## "Closed" set in a metric space

1. The problem statement, all variables and given/known data
1) Let (X,d) be a metric space. Prove that a "closed" ball {x E X: d(x,a) ≤ r} is a closed set. [SOLVED]

2) Suppose that (xn) is a sequence in a metric space X such that lim xn = a exists. Prove that {xn: n E N} U {a} is a closed subset of X.

3. The attempt at a solution
Let B(r,a)={x E X: d(x,a) < r} denote the open ball of radius r about a.
Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.
Definition: Let F be a subset of X. F is called closed iff whenever (xn) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F)
Theorem: F is closed in X iff Fc is open.

I know the definitions, but I just don't know out how to construct the proofs rigorously...

May someone kindly help me out?
Any help is appreciated!

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 Recognitions: Homework Help for 2) you could try picking any sequence of points and show the limit of the sequence is within the set...
 Recognitions: Homework Help for 1) you could try looking at the complement and showing it is open...

## "Closed" set in a metric space

There is a very easy way: prove that, in any metric space and relative to the topology it generates, the distance function is always continuous (here, you need to prove only that it's continuous with one of the arguments fixed).

 Quote by lanedance for 2) you could try picking any sequence of points and show the limit of the sequence is within the set...
So I start the proof by saying that "Let (yk) be any sequence in the set {xn: n E N} U {a}". How can we PROVE that the limit of ANY sequence is within the set?

 Quote by lanedance for 1) you could try looking at the complement and showing it is open...
How can we PROVE that {x E X: d(x,a) > r} is open?

 Quote by JSuarez There is a very easy way: prove that, in any metric space and relative to the topology it generates, the distance function is always continuous (here, you need to prove only that it's continuous with one of the arguments fixed).
I haven't learnt this theorem yet, so it's best for me to do the proofs using first principoles.

Thank you!

 Then the crucial first principle here is the triangle inequality.
 1) OK, for this one, I proved that the complement is open by using the definition of open set and the triangle inequality, so this problem is solved. But I'm still interested if there is a way to solve this problem directly using the definition of "closed" set, or perhaps proof by contradiction. Does anyone have an alternative proof? 2) So now I'm left stuck with this one... Thanks for any help!

 I'm still interested if there is a way to solve this problem directly using the definition of "closed" set
Let $x_n \in \bar{B_r\left(a\right)}$ be a convergent sequence in the closed ball $\bar{B}_r\left(a\right)$. Now prove that its limit satisfies $d\left(a,x\right) \leq r$ (by using the definition of limit and the triangle inequality).

For 2), use the sequential definition of closed set. A sequence in the set {xn: n E N} U {a} must be what?

 Quote by JSuarez Let $x_n \in \bar{B_r\left(a\right)}$ be a convergent sequence in the closed ball $\bar{B}_r\left(a\right)$. Now prove that its limit satisfies $d\left(a,x\right) \leq r$ (by using the definition of limit and the triangle inequality). For 2), use the sequential definition of closed set. A sequence in the set {xn: n E N} U {a} must be what?
1) Suppose xn->x. For all ε>0, there exists N s.t. n>N=>d(xn,x)<ε.
Now d(x,a) ≤ d(x,xn)+d(xn,a)< ε+ r by the triangle ineqaulity.
But how to prove that d(x,a) ≤ r???

2) What do you mean by the "sequential definition of closed set"? Is this the definition of closed set that I outlined in my first post?

Thanks!

 But how to prove that d(x,a) ≤ r?
Well, you proved that d(x,a) ≤ ε + r, for arbitrary ε>0; is then possible that d(x,a)>r?

Regarding 2), yes. What is a convergent sequence in the set you want to prove that is closed?

 1) OK, I got it. So now I've seen two different proofs of it. 2) "Suppose that (xn) is a sequence in a metric space X such that lim xn = a exists. Prove that {xn: n E N} U {a} is a closed subset of X." But I still have no idea how to proof this one. So me start the proof as follows: Let B = {xn: n E N} U {a} Let (yk) be a sequence in B converging to c, we must show that c E B. But I have no idea how to show that c E B. Can someone help me, please?
 You must look at the problem carefully: what is your set B? What is a convergent sequence in that set? If yn is a convergent sequence in B, what must be its limit?

 Quote by JSuarez If yk is a convergent sequence in B, what must be its limit
I don't think there is a limit that it MUST be...the limit can be a number of different things...(yk) need not converge to a, it can converge to something else. How can we divide this problelm into the different cases (without missing any)? And how many different cases are there?

Also, are you referring to the theorem:
"a sequence (xn) converges to a iff every subsequence of (xn) also converges to a"?

Thanks!

 I don't think there is a limit that it MUST be...the limit can be a number of different things...(yk) need not converge to a, it can converge to something else.
Well, there are two cases: yk can have only a finite number of distinct terms, or an infinite one. What are the possibilities for the limit?

 "a sequence (xn) converges to a iff every subsequence of (xn) also converges to a"?
Yes, that's a key result here.

 But here I think we're dealing with something slightly different from a "subsequence" of (xn). We are taking terms from the set B = {xn: n E N} U {a}, so something like y1=x1,y2=x1,y3=x2,y4=x1,y5=a,y6=x1..., is possible, while for subsequences of (xn), we can't take the same term over and over again. Also, in the set B = {xn: n E N} U {a}, there is an extra element "a" which we're allowed to take. Thus, (yk) is actually a lot different from being a "subsequence" of (xn), isn't it? So how can we justify our claims formally?? Thanks for explaining!

 so something like y1=x1,y2=x1,y3=x2,y4=x1,y5=a,y6=x1..., is possible
That's why I said we have two cases; this is the first one. Now, for a sequence like that to be convergent in B, what must be its form, given that its limit must be unique?

More, you are not taking advantage of the fact that B a the set whose elements are the terms of a convergent sequence, plus its limit.

 An alternative definition of the closure of a set E is that for all x in the closure of E, there exists a sequence {x_n} in E converging to x. Note also that if a sequence converges, then every subsequence converges to the same point. Putting those two facts together properly will give you what you need for part 2.