Proposition 12

If X is a metrizable topological space, the following statements are equivalent:

a) X is of countable type. b) X has a countable dense subset. c) *X is homeomorphic to a subspace of the cube $\mathbf{I}^{\mathbf{N}}$, where $\mathbf{I}$ is the interval $[0, 1]$ in $\mathbf{R}$.

From the preceding remarks it is clear that c) implies a); a) implies b), for if $(U_n)$ is a countable base of the topology of X and $a_n$ is a point of $U_n$, the $a_n$ form a dense subset of X. Finally let us show that b) implies c). Let $(a_n)$ be a dense sequence of points of X, and for each $x \in X$ let $\varphi(x)$ be the point $(d(x, a_n))_{n \in \mathbf{N}}$ of $\mathbf{I}^{\mathbf{N}}$ ($d$ being a metric compatible with the topology of X, with respect to which the diameter of X is $\leq 1$). We shall show that $\varphi$ is a homeomorphism of X onto a subspace of $\mathbf{I}^{\mathbf{N}}$. Indeed, $\varphi$ is continuous, because each of the functions $x \to d(x, a_n)$ is continuous; and $\varphi$ is injective, because each point of X is the limit of some subsequence of $(a_n)$ (Proposition 8). Let B be a ball with centre $x_0$ and radius $r$ in X, and let $n$ be an integer such that

The image under $\varphi$ of the set W of points $x \in X$ such that

is by definition a neighbourhood of $\varphi(x_0)$ in $\varphi(X)$. But for each $x \in W$ we have $d(x, a_n) < d(x_0, a_n) + \frac{1}{3} r < \frac{2}{3} r$, whence

which shows that W is a neighbourhood of $x_0$ contained in B, and therefore that $\varphi$ is a homeomorphism of X onto $\varphi(X)$.

Note that for an arbitrary topological space X, property b) does not necessarily imply the existence of a countable base, even if X is compact and every point of X has a countable fundamental system of neighbourhoods (Exercise 13; cf. Chapter I, § 1, Exercise 7).