All Second Countable Topological Spaces are First Countable

All Second Countable Topological Spaces are First Countable

Recall from the First Countable Topological Spaces page that a topological space $(X, \tau)$ is said to be first countable if every $x \in X$ has a countable local basis. (Remember that a local basis for $x \in X$ is a collection $\mathcal B_x$ of open neighbourhoods of $x$ such that for all $U \in \tau$ with $x \in U$ there exists a $B \in \mathcal B_x$ such that $x \in B \subseteq U$).

Also recall from the Second Countable Topological Spaces page that a topological space $(X, \tau)$ is said to be second countable if this space has a countable basis. (Remember that a set $\mathcal B$ is a basis for the topology $\tau$ if for every $U \in \tau$ there exists subset $\mathcal B^* \subseteq \mathcal B$ such that $U = \bigcup_{B \in \mathcal B^*} B$).

We will now prove a relatively simple but substantial theorem which will show us that every second countable topological space is first countable.

 Theorem 1: Let $(X, \tau)$ be a second countable topological space. Then $(X, \tau)$ is also a first countable topological space.
• Proof: Let $(X, \tau)$ be a second countable topological space. Then there exists a countable basis $\mathcal B$ of the topology $\tau$.
(1)
\begin{align} \quad \mathcal B_x = \{ B \in \mathcal B : x \in B \} \end{align}
• Then the local basis of $x$ defined above is a subset of the basis $\mathcal B$ for each $x \in X$, i.e., for all $x \in X$, $\mathcal B_x \subseteq \mathcal B$. However, the subset of a countable set is countable. Therefore each $\mathcal B_x$ is countable.
• Therefore every $x \in X$ has a countable local basis $\mathcal B_x$. Hence $(X, \tau)$ is a first countable topological space. $\blacksquare$