First Countable Topological Spaces

First Countable Topological Spaces

Recall from the Local Bases of a Point in a Topological Space page that if $(X, \tau)$ is a topological space and $x \in X$ then a local basis of $x$ is a collection $\mathcal B_x$ of open neighbourhoods of $x$ such that for each $U \in \tau$ with $x \in U$ there exists a $B \in \mathcal B_x$ such that:

\begin{align} \quad x \in B \subseteq U \end{align}

We are now ready to define a special type of topological spaces known as first countable topological spaces.

Definition: Let $(X, \tau)$ be a topological space. Then $(X, \tau)$ is said to be a First Countable topological space if every point $x \in X$ has a countable local basis.

For example, consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}$. We claim that this topological space is first countable.

To prove this, let $x \in X$. Then any open set containing $x$ must contain an open interval $(a, b)$ containing $x$, i.e., $x \in (a, b)$.

Define the local basis for $x$ as:

\begin{align} \quad \mathcal B_x = \left \{ B_{x_n} = \left ( x - \frac{1}{n}, x + \frac{1}{n}\right ) : n \in \mathbb{N} \right \} \end{align}

So, for each $U \in \tau$ with $x \in U$ we have that $x \in (a, b) \subseteq U$ and furthermore, there exists an $n \in \mathbb{N}$ such that:

\begin{align} \quad x \in \left ( x - \frac{1}{n}, x + \frac{1}{n} \right ) \subseteq (a, b) \subseteq U \end{align}

Note that if $d = \min \{ \mid x - a \mid, \mid x - b \mid \} > 0$ then $n \in \mathbb{N}$ is chosen such that $\frac{1}{n} < d$. So $\mathcal B_x$ is a local basis of $x$. Furthermore, each of these local bases are countable because there exists a bijection $f : \mathbb{N} \to \mathcal B_x$ defined by $f(n) = B_{x_n}$. Since $x \in \mathbb{R}$ was arbitrary, we see that every $x \in \mathbb{R}$ has a countable local basis and so $(\mathbb{R}, \tau)$ is a first countable topological space.

In fact, all metric spaces $X$ are first countable topological spaces. For each $x \in X$ we can take the local basis of $x$, $\mathcal B_x$ to be the set of all balls centered at $x$ with radius $\frac{1}{n}$

\begin{align} \quad \mathcal B_x = \left \{ B \left ( x, \frac{1}{n} \right ) : n \in \mathbb{N} \right \} \end{align}

(This is precisely what we did to get a countable basis for each $x \in \mathbb{R}$ of the metric space $\mathbb{R}$ with the usual metric.)

For another example, if $X$ is a finite set then $(X, \tau)$ must be a first countable topological space. To prove this, we note that if $X$ is finite then $\mid X \mid = n$ for some $n \in \mathbb{N}$. Furthermore, we have that $\mid \tau \mid \leq \mid \mathcal P(X) \mid = 2^n$. Therefore the number of open sets (members of $\tau$) is finite. For each $x \in X$ if $\mathcal B_x$ is a local basis of $x$ then $\mathcal B_x \subseteq \tau$ since $\mathcal B_x$ is a collection of open neighbourhoods containing $x$. Therefore each $\mathcal B_x$ is finite and is also countable. Hence, every $x \in X$ has a countable local basis, so $(X, \tau)$ must be a first countable topological space.

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