Is A+B Closed for Two Closed Sets in Real Analysis?

[AdrianZ]

Homework Statement

Suppose A is a compact set and B is a closed subset of R^k. then A+B is closed in R^k. show that A+B for two closed sets is not necessarily closed by a counter-example.

well, since A is a compact set and there's a theorem in Rudin's mathematical analysis chapter 2 stating that compact sets of metric spaces are closed, we can conclude that A is closed. that means every limit point of A is in A. B is closed by assumption, and that means every limit point of B is in B. now here is where I get confused. is A in R^k? I think it should be because if not then how A+B is defined? I want to define A+B as the set {a+b: A,B are in R^k} but the problem has ambiguity and doesn't precisely say what A is. Do we need to know what A is? or we can solve it without knowing what A is? It's easy to solve it if A is in R^k, but I don't know what I should do if A is not in R^k.

 

[Dick]

A is in R^k. As you said, if it's not then A+B doesn't make much sense.

 

[hunt_mat]

Dick, perhaps it the set defined as A+B:=\{a+b|a\in A,b\in B\}

 

[AdrianZ]

A is in R^k. As you said, if it's not then A+B doesn't make much sense.

Yea. probably. I guess that's the only way that it can make sense.

Dick, perhaps it the set defined as A+B:=\{a+b|a\in A,b\in B\}

Yea but how do you define a+b? imagine that a is a scalar quantity in R and b is a vector in R³. how do you define a+b? the question is not how A+B is defined, it's how a+b is defined.

Assuming that A is a subset of R^k and Assuming lima=p and limb=q, can I say that A+B is closed because for every a+b we have lim(a+b) = lima + limb=p+q which is an element of A+B? (because p is in A and q is in B since A,B are closed sets)

 

[micromass]

Assuming that A is a subset of R^k and Assuming lima=p and limb=q, can I say that A+B is closed because for every a+b we have lim(a+b) = lima + limb=p+q which is an element of A+B? (because p is in A and q is in B since A,B are closed sets)

No. You must prove that if (a_n+b_n) converges to z, then z in A+B. But you don't know if (a_n)_n or if (b_n)_n converges...
So you can't assume lim(a)=p and lim(b)=q...

HOWEVER! What does compactness of A tell you about the sequence (a_n)_n?? What does it have?

 

[AdrianZ]

No. You must prove that if (a_n+b_n) converges to z, then z in A+B. But you don't know if (a_n)_n or if (b_n)_n converges...
So you can't assume lim(a)=p and lim(b)=q...

HOWEVER! What does compactness of A tell you about the sequence (a_n)_n?? What does it have?

There is a theorem in Rudin that says if p is a limit point of E, then there is a sequence in E that converges to p. now if a+b is a point of A+B, then by definition of A+B, a is a point of A that means there's a sequence in A that converges to a, also b is a point of B that means there's a sequence in B that converges to b. now can't I say that a+b is also a limit point because of the argument I said earlier?

compactness of A tells me that for each of A's open covers there exists a finite sub-cover.

 

[micromass]

What does compactness say about sequences?? Apply theorem 2.37 to \{a_n~\vert~n\}

Hmm, actually I'm referring to theorem 3.6. But did you see that already?

 

[AdrianZ]

Yes, but that only proves that points of the form a+b are limit points of A+B. You also need to show the converse, that a limit point of A+B has the form a+b.

Now I see. and to prove the converse, A and B must be complete but we don't know if they are complete or not. am I right?

What does compactness say about sequences?? Apply theorem 2.37 to \{a_n~\vert~n\}[/QUOTE]

theorem 2.37 says if E is an infinite subset of a compact set K, then E has a limit point in K. since {a(n)|n} is a subset of A and A is compact and also {a(n)|n} is an infinite set then the theorem assures us that a(n) has a limit point in A. so? I'm not quite understanding your next step.

 

[micromass]

Now I see. and to prove the converse, A and B must be complete but we don't know if they are complete or not. am I right?

Right. Well, we do know that A is complete since it is compact, but we know nothing of B.

What does compactness say about sequences?? Apply theorem 2.37 to \{a_n~\vert~n\}
theorem 2.37 says if E is an infinite subset of a compact set K, then E has a limit point in K. since {a(n)|n} is a subset of A and A is compact and also {a(n)|n} is an infinite set then the theorem assures us that a(n) has a limit point in A. so? I'm not quite understanding your next step.

I meant theorem 3.6 (which follows from 2.37).

Let me reformulate.

We have to show that if z is a limit point of A+B, that then z is in A+B, ok?
Well, if z is a limit point, then there exists a sequence a_n+b_n\rightarrow z.
You have already shown that if (a_n)_n and (b_n)_n converge, that z is indeed in A+B.

The problem is that (a_n)_n doesn't need to converge. That is where compactness comes in. Compactness tells us that there exists a subsequence (a_{k_n})_n which DOES converge! Can you know
1) show that also (b_{k_n})_n converges??
2) modify your argument above to conclude that z is in A+B?

 

[hunt_mat]

Yea. probably. I guess that's the only way that it can make sense.



Yea but how do you define a+b? imagine that a is a scalar quantity in R and b is a vector in R³. how do you define a+b? the question is not how A+B is defined, it's how a+b is defined.

Assuming that A is a subset of R^k and Assuming lima=p and limb=q, can I say that A+B is closed because for every a+b we have lim(a+b) = lima + limb=p+q which is an element of A+B? (because p is in A and q is in B since A,B are closed sets)

I am assuming both A and B are subsets in R^k, so a+b would be simple vecor addition.

 

[AdrianZ]

I meant theorem 3.6 (which follows from 2.37).

You mean theorem 3.6 part a?

We have to show that if z is a limit point of A+B, that then z is in A+B, ok?

Yea. and we have to prove this regardless of assuming that a(n) and b(n) are convergent sequences.

Well, if z is a limit point, then there exists a sequence a_n+b_n\rightarrow z.
You have already shown that if (a_n)_n and (b_n)_n converge, that z is indeed in A+B

The problem is that (a_n)_n doesn't need to converge. That is where compactness comes in. Compactness tells us that there exists a subsequence (a_{k_n})_n which DOES converge!

Actually I looked at the theorem you referred me to. It states that "if {p_n} is a sequence in a compact metric space X, then some subsequence of {p_n} converges to a point of X". it doesn't precisely say if that point is a particular point of X or can be any point of X.maybe every sequence in A is convergent but converges to only one point of A? Isn't that possible? then there are points of A that there is no sequence that converges to them and my argument fails again.

1) show that also (b_{k_n})_n converges??

I don't know how to do that. the problem tells us only that it's closed.

This is the result of my new thoughts about the problem. assume that a+b is a limit point of A+B. that means every neighborhood of (a+b) contains an infinitely many points of A+B. if it contains an infinitely many points of A+B then it also contains an infinitely many points of a & b by definition of A+B (Is it true? I guess It's true because if it doesn't, let's say if it contains only a finitely many points of A, then a+b's won't be infinitely many). therefore a+b is also a limit point of A and B and that means I can find sequences in A and B that converge to the points a and b in A and B. that means I can again use my previous argument and deduce that a+b is in A+B. is this a valid argument?

I am assuming both A and B are subsets in R^k, so a+b would be simple vecor addition.

Yea but my first question was whether A is a subset of R^k or not. Thanks for the help anyways.

 

[micromass]

You mean theorem 3.6 part a?

Yes.

Actually I looked at the theorem you referred me to. It states that "if {p_n} is a sequence in a compact metric space X, then some subsequence of {p_n} converges to a point of X". it doesn't precisely say if that point is a particular point of X or can be any point of X.maybe every sequence in A is convergent but converges to only one point of A? Isn't that possible? then there are points of A that there is no sequence that converges to them and my argument fails again.

I don't see what you're worrying about here. You know that (a_n)_n has a convergent subsequence (a_{k_n})_n. And you still know that

a_{k_n}+b_{k_n}\rightarrow z

(with z being our limit point). So can you show b_{k_n} to be convergent as difference of (a_{k_n}+b_{k_n})_n and (a_{k_n})_n??

This is the result of my new thoughts about the problem. assume that a+b is a limit point of A+B.

It already breaks down here. You do not know that a limit point of A+B is of the form a+b. Showing that it is of the form a+b is exactly what you need to show!

 

[AdrianZ]

I don't see what you're worrying about here. You know that (a_n)_n has a convergent subsequence (a_{k_n})_n. And you still know that

a_{k_n}+b_{k_n}\rightarrow z

(with z being our limit point). So can you show b_{k_n} to be convergent as difference of (a_{k_n}+b_{k_n})_n and (a_{k_n})_n??

Ah right. then b_{k_n} = (a_{k_n}+b_{k_n})_n - (a_{k_n})_n is convergent too. Now we're done. right?
but we have nowhere used the fact that A,B are subsets of R^k. Does that mean we can generalize our problem to other metric spaces as well provided we can define A+B?

It already breaks down here. You do not know that a limit point of A+B is of the form a+b. Showing that it is of the form a+b is exactly what you need to show!

Okay right. Thanks for pointing that out.

 

[micromass]

Ah right. then b_{k_n} = (a_{k_n}+b_{k_n})_n - (a_{k_n})_n is convergent too. Now we're done. right?
but we have nowhere used the fact that A,B are subsets of R^k. Does that mean we can generalize our problem to other metric spaces as well provided we can define A+B?

Yes, this is also true in the more general case! In fact, I think that this is true for arbitrary topological groups, but that requires a quite different proof, as we cannot work with sequences there.

 

[AdrianZ]

Yes, this is also true in the more general case! In fact, I think that this is true for arbitrary topological groups, but that requires a quite different proof, as we cannot work with sequences there.

I'm reading the whole topic now to come up with a full argument. first I've shown that if a(n) and b(n) converge to p and q, then p+q is a limit point of A+B which is of the form a+b and is indeed in A+B.
later, we've tried to show that if z is a limit point, then we can find two convergent sequences in A,B that their sum converges to z. but we don't know what a(n) and b(n) converge to. we only know that their sum converges to z, but we don't know what each of these sequences converge to alone. right?

 

[micromass]

I'm reading the whole topic now to come up with a full argument. first I've shown that if a(n) and b(n) converge to p and q, then p+q is a limit point of A+B which is of the form a+b and is indeed in A+B.
later, we've tried to show that if z is a limit point, then we can find two convergent sequences in A,B that their sum converges to z. but we don't know what a(n) and b(n) converge to. we only know that their sum converges to z, but we don't know what each of these sequences converge to alone. right?

That would be a good summary, yes!

Furthermore, we don't even know if a(n) and b(n) converge! They might even diverge

 

[AdrianZ]

Furthermore, we don't even know if a(n) and b(n) converge! They might even diverge

That's exactly what I was thinking about. How have we taken care of that case? I mean we bothered ourselves to show that a(n) converges and then concluded that b(n) converges too. but why did we do that if they can both be divergent and yet their sum converges to z?

 

[micromass]

That's exactly what I was thinking about. How have we taken care of that case? I mean we bothered ourselves to show that a(n) converges and then concluded that b(n) converges too. but why did we do that if they can both be divergent and yet their sum converges to z?

No, we never showed that a(n) converges. We showed that a subsequence of a(n) converges! And it is sufficient to do everything with this subsequence!

 

[AdrianZ]

No, we never showed that a(n) converges. We showed that a subsequence of a(n) converges! And it is sufficient to do everything with this subsequence!

Yes, I know what you mean. but when did we show that a(n)+b(n)->z even if a(n) and b(n) diverge?

 

[micromass]

Yes, I know what you mean. but when did we show that a(n)+b(n)->z even if a(n) and b(n) diverge?

We assumed a(n)+b(n)-> z.

We took z to be a limit point, thus there existed a sequence in A+B that converges to z. This sequence can be taken as a(n)+b(n)...

 

[AdrianZ]

We assumed a(n)+b(n)-> z.

We took z to be a limit point, thus there existed a sequence in A+B that converges to z. This sequence can be taken as a(n)+b(n)...

and how does it tells us that a(n) and b(n) may diverge as well? I've fully understood the case where a(n) and b(n) are convergent, but don't know how what we've done tells that a(n)+b(n) still converges to z even if both of them diverge.
(btw, you got 5K posts now, congratulations.xP)

 

[micromass]

and how does it tells us that a(n) and b(n) may diverge as well? I've fully understood the case where a(n) and b(n) are convergent, but don't know how what we've done tells that a(n)+b(n) still converges to z even if both of them diverge.
(btw, you got 5K posts now, congratulations.xP)

But we have taken a(n)+b(n) to converge to z! That was the definition of a(n)+b(n)!

In fact, the only thing we know about a(n)+b(n) is that it converges to z!

 

[AdrianZ]

But we have taken a(n)+b(n) to converge to z! That was the definition of a(n)+b(n)!

In fact, the only thing we know about a(n)+b(n) is that it converges to z!

True. let me see if now I've understood the whole argument completely.

you mean we've assumed that z is a limit point. therefore there exists a non-constant sequence in A+B like (a+b)(n) that converges to z. but (a+b)(n)=a(n)+b(n) (cuz they are functions). therefore we can say that there exists a non-constant sequence a(n)+b(n) that converges to z. and that's why the definition makes sense. 'cuz I'm wondering why such a definition makes sense.

 

[micromass]

That seems ok...