T2 Space is Preserved under Closed Bijection
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 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 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 $T_2$ (Hausdorff) space.
Hence $T_B$ must after all be a $T_2$ (Hausdorff) space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces