topology: embedding Billingsley writes in his book convergence of probability measures:
" If $S$ can be embedded as an open set in a some complete metric space, then it is topologically complete".
I have taken one course in analysis (I am not from maths background
) which covers chapters 1-6 and some parts of chapter 7, and I can make nothing out of this statement. I guess he is looking for some sort of map?
 A: First just to handle some definitions:


*

*By an embedding of a topological space $X$ into a topological space $Y$ we mean a (one-to-one) function $f : X \to Y$ such that if you consider it as a mapping into $f [X]$, the range of $f$, then it is a homeomorphism onto this subspace of $Y$.

*A topological space $X$ is called topologically complete if there is a complete metric $d$ on $X$ which yields the same open sets (i.e., the topology can be induced by a complete metric on that same space).


So the meaning of this statement is essentially this

If $S$ is a topological space which is homeomorphic to an open subspace of a complete metric space $Y$, then the topology on $S$ is induced by a complete metric (you can give $S$ a complete metric which yields the same open sets).

Actually, a stronger statement is also true:

If $S$ is a topological space which is homeomorphic to a G$_\delta$ subspace (a countable intersection of open sets) of a complete metric space $Y$, then the topology on $S$ is induced by a complete metric.


To give a bit of an example: Consider the open interval $S = (0,1)$ with the usual subspace topology.  As $S$ is itself an open subspace of the real line $\mathbb{R}$, it follows by the statement that $S$ is topologically complete.  But we know that the usual metric on $S$ is not complete: the Cauchy sequence $\langle \frac1n \rangle_{n \in \mathbb{N}}$ does not converge to any point in $S$.  However we can give $S$ a different metric -- namely $$d^\prime (x,y) = \left| \tan \left( \frac{2\pi x - \pi}2 \right) - \tan \left( \frac{2\pi y - \pi}2 \right) \right|$$
and show that this metric is both complete, and yields the same open sets.  It is not too hard to see that the sequence mentioned above is not Cauchy with respect to the $d^\prime$ metric.
