The Hausdorff Property under Homeomorphisms on Topo. Sps.

# The Hausdorff Property under Homeomorphisms on Topological Spaces

Recall from the Hausdorff Topological Spaces page that a topological space $(X, \tau)$ is said to be Hausdorff if for every distinct pair of points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that:

(1)
\begin{align} \quad U \cap V = \emptyset \end{align}

We will now see that the Hausdorff property is preserved under homeomorphisms.

 Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a homeomorphism. If $X$ is a Hausdorff space then $Y$ is Hausdorff.
• Proof: Let $X$ be a Hausdorff space and let $f : X \to Y$ be a homeomorphism between $X$ and $Y$.
• Let $x, y \in Y$. Since $f$ is a homeomorphism, $f$ is bijective, and so $f^{-1}(x)$ and $f^{-1}(y)$ are distinct points in $X$.
• Since $X$ is Hausdorff, there exists open neighbourhoods $U$ of $f^{-1}(x)$ and $V$ of $f^{-1}(y)$ such that:
(2)
\begin{align} \quad U \cap V = \emptyset \end{align}
• Since $f$ is an open map, $f(U)$ is an open neighbourhood of $x$ and $f(V)$ is an open neighbourhood of $y$. Furthermore, we see that:
(3)
\begin{align} \quad f(U) \cap f(V) = \emptyset \end{align}
• If not, i.e., if there existed an $x \in f(U) \cap f(V)$ then this would imply that there exists a $z \in U$ and a $w \in V$ such that $f(z) = x$ and $f(w) = x$ which contradicts $f$ being injective.
• So, for all points $x, y \in Y$ there exists open neighbourhoods $f(U)$ of $x$ and $f(V)$ of $y$ such that $f(U) \cap f(V) = \emptyset$. Therefore, $Y$ is a Hausdorff space. $\blacksquare$