1 Property is Preserved under Closed Bijection

If $T_A$ is a $T_1$ space, then so is $T_B$.

Proof:
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 $S_A$.
Hence every point in $S_B$ is closed.
That is, $T_B$ is a $T_1$ space.
{{qed}}
Theorem: T2 Property is Preserved under Closed Bijection

If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$.

Proof:
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, there exists at least one pair of points $a$ and $b$ for which all the open sets containing $a$ and $b$ are not disjoint.
As $\tau_B$ is a topology, it follows that $W_B = U_B \cap V_B$ is also an open set.
Let $U_A = \phi^{-1} \sqbrk {U_B}, V_A = \phi^{-1} \sqbrk {V_B}, W_A = \phi^{-1} \sqbrk {W_B}$.
From Preimage of Intersection under Mapping, we have:
:$\phi^{-1} \sqbrk {U_B \cap V_B} = \phi^{-1} \sqbrk {U_B} \cap \phi^{-1} \sqbrk {V_B}$
that is:
:$U_A \cap V_A = W_A$
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.
As $\phi$ is continuous, all of $U_A, V_A, W_A$ are open in $T_A$.
From the bijective nature of $\phi$, we have that:
:$a \in U_B \implies \map {\phi^{-1} } a \in U_A$
:$b \in V_B \implies \map {\phi^{-1} } b \in V_A$
Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.
Thus we have that:
:$\exists x, y \in S_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A \cap V_A \ne \O$
contradicting the fact that $T_A$ is a $T_2$ space.
Hence $T_B$ must after all be a $T_2$  space.
{{qed}}
Theorem: 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 under Expansion}}
=== $T_2$ Property is Preserved under Expansion ===
{{:T2 Property is Preserved under Expansion}}
=== $T_{2 \frac 1 2}$ Property is Preserved under Expansion ===
{{:T2.5 Property is Preserved under Expansion}}

Proof:
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 $\struct {S, \tau_2}$.
From Identity Mapping to Expansion is Closed, we have that $I_S$ is closed.
We also have Identity Mapping is Bijection.
So we can directly apply:
:$T_0$ Property  is Preserved under Closed Bijection
:$T_1$ Property is Preserved under Closed Bijection
:$T_2$ Property is Preserved under Closed Bijection
:$T_{2 \frac 1 2}$ Property is Preserved under Closed Bijection
as appropriate.
Hence the result.
{{qed}}
Theorem: 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.

Proof:
$\tau_1 \subseteq \tau_2$ 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 $\struct {S, \tau_2}$.
So, by definition, $I_S \sqbrk A = A$ is closed in $\struct {S, \tau_2}$.
So for all $A \subseteq S$ closed in $\struct {S, \tau_1}$, it holds that $I_S \sqbrk A$ is closed in $\struct {S, \tau_2}$.
So by definition of closed mapping, the result follows.
{{qed}}
Theorem: 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.

Proof:
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 Subset is Subset:
:$A \subseteq K \iff A \cap K = A$
As $K$ is itself closed in $T$, it follows that so is $A \cap K = A$ from Topology Defined by Closed Sets.
Similarly, so is $B \cap K = B$.
Because $T$ is a $T_4$ space:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
As $A, B \subseteq K$:
:$A \subseteq U \cap K, B \subseteq V \cap K: \paren {U \cap K} \cap \paren {V \cap K} = \O$
From the definition of topological subspace, both $U \cap K$ and $V \cap K$ are open in $K$.
Thus the $T_4$ axiom is satisfied in $K$.
{{qed}}
Theorem: 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.

Proof:
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/Definition 3}}
From $T_1$ Property is Hereditary, every subspace of a $T_1$ space is also a $T_1$ space.
Hence the result.
{{qed}}
Theorem: Little Picard Theorem

A non-constant entire function $f: \C \to \C$ omits at most one complex value $a \in \C$.

Proof:
{{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 modular function.
{{AimForCont}} that an entire function $f: \C \to \C$ omits two different complex values.
Assume for simplicity that the omitted values are $0$ and $1$.
This can always be achieved by postcomposing $f$ with a suitable complex affine map.
Because $\C$ is simply connected and $\phi$ is a covering, we can lift $f$ to a holomorphic function:
:$F: \C \to \Bbb D$
with $\phi \circ F = f$.
By Liouville's Theorem (Complex Analysis), $F$ is constant.
Therefore $f$ is also constant.
Hence the result.
{{qed}}
{{Namedfor|Charles Émile Picard|cat = Picard}}
Category:Complex Analysis
jgwqq5efht6ageqvsiyxg1eifl1yash
Theorem: Completely Regular Space is Urysohn

Let $\struct {S, \tau}$ be a completely regular space.
Then $\struct {S, \tau}$ is also an Urysohn space.

Proof:
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 $\set y$ are closed.
So by definition of a $T_{3 \frac 1 2}$ space, there exists an Urysohn function for $\set x$ and $\set y$.
This is exactly the definition of an Urysohn space.
{{qed}}
Theorem: Urysohn Space is T2.5

Let $\struct {S, \tau}$ be an Urysohn space.
Then $\struct {S, \tau}$ is also a $T_{2 \frac 1 2}$ space.

Proof:
Let $T = \struct {S, \tau}$ be an Urysohn space.
{{Recall|Urysohn Space|Urysohn space}}
{{:Definition:Urysohn Space}}
{{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space}}
{{:Definition:T2.5 Space/Definition 1}}
{{proof wanted|Then we do some stuff.}}
Thus:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
which is precisely the definition of a $T_{2 \frac 1 2}$ space.
{{finish}}
Theorem: Perfectly Normal Space is Completely Normal

Let $T = \struct {S, \tau}$ be a perfectly normal space.
Then $T$ is also a completely normal space.

Proof:
Let $T = \struct {S, \tau}$ be a perfectly normal space.
{{Recall|Perfectly Normal Space|perfectly normal space|index = 1}}
{{:Definition:Perfectly Normal Space/Definition 1}}
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
From Perfectly $T_4$ Space is $T_5$, $T$ is a $T_5$ space.
The result follows.
{{qed}}
Theorem: Open Set in T3 Space is Union of Regular Open Sets

Let $T = \struct {S, \tau}$ be a $T_3$ space.
Then every open set of $T$ can be expressed as the union of regular open sets.

Proof:
Let $\BB = \set {B \subseteq S: B^{- \circ} = B}$.
In other words, $\BB$ is the collection of regular open sets contained in $S$.
It follows immediately from the definition of a basis that our theorem is proved if we can show that:
:$\BB$ is a cover for $S$
:$\forall U, V \in \BB: \forall x \in U \cap V: \exists W \in \BB: x \in W \subseteq U \cap V$
First we show that $\BB$ is a cover for $S$. 
Since $S$ is open:
:$S^\circ = S$
Since $S$ is closed:
:$S^- = S$
Therefore:
:$S^{- \circ} = S$
and so:
:$S \in \BB$
From Set is Subset of Union:
:$S \subseteq \bigcup \BB$
Thus $\BB$ covers $S$.
{{qed|lemma}}
Next we demonstrate condition $\paren {\text B 2'}$ for $\BB$ to be a basis.
Let $U, V \in \BB$, $x \in U \cap V$.
We will show that for some $W \in \BB$:
:$x \in W \subseteq U \cap V$
By General Intersection Property of Topological Space, $U \cap V \in \tau$. 
Since $T$ is a $T_3$ space, there is a closed neighborhood $N_x$ around $x$ that is contained in $U \cap V$:
:$\exists N_x: \relcomp S {N_x} \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \paren {U \cap V}$
Let $W := {N_x}^{- \circ}$. 
Then by Interior of Closure is Regular Open, $W^{- \circ} = W$ and $W \in \BB$.
On the other hand, since $N_x$ is closed:
:${N_x}^- = N_x$, and thus $W = {N_x}^\circ$
By definition of interior, we have:
:$Q \subseteq W \subseteq N_x$
Therefore we have:
:$x \in W \subseteq N_x \subseteq U \cap V$
as desired.
It follows that $\BB$ is a basis for $\tau$.
{{qed}}
Theorem: Regular Space is Semiregular

Let $\struct {S, \tau}$ be a regular space.
Then $\struct {S, \tau}$ is also a semiregular space.

Proof:
Let $T = \struct {S, \tau}$ be a regular space.
{{Recall|Regular Space|index = 3}}
{{:Definition:Regular Space/Definition 3}}
{{Recall|Semiregular Space|semiregular space}}
{{:Definition:Semiregular Space}}
We also have that an open set in a $T_3$ space is the union of regular open sets.
That is, from the definition of basis, the regular open sets of $T$ form a basis for $T$.
Hence the result.
{{qed}}
Theorem: Compact Space is Sigma-Compact

Every '''compact space''' is also a '''$\sigma$-compact space'''.

Proof:
By definition, a $\sigma$-compact space is the union of countably many compact sets.
A compact space is the union of exactly one compact space.
Hence the result.
{{qed}}