Heredity of the Hausdorff Property on Topological Subspaces

# Heredity of the Hausdorff Property on Topological Subspaces

Recall from the Hereditary Properties of Topological Spaces page that if $(X, \tau)$ is a topological space that a property of $X$ is said to be hereditary if for all subsets $A \subseteq X$ we have that the topological subspace $(A, \tau_A)$ also has that property (where $\tau_A$ is the subspace topology on $A$).

We will now show that second countability is hereditary.

Theorem 1: The Hausdorff property is hereditary, that is, if $(X, \tau)$ is a Hausdorff topological space and $A \subseteq X$ then $(A, \tau_A)$ is a Hausdorff topological space where $\tau_A = \{ A \cap U : U \in \tau \}$ is the subspace topology on $A$. |

**Proof:**Let $(X, \tau)$ be a Hausdorff topological space and let $x, y \in A$. Then $x, y \in X$ since $A \subseteq X$.

- So since $X$ is Hausdorff there exists open neighbourhoods $U$ of $x$ and $V$ of $Y$ such that:

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

- Since $U$ and $V$ contain $x$ and $y$, we have that $A \cap U$ and $A \cap V$ are open neighbourhoods of $x$ and $y$ in $A$. Moreover we see that:

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

- So for all $x, y \in A$ there exists open neighbourhoods $A \cap U$ and $A \cap V$ of $x$ and $y$ respectively such that $(A \cap U) \cap (A \cap V) = \emptyset$. So $(A, \tau_A)$ is Hausdorff. This shows that the Hausdorff property is hereditary. $\blacksquare$