T3 Space is Preserved under Homeomorphism

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Theorem

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.

Let $\phi: T_A \to T_B$ be a 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 mutandis, 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.

$\blacksquare$


Sources

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