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Topic: A compact => A separable?
Replies: 2   Last Post: Dec 25, 2010 11:20 PM

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sto

Posts: 82
Registered: 3/27/08
A compact => A separable?
Posted: Dec 22, 2010 11:01 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply


Let (M,d) be a metric space and A,B subsets of M.
Denote by [A] the closure of A.

Define A to be dense in B if B in [A].
Define A to be separable if it contains a countable dense subset.

I'm having trouble with the proof of all three of these implications:

A countable => A separable
A compact => A separable
B in A and A separable => A separable

First of all, if A is countable does this imply that it is separable? I
think it does. If A is countable, then it contains a countable subset
(itself) and, since A in [A], that subset is dense.

For the second implication, if A is compact then assume that there
exists an e > 0 such that A *cannot* be covered by any finite union of
open balls each ball having radius equal to e. Then given any x1 in A I
can choose an x2 in A such that x2 is not in the open ball B(x1,e). By
assumption, I can do this again to obtain an x3 in A such that x3 not in
union{ B(x1,e), B(x2,e) }. Proceeding in this manner I obtain an
infinite sequence {x1,x2,x3,...} having the property that d(xi,xj) > e
for all i,j, i.e. the sequence in not Cauchy and therefore does not
converge. But because A is compact, it is sequentially compact so this
is a contradiction. It follows that for any natural number n, there
exists a finite set of numbers Y_n = {y1, y2, ..., y_k} *in M* such
that the finite union of open balls union(i=1,k, B(y_i,1/n) ) covers A.
Since each of these unions is finite, I can take the union of all of
them to form the countable set Y = union(n=1, oo, Y_n). The set Y is
dense in A (every point in A is the limit of a sequence in Y).

The whole problem with this proof is that although Y is a countable set
that is dense in A, it is not necessarily a *subset* of A. It seems to
me that unless Y is a subset of A, A has not been proved separable (by
the definition of separability).


As far as

B in A and A separable => B separable

goes I run into the same problem as above: if A is separable, then it
has a countably dense subset Y. Each x in B, since it is also in A, is
then the limit of a sequence of points in Y, *but* who says Y is a
subset of B???

If anyone can sort this out, maybe I skip having to pull an allnighter
on Christmas for a change.
-sto



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