Example of second countable space in topology

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 [also recall that a local basis for a point $x \in X$ is a collection $\mathcal B_x$ of open sets such that for every open neighbourhood of $x$, $U \in \tau$ with $x \in U$ we have that there exists a $B \in \mathcal B_x$ such that $x \in B \subseteq U$].

We will now look at another type of topological space called second countable topological spaces.

Example 1

If $X$ is any nonempty finite set with $n$ elements then the topological space $[X, \tau]$ is always second countable since any basis $\mathcal B$ of $\tau$ is a subset of $\tau$ and the size of $\mathcal B$ is bounded above:

\begin{align} \quad \mid \mathcal B \mid \leq \mid \tau \mid \leq \mid \mathcal P [X] \mid = 2^n \end{align}

For another example, let $[X, \tau]$ be a topological space such that $X$ is infinite and $\tau$ is a nested topological space. Then we have that each of the subsets $U_i \in \tau$ are nested in the manner:

\begin{align} \quad U_1 \subset U_2 \subset ... \subset U_n \subset ... \end{align}

We can clearly define a bijection $f : \mathbb{N} \to \tau$ defined for each $n \in \mathbb{N}$ by $f[n] = U_n$. So if $\tau$ induces the nesting above then $\tau$ is countable and so any subset $\mathcal B \subseteq \tau$ is also countable.