Corollary

Let X be a metrizable, compact and nonempty topological space. There exists a continuous and surjective mapping from $\{0,1\}^{\mathbf{N}}$ into X.

Every compact and metrizable topological space is homeomorphic to a necessarily closed subspace of the topological space $\mathbf{I}^{\mathbf{N}}$ (TG, IX, p. 18, prop. 12 and p. 21, prop. 16). Let A be a closed subspace of $\mathbf{I}^{\mathbf{N}}$ and $h$ a homeomorphism of A onto X.

Let K = $\{0,1\}^{\mathbf{N}}$ and, for $\alpha = (a_n)\in$ K, put $f(\alpha ) =$ $\sum^{\infty}_{n=0}a_n2^{-n-1}$; we have $f(\alpha )\in \mathbf{I}$. The mapping $f: K\rightarrow \mathbf{I}$ thus resulting is surjective (TG, IV, p. 42) and continuous. Indeed, if two elements $\alpha$ and $\beta$ of K have the same coordinates of index $< n$, we have $|f(\beta )-f(\alpha )|\leqslant 2^{-n}$. Let $g$ denote the mapping $(\alpha_n)\mapsto (f(\alpha_n))$ of $K^{\mathbf{N}}$ into $\mathbf{I}^{\mathbf{N}}$; it is continuous and surjective. The topological space $K^{\mathbf{N}}$ is compact (TG, I, p. 63, th. 3), metrizable (TG, IX, p. 15, cor. 2) and totally disconnected (TG, I, p. 84, prop. 10) and the same is true of its closed subspace $\overset{-1}{g}(A)$; the latter is nonempty since the mappings $g$ and $h$ are surjective.

Then, the space $\overset{-1}{g}(A)\times K$ is homeomorphic to $\{0,1\}^{\mathbf{N}}$ (prop. 14), since it is a compact, metrizable, totally disconnected topological space without isolated point and the mapping $(x, y)\mapsto h(g(x))$ of $\overset{-1}{g}(A)\times K$ into X is continuous surjective.