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Subset of separable space is separable

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Homework Statement


Show that if X[itex]\subset[/itex]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 [itex]\cap[/itex] E = Ø.]

Homework Equations


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

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 [itex]\subset[/itex] M, [itex]\forall[/itex]x[itex]\in[/itex]X, x[itex]\in[/itex]M. Thus, since M is dense in E, [itex]\forall[/itex]x[itex]\in[/itex]X, [itex]\forall[/itex][itex]\epsilon[/itex]>0, [itex]\exists[/itex]e[itex]\in[/itex]E st d(x,e)<[itex]\epsilon[/itex]. 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!
 
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Answers and Replies

  • #2
Dick
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Homework Helper
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Homework Statement


Show that if X[itex]\subset[/itex]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 [itex]\cap[/itex] E = Ø.]

Homework Equations


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

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 [itex]\subset[/itex] M, [itex]\forall[/itex]x[itex]\in[/itex]X, x[itex]\in[/itex]M. Thus, since M is dense in E, [itex]\forall[/itex]x[itex]\in[/itex]X, [itex]\forall[/itex][itex]\epsilon[/itex]>0, [itex]\exists[/itex]e[itex]\in[/itex]E st d(x,e)<[itex]\epsilon[/itex]. 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.
 
  • #3
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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.
 
  • #4
Dick
Science Advisor
Homework Helper
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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.
Then the first step is to figure why you think your proof is complete. Take a concrete example. Q (the rational numbers) is a countable dense set of R (the real numbers) so R is separable. Now take the set R-Q (the irrational numbers). You need to show that's separable too and contains a countable dense set. But none of those points can be rational, so they can't be elements of your original countable dense set Q. Try running through the steps of your proof with that example.
 

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