Proposition 4

If a topological space X is completely regular, it is homeomorphic to a subspace of a cube.

Let $(f_i)_{i \in I}$ denote the family of all continuous mappings of X into K = [0, 1], and let g denote the mapping $x \mapsto (f_i(x))$ of X into

(*) See A. Tychonoff, Math. Ann., 102, (1930), p. 553.

K¹. If $x, y$ are any two distinct points of $X$, it follows from axioms (H) and (O_IV) that there is an index $i$ such that $f_i(x) \neq f_i(y)$, and therefore $g$ is a one-to-one mapping of $X$ into $K^1$. Moreover, it is immediate that $g$ is an isomorphism of the coarsest uniformity on $X$ for which all the $f_i$ are uniformly continuous, onto the uniformity induced on $g(X)$ by the product uniformity of $K^1$; a fortiori, $g$ is a homeomorphism of $X$ onto $g(X)$.