The following is from Munkres Topology:
A space X is completely regular if and only if it is homeomorphic to a subspace of $[0,1]$$^J$ for some $J$.
Note: $[0,1]$$^J$ is given the product topology.
I realize that $[0,1]$$^J$ is a completely regular space and any subspace of a completely regular space is also completely regular.
However, why can we conclude for a space X that is homeomorphic to $[0,1]$$^J$ that it must also be completely regular?
I don't find the answer that homeomorphisms "preserve topological properties" convincing enough and am looking for something more specific.
Thanks!
Best Answer
Any product of completely regular spaces is completely regular. So $[0,1]^J$ is completely regular for any $J$.
A subspace of a completely regular space is completely regular so any subspace of $[0,1]^J$ is completely regular.
If $X$ embeds into $[0,1]^J$, it is homeomorphic to a subspace of it, and a space homeomorphic to a completely regular space is completely regular. This is obvious from the definitions: we can move a point $x$ and a closed set not containing it by the homeomorphism to the completely regular space, get our real-valued function there and compose with the homeomorphism to get a separating function on the original space.
So if $X$ embedds into $[0,1]^J$, it has to be completely regular. The reverse follows from the embedding theorem (separating points and point/closed sets, using all functions from $X$ to $[0,1]$, or those connected to a minimal base, say, get a smaller index set $J$).