# A space is completely regular if and only if it is homeomorphic to a subspace of $[0,1]^ J$ for some $J$

This is a theorem from Topology by James Munkres.

Theorem 34.3 A space is completely regular if and only if it is homeomorphic to a subspace of $[0,1]^ J$ for some $J$

The book merely states that this theorem is an immediate corollary of the fact that the existence of a family of continuous functions that separate points from closed sets in a space $X$ is equivalent, for a space $X$ in which one-point sets are closed, to the requirement that $X$ be completely regular.

I have proved the right direction, that is, if a space is completely regular then it is homeomorphic to a subspace of $[0,1]^ J$ for some $J$. However, I don't see how I can show that a space $X$ that is homeomorphic to a subspace of a cube is completely regular.

One more question after looking at the book again regarding the third paragraph. The existence of a family of continuous functions that separate points from closed sets in a space X, implies, for a T1 space X, that X is completely regular. I see that in this case, for any point $x \in X$ and each closed set $A$ not containing $x$, $A^{c}$ is an open neighborhood of $x$ so there is a continuous map $f_{\alpha}$ in the family that is positive at $x$ and vanishes outside $A$.

But the definition of completely regular in this text says that a space $X$ is T1 and for each point $x_0$ and each closed set $A$ not containing $x_0$, there is a continiuous function $f: X \to [0,1]$ such that $f(x_0)=1$ and $f(A)={0}$. However, the fact above only guarantees that $f_{\alpha}(x) \gt 0$ and not $f_{\alpha}: X \to [0,f_{\alpha}(x)]$. So how does the existence of such a family guarantee that $X$ be completely regular?

Any solutions, hints or suggestions would be appreciated.

If $Y$ is completely regular, then $Y^J$ is completely regular for all $J$ - more generally, arbitrary products of completely regular spaces are completely regular. In particular, $[0,1]^J$ is completely regular.
• Clamping and scaling: If $f(x_0) > 0$ and $f(A) \subset \{0\}$, consider $$g(x) = \max \left\{ 0,\min \left\{ 1, \frac{f(x)}{f(x_0)}\right\}\right\}.$$ Then $g$ is continuous - the maximum resp. minimum of two continuous functions is continuous - with $0 \leqslant g(x) \leqslant 1$ for all $x$, $g(x_0) = 1$ and $g(A) \subset \{0\}$. – Daniel Fischer Dec 8 '14 at 18:50