📂Topology

First Countability and Second Countability

Definition ¹

Suppose a topological space $X$ is given.

1. If a countable local basis exists for every point $x \in X$, then it is called a first-countable space.
2. If $X$ has a countable basis, then it is called a second-countable space.

Explanation

One can think of it as creating a new branch of countability through the concept of basis and local basis.

Example That Fails to Be First-Countable

The finite complement space $\left( \mathbb{R} , \mathscr{T}_{f} \right)$ fails to be first-countable, and needless to say, it fails to be second-countable as well.

Intuitive Understanding

Though it is not an exact description, first-countability can be accepted with the feeling that at every point there exist countably many open sets. On the other hand, second-countability resembles the concept of separability in the sense that a countable set encompasses the whole. Unlike first category & second category, first-countability and second-countability are not negations of each other but hold an inclusion relation. This can be confirmed without difficulty by thinking about the relationship between basis and local basis. Meanwhile, although we said earlier that second-countability resembles the concept of separability, in fact not only that, one can even show that it is truly separable.

Theorem

• [1]: Every second-countable space is a first-countable space.
• [2]: Every second-countable space is separable.

Proof

A Second-Countable Space Is First-Countable

If $X$ is a second-countable space, then $X$ will have a countable basis $\mathscr{B}$.

Relationship between basis and local basis: If $\mathscr{B}$ is a basis of $X$, then $\mathscr{B}_{x} := \left\{ B \in \mathscr{B} \ | \ x \in B \right\}$ is a local basis at $x \in X$.

Since $\mathscr{B}_{x} = \left\{ B \in \mathscr{B} \ | \ x \in B \right\}$ is countable for every $x \in X$, $X$ is first-countable.

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A Second-Countable Space Is Separable

If $X$ is a second-countable space, then $X$ will have a countable basis $\mathscr{B}$. For every nonempty empty set $B \in \mathscr{B}$, choose $x_{B} \in B$ and define $D : = \left\{ x_{B} \in B \ | \ \emptyset \ne B \in \mathscr{B} \right\}$. Since $D$ is a set formed by picking one element each from the countable basis $\mathscr{B}$, $D$ is also countable, and the proof is complete if we show that $\overline{D} = X$.

Let $U$ be an open set containing $x \in X \setminus D$.

Since $\mathscr{B}$ is a basis of $X$, there will exist $B \in \mathscr{B}$ satisfying $x \in B \subset U$. Because $x_{B} \in B \cap D$ and $x \notin D$, $$ D \cap (B \setminus \left\{ x \right\} ) \ne \emptyset $$ holds. As stated earlier, since $B \subset U$, $$ D \cap (U \setminus \left\{ x \right\} ) \ne \emptyset $$ still holds. By the definition of a limit point, $x$ is a limit point of $D$ and $x \in \overline{D}$, so $X \setminus D \subset \overline{D}$. Of course, since $D \subset \overline{D}$, in order to satisfy both simultaneously, it must be that $X = \overline{D}$.

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