Sum of a closed set and a compact set, closed?

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SUMMARY

The discussion centers on proving that the sum of a compact set X and a closed set Y, both consisting of real numbers, is closed. The key argument is that while the sum of two closed sets is not necessarily closed, the compactness of set X ensures boundedness, preventing limit points from being excluded from the sum. The standard definition of the sum of two sets, X + Y = {x + y : x ∈ X and y ∈ Y}, is crucial for the proof. The conclusion reached is that if X is compact and Y is closed, then X + Y is indeed closed.

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utleysthrow
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Homework Statement


I am trying to prove that, if X is compact and Y is closed, X+Y is closed. Both X and Y are sets of real numbers.


Homework Equations





The Attempt at a Solution


I know that a sum of two closed sets isn't necessarily closed. So I presume the key must be the difference between the compact set and the closed set, namely the boundedness of a compact set. But I'm not sure how to exploit that fact for a proof.
 
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Think about how the sum of two closed sets can fail to be closed, then explain why this can't happen if one of them is compact
 
Office_Shredder said:
Think about how the sum of two closed sets can fail to be closed, then explain why this can't happen if one of them is compact

If X=N and Y={-n+1/n} where n is a natural number, then both are closed but the sum X+Y={1/n} is not, since it doesn't include 0 which is a limit.

If one of them, say Y, is compact, this doesn't happen because {-n+1/n} has to be bounded by the definition of compactness, which means not all X=N is "canceled out" by the -n's in Y...

A+B would look something like 1, 1/2, 1/3 ... 4, 5, 6, ... which is still closed.

I feel like I only have a superficial understanding of this. I know it has to do with boundedness, but can't quite point to what it is without using a specific example... What am I missing?
 
Hi!

Your definition of the set X + Y doesn't correspond to my understanding. AFAIK, the standard definition is

X + Y = {x + y : x \in X and y \in Y}

With this definition, your set X + Y is still closed (since it doesn't contain the limit point 0), but it does contain points other than 1/n. Just wanted to mention this in case it helps.

Do you know about topological groups? I ask because (R, +), with R having the usual topology, is such a group. I probably could give you some hints on your problem if you know the basic definitions and theorems in this subject. Even if not, try proving that the complement of the set X + Y is open in R.

Let me know if any of the above is unclear, or if you want more help.

Petek
 
Petek said:
Hi!

Your definition of the set X + Y doesn't correspond to my understanding. AFAIK, the standard definition is

X + Y = {x + y : x \in X and y \in Y}

With this definition, your set X + Y is still closed (since it doesn't contain the limit point 0), but it does contain points other than 1/n. Just wanted to mention this in case it helps.

Do you know about topological groups? I ask because (R, +), with R having the usual topology, is such a group. I probably could give you some hints on your problem if you know the basic definitions and theorems in this subject. Even if not, try proving that the complement of the set X + Y is open in R.

Let me know if any of the above is unclear, or if you want more help.

Petek

X=N

Y=\left\{ -n+\frac{1}{n} : n \in N \right\}

X+Y=\left\{\frac{1}{n} : n \in N \right\}

Thank you for your reply. Closed set, as far as I know, contains all of its limit points. Using that definition, my X+Y is not closed because 0 is a limit point of X+Y, but it is not contained in the set. Right?

I haven't learned about topological groups, but I'll try working with the complement of X+Y.
 
X+Y will contain for example the point 2+(-1+1)=2 The n you picked from X and the n you picked from Y don't have to be the same

So suppose you have a sequence in X+Y that does not converge to a limit point. Think about how you can use this sequence to talk about sequences in X and Y
 
Office_Shredder said:
X+Y will contain for example the point 2+(-1+1)=2 The n you picked from X and the n you picked from Y don't have to be the same

So suppose you have a sequence in X+Y that does not converge to a limit point. Think about how you can use this sequence to talk about sequences in X and Y

Oh, right. I didn't catch that...
 
Now that I think about it... X+Y is still NOT closed even if it contains points other than 1/n, isn't it? That doesn't change the fact that 0 is still a limit point of X+Y, which is not contained in the set. Finding one limit point not in the set is all it takes to make the set NOT closed, no?
 
Yes, your set isn't closed. And finding one limit point not contained in the set makes it not closed. But you might be getting distracted from the original problem. If L is a limit point of X+Y then there is are sequences {xi} and {yi} such that xi+yi converges to L. If X is compact, what can you say about the sequence {xi}?
 
  • #10
If X is compact, the seq {xi} should converge to some x \in X
 
  • #11
utleysthrow said:
If X is compact, the seq {xi} should converge to some x \in X

Not right. Can you rephrase that? [0,1] is compact. The sequence {0,1,0,1,0,1...} does not converge.
 
  • #12
Dick said:
Not right. Can you rephrase that? [0,1] is compact. The sequence {0,1,0,1,0,1...} does not converge.

If X is compact... there is a subsequence of {xi} that converges to a limit x in X?
 
  • #13
utleysthrow said:
If X is compact... there is a subsequence of {xi} that converges to a limit x in X?

Much better. So there is a subsequence of xi+yi, which converges to L, call it xj+yj such that xj converges to M which is an element of X. Now what about yj? What must it converge to?
 
  • #14
I would say that yj has to converge to L-M... but I'm not sure how to justify that. Does it simply follow from the fact that xj+yj converges to L, and xj converges to M, so (xj+yj)-xj=yj converges to L-M?
 
  • #15
And if that is correct, it would follow that L-M is in Y since Y is a closed set. Then M+(L-M)=L is in X+Y
 
  • #16
Sure it does. Use epsilons and deltas if you have to. But if yj converges to L-M, then is L-M in Y? Y is closed, right?
 
  • #17
utleysthrow said:
And if that is correct, it would follow that L-M is in Y since Y is a closed set. Then M+(L-M)=L is in X+Y

Yes. So X+Y is closed. Right?
 
  • #18
Right. I have kind of a silly question though... how can we assume that xi+yi converges to anything?
 
  • #19
We're assuming that X+Y is not closed. So there's a value
L such that xi+yi converges to L (xi+yi is some sequence of points in X+Y). We're assuming we pick a sequence that converges
 
Last edited:
  • #20
utleysthrow said:
Right. I have kind of a silly question though... how can we assume that xi+yi converges to anything?

Go back to post 9. We are taking L to be a limit point of X+Y because we want to show X+Y contains all of it's limit points. What's the definition of 'limit point'?
 

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