A compact => A separable?
sto
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
Arturo Magidin
> 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].
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.
> 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
This is not what you say below.
>
> First of all, if A is countable does this imply that it is separable?
Yes.
> I
> think it does. If A is countable, then it contains a countable subset
> (itself) and, since A in [A], that subset is dense.
A contained in [A]. Yes. This is a correct argument.
> 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.
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.
> 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.
This is needlessly complicated. Just apply the definition of "compact"
to the cover mentioned above.
> It follows that for any natural number n, there
> exists a finite set of numbers Y_n = {y1, y2, ..., y_k} *in M*
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.
> 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,
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.
>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).
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.
> 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).
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.
>
> 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???
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
sto
> On Dec 22, 10:01 pm, sto<s...@address.invalid> wrote:
>> 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].
>
> 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.
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].
>
>> 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
>
> This is not what you say below.
This should have read
B subset A and A separable => B separable
I clearly missed a night of sleep somewhere.
>
>>
>> First of all, if A is countable does this imply that it is separable?
>
> Yes.
>
>> I
>> think it does. If A is countable, then it contains a countable subset
>> (itself) and, since A in [A], that subset is dense.
>
> A contained in [A]. Yes. This is a correct argument.
>
>> 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.
>
> 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.
>
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.
>> 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.
>
> This is needlessly complicated. Just apply the definition of "compact"
> to the cover mentioned above.
>
>
>> It follows that for any natural number n, there
>> exists a finite set of numbers Y_n = {y1, y2, ..., y_k} *in M*
>
> 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.
>
>> 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,
>
> 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.
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
>
> 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.
>
Ok that makes perfect sense now.
Thanks.
> --
> Arturo Magidin