Heredity of First Countability on Topological Subspaces

# Heredity of First 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 prove that first countability on topological subspaces is hereditary, that is, if $(X, \tau)$ is a first countable topological space then any topological subspace $(A, \tau_A)$ where $A \subseteq X$ is also first countable.

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

**Proof:**Let $(X, \tau)$ be a first countable topological space and let $A \subseteq X$. Since $X$ is first countable we have that for every $a \in A \subseteq X$ ($a \in X$) has a countable local basis, call it $\mathcal B_a$.

- For each $a \in A$ we claim that the following collection is a countable local basis for $a \in A$ in $A$:

\begin{align} \quad \tilde{\mathcal B_{a}} = \{ A \cap B : B \in \mathcal B_a \} \end{align}

- Clearly the collection above is countable because $\mathcal B_a$ is countable.

- Recall that a collection of sets is a local basis of $a \in A$ if for all open sets in $A$ there exists a set in the local basis that is fully contained in this open set.

- Let $U \subseteq A$ be any open set in $A$. Since $U$ is open in $A$ this implies that there exists an open set $V$ in $X$ such that:

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

- Since $V$ is an open set in $X$ and $\mathcal B_a$ is a local basis of $a$ in $X$ we have that there exists a set $B \in \mathcal B_a$ such that:

\begin{align} \quad a \in B \subseteq V \end{align}

- Hence we see that:

\begin{align} \quad a \in A \cap B \subseteq A \cap V = U \end{align}

- The set $A \cap B$ is an open set in $A$ (with the subspace topology $\tau_A$), so, for all open sets $U$ in $A$ and for all $a \in U$ there exists an element $A \cap B \in \tilde{\mathcal B_a}$ such that $a \in A \cap B \subseteq U$, so $\tilde{\mathcal B_a}$ is a local basis of $a$ in $A$. Hence $\tilde{\mathcal B_a}$ is a countable local basis of $a$ in $A$, so first countability is hereditary. $\blacksquare$