Subset of separable space is separable

Homework Statement

Show that if X$\subset$M and (M,d) is separable, then (X,d) is separable. [This may be a little bit trickier than it looks - E may be a countable dense subset of M with X $\cap$ E = Ø.]

Homework Equations

No equations, but there are relevant definitions. Per our book:
A metric space (M,d) is separable if $\exists$ a countable dense E $\subset$ M.
E$\subset$M is dense in M if $\forall$x$\in$M and $\forall$ $\epsilon$ > 0, $\exists$ e $\in$ E st d(x,e)< $\epsilon$

The Attempt at a Solution

My best attempt was doomed from the start, because I don't quite understand the hint. My thought process went as follows:
since X $\subset$ M, $\forall$x$\in$X, x$\in$M. Thus, since M is dense in E, $\forall$x$\in$X, $\forall$$\epsilon$>0, $\exists$e$\in$E st d(x,e)<$\epsilon$. At this point, I was done, because the set of e's satisfying the above, is a subset of E, a countable set. So a subset of a countable set is dense in X, and X is separable. This is incorrect, but I cannot see why.
Any help clearing up the confusion would be greatly appreciated.
Thanks!

Last edited:

Dick
Homework Helper

Homework Statement

Show that if X$\subset$M and (M,d) is separable, then (X,d) is separable. [This may be a little bit trickier than it looks - E may be a countable dense subset of M with X $\cap$ E = Ø.]

Homework Equations

No equations, but there are relevant definitions. Per our book:
A metric space (M,d) is separable if $\exists$ a countable dense E $\subset$ M.
E$\subset$M is dense in M if $\forall$x$\in$M and $\forall$ $\epsilon$ > 0, $\exists$ e $\in$ E st d(x,e)< $\epsilon$

The Attempt at a Solution

My best attempt was doomed from the start, because I don't quite understand the hint. My thought process went as follows:
since X $\subset$ M, $\forall$x$\in$X, x$\in$M. Thus, since M is dense in E, $\forall$x$\in$X, $\forall$$\epsilon$>0, $\exists$e$\in$E st d(x,e)<$\epsilon$. At this point, I was done, because the set of e's satisfying the above, is a subset of E, a countable set. So a subset of a countable set is dense in X, and X is separable. This is incorrect, but I cannot see why.
Any help clearing up the confusion would be greatly appreciated.
Thanks!

The problem statement already warned you what could go wrong with an approach like that. Suppose X is M-E. Your countable dense set, call it F, has to be a subset of M-E. But M-E doesn't contain any elements of E, so you can't use them. You have to prove M-E contains a different countable dense set. Try to construct one.

Yes, I know the hint clearly stated that this could cause issues. I said that above. I asked *why* this hint is there. The fact that I believe my proof is complete, means that I do not understand the hint. I don't understand at all what you mean by trying to construct a countable dense set that is in M-E. I have no idea what M is, all I know is that it contains 1 countable dense set.

Dick