Heredity of Second Countability on Topological Subspaces

# Heredity of Second Countability 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: Second countability is hereditary, that is, if $(X, \tau)$ is a second countable topological space and $A \subseteq X$ then $(A, \tau_A)$ is a second countable topological space where $\tau_A = \{ A \cap U : U \in \tau \}$ is the subspace topology on $A$. |

**Proof:**Let $(X, \tau)$ be a second countable topological space and let $A \subseteq X$. Since $X$ is second countable we have that there exists a countable basis $\mathcal B$ of $X$.

- We claim that the following collection is a countable basis for $A$:

\begin{align} \quad \mathcal B_A = \{ A \cap B : B \in \mathcal B \} \end{align}

- Clearly $\mathcal B_A$ is countable since $\mathcal B$ is countable, so we only need to show that $\mathcal B_A$ is a basis of the topology $\tau_A$ on $A$.

- To prove this we must show that every open set in $A$ (with respect to the subspace topology $\tau_A$ on $A$) is a union of a collection of sets from $\mathcal B_A$. Let $U \subseteq A$ be an open set in $A$. Since $U$ is open in $A$ and $A \subseteq X$ we have that there exists an open set $V$ in $X$ such that:

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

- Since $V$ is open in $X$ and $\mathcal B$ is a basis of the topology $\tau$ on $X$ we have that there exists a subcollection $\mathcal B^* \subseteq \mathcal B$ such that:

\begin{align} \quad V = \bigcup_{B \in \mathcal B^*} B \end{align}

- Hence we see that:

\begin{align} \quad U = A \cap \left ( \bigcup_{B \in \mathcal B^*} B \right ) = \bigcup_{B \in \mathcal B^*} (A \cap B) \end{align}

- But $A \cap B \in \tau_A$ for all $B \in \mathcal B^* \subseteq \mathcal B$. So each open set $U$ in $A$ can be expressed as a union of open sets in $\mathcal B_A$. Hence $\mathcal B_A$ is indeed a basis of the topology $\tau_A$ on $A$. Hence $\mathcal B_A$ is a countable basis of $\tau_A$, so first countability is hereditary. $\blacksquare$