# A compact => A separable?

Discussion in 'Undergraduate Math' started by sto, Dec 23, 2010.

1. ### stoGuest

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

sto, Dec 23, 2010

2. ### Arturo MagidinGuest

You mean "B is contained in [A]". "B in [A]" can easily be
misunderstood as to mean that B is an *element* of the set [A], which
would not be the case for most metric spaces.
This is not what you say below.
A contained in [A]. Yes. This is a correct argument.
This is impossible, since you are assuming that A is compact: place an
open ball of radius e>0 around *EVERY* element of A; this covers A, so
there is a finite subcover, giving you the finite union of open balls
you are looking for.
This is needlessly complicated. Just apply the definition of "compact"
to the cover mentioned above.

No, you can take them in A. See above. And better write Y_n={y_n1,
y_n2,...,y_{n,k_n}}, so that you don't use the same name for possibly
different points.
You mean each of the Y_n, not each of the unions; the union may very
well be infinite (if any of the balls are infinite); think A=R, the
real numbers.
Better: given any a in A, you want to show that every open ball with
center in a must intersect Y; let e>0, and let N be such that 1/N < e.
Since the balls of radius 1/N and centers in y_N1,...,y_Nk_N cover A,
there exists an i such that the ball with radius 1/N and center at
y_Ni contains a. Then the ball with center in a and radius e contains
y_Ni, hence intersects Y, as desired.
Your argument was needlessly complicated, but you could have gotten
the necessary conclusion if instead of just saying "not covered by a
finite union of open balls each of radius e" you had said "not covered
by a finite union of open balls, each centered at a point of A and
each of radius e". Then you could have said your sets Y_n were subsets
of A to begin with.
This is not what you said above. Above your conclusion was "A
separable". Do you mean this, or did you mean something else?

But: don't use the definition of limits. Think about closure.

Let Y be a countable dense subset of A. Let n>0. I claim that the
union of the open balls of radius 1/n centered at elements of Y cover
A (that is, A is contained in this union). For, if this were not the
case, then there would exist an element a in A such that the open ball
of radius 1/2n and center at a contains no elements of Y, hence a is
not in the closure of Y, hence Y is not dense, contradicting our
assumption.

Let Y = {y1, y2. y3. ...}

Then, for each n in N, the union of the balls B(y_i/1/n) contains B.

Let Y_n = {y_i in Y : B(y_i,1/n) intersects B}. Note that Y_n is
countable, since it is a subset of Y.

Now, pick an element of B in each B(y,1/n) for each y in Y_n, for each
n>0; call this set Z. This is a countable subset of B.

I claim that this set is dense in B. Let b in B, let e>0. Find n such
that 1/n < e/2. Then there exists some y in Y such that y is in B(b,1/
n). In particular, y is in Y_n, and therefore there exists a b' in Z
such that b is in B(y,1/n).

Then d(b,b') <= d(b,y) + d(y,b') < 1/n + 1/n = 2/n < 2(e/2) = e.

Thus, there is an element of Z in B(b,e) for every b in B, for every
e>0. Therefore, Z is dense in B, and since Z is countable, B is
separable.

Arturo Magidin, Dec 23, 2010

3. ### stoGuest

I don't know why I keep writing that. I even went back and proofread
the post to correct that terminology, but I guessed I missed several
instances. Of course I meant B subset [A].
B subset A and A separable => B separable
I clearly missed a night of sleep somewhere.
Ok. I think this fixes my problem. I guess it did not occur to me to
choose a cover of opens balls centered on the elements A to begin with.
What I meant here is that each of the unions is comprised of a finite
number of open balls, so that each of the Y_n in turn contains a finite
number of points. Arguably not a coherent way of saying that.
B subset A and A separable => B separable
is what I meant
Ok that makes perfect sense now.
Thanks.

sto, Dec 26, 2010