# Homework Help: Sequences, sets and cluster points

1. Feb 24, 2012

### zbr

Hello all,

I am having trouble with a homework problem. The problem is as such:

Let a = {zn = (xn,yn) be a subset of ℝ2 and zn be a sequence in ℝ2 such that xn ≠ xm and yn ≠ ym for n≠m.
Let Ax and Ay be the projections onto the x and y axis (i.e. Ax = {xn} and Ay = {yn}. Assume that the cluster points of Ax = {0,1} and the cluster points of Ay = {0,1}.

a) Prove that the cluster points of A (denoted A') are a subset of {(0,0), (0,1), (1,0), (1,1)}.
b) Can A' = ∅? If yes, give an example; if no provide a proof.
c) Assume that A is bounded. Can A' consist of a single point? If yes, give an example; if no provide a proof.

Here is what I have so far:
a) Here I am confused as to why this statement is not an equality. For example, if we consider the point (0,0), then for every r A∩Br(punctured) is non empty since 0 is a cluster point of both the x and y projections. Furthermore, shouldn't this set include more points i.e. (1,y) where y is in yn? I am most likely thinking of this wrong... and my confusion here is preventing me from attempting the other two parts of the problem. Any and all help is much appreciated!

2. Feb 24, 2012

### Dick

Think about the sequence (0,1), (1,0), (0,1), (1,0), ... Is (0,0) a cluster point of that?

3. Feb 24, 2012

### zbr

In this case (0,0) would not be a cluster point of the sequence, but I would also argue that 0 nor 1 is a cluster point of the projections. This is because if we consider the projection on the x-axis:
Ax ∩ B0.5(0)(punctured) is empty since we have a punctured ball centered around 0 with radius 0.5 and there are no other points apart from 0 in this radius in the set Ax

Am I visualizing this wrong?

4. Feb 24, 2012

### Dick

No, you aren't. 0 isn't strictly a cluster point. But can't you think how to change my sequence so it is?

5. Feb 24, 2012

### zbr

Well here i am thinking:

(0,1), (0,2), (1,0), (2,0), (0,1/2), (0,3/2), (1/2,0), (3/2,0), (0,1/3), (0,4/3), (1/3,0), (4/3,0)...

In this cae both the x and y projections would have cluster points at 0 and 1, and here I'm pretty sure that (0,0) is a cluster point, but then the point (0,1) wouldn't be a cluster point and nor would (1,0) nor would (0,1)... How about the point (1,1) is that a cluster point? I am fairly sure it is not but again my visualization could be wrong here...

But back to the proof for part a) to show that A' is a subset of {(0,0), (0,1), (1,0), (1,1)}, I have started out by saying that if x in A is not one of these points then it is not a cluster point, from here do I then need to simply show that these points can be cluster points?

6. Feb 24, 2012

### Dick

I'm having a hard time visualizing what you intend by that sequence. Let's just try and straighten out the example where (0,0) is not a cluster point first before continuing. Define a sequence whose nth term is (1/n,1/n). Now add that to the sequence I suggested. What are the cluster points for the ordered pairs and the projections?

7. Feb 24, 2012

### zbr

My apologies for making a confusing example.

If I am understanding you correctly, I believe you mean the following sequence:
(0,1), (1,1), (1,0), (1/2,1/2), (0,1), (1/3,1/3), (1,0), (1/4,1/4), ...
So here the cluster points of the projections would be 0 for both cases. But wouldn't (0,0) be a cluster point for our set A here? Since, if you take any punctured ball centered around (0,0) you can find points inside that ball that are also in A, namely (1/N,1/N) where N > 1/(radius of the ball)?

To clarify, for the example I gave earlier, I meant it for the last 4 nth terms to be as follows:
(0,1/n), (0, 1/n + 1), (1/n,0), (1/n + 1,0)

8. Feb 24, 2012

### Dick

Let's change that to the pattern (1 + 1/n,1/n), (1/n, 1/n + 1), and repeat that pattern as n increases, that's close to what I meant. Now what are the cluster points?

9. Feb 24, 2012

### zbr

Yes, ok so here the cluster points of the projections would be {0,1} for both projections.
The cluster points of the sequence then would be (0,1) and (1,0) but not (1,1) or (0,0). Ok so that clarify's the equality situation, thanks!

However, for the roof I am still a bit stumped... Essentially I am not sure exactly how to start to show that a cluster point of A has to be one of those four points. Initially I thought to show that any of those four points can be cluster points, but I don't think that is the right way to go with it...

10. Feb 24, 2012

### Dick

Prove it by contradiction. Suppose (a,b) is a cluster point where either a or b is not in {0,1}. What can you say about the projection?

11. Feb 24, 2012

### zbr

Ok here is what I have:

Assume that (a,b) is a cluster point with one of a or b not being 0 or 1.
Then we have a A ∩ Br(a,b)(punctured) ≠ ∅ for all r.
This implies that Ax ∩ Br(a)(punctured) ≠ ∅ and Ay ∩ Br(b)(punctured) ≠ ∅ which implies that x is a cluster point of Ax and b is a cluster point of Ay, but one of x or y is not 0 or 1 so we have a contradiction. Therefore A' is a subset of {(0,0), (0,1), (1,0), (1,1)}.

Is my reasoning correct?

12. Feb 24, 2012

### Dick

Actually, this has got me all worried now. Take the pattern (1 + 1/n,1/n), (1/n, 1/n + 1), (2, 1/n). Then with your punctured disk reasoning about cluster points, 2 is not a projected cluster point. But (2,0) is a cluster point of the sequence. Agree? With your strict definition of cluster point, a) would seem to me to be false.

Last edited: Feb 24, 2012
13. Feb 24, 2012

### zbr

I would agree with you that 2 is not a projected cluster point and that (2,0) is a cluster point of the sequence... but wouldn't (2,0) be a cluster point using any disk, regardless of it being punctured or not?

The definition that we have been given in class of cluster points is as such:
x in ℝd is a cluster point of A (a subset of ℝd) if for every positive r A ∩ Br(x)(punctured) ≠ ∅

14. Feb 24, 2012

### Dick

That's exactly what I think. So we agree then. Given that definition of cluster point, the theorem in a) is false. So there is no point in trying to prove it. Somebody wasn't thinking very hard when they made up this exercise. It's pretty sloppy. You can make another definition of a limit point as one where every disk (nonpunctured) contains an infinite number of elements in a sequence. Then you can do it. I'd ask about this.

Last edited: Feb 24, 2012
15. Feb 24, 2012

### zbr

There is an issue with the example you provided, it does not satisfy the property that xm ≠ xn for m ≠ n, so that example doesn't work for this question... so I believe the proof that I gave earlier is the correct way to procede with the first part of that question.

16. Feb 24, 2012

### Dick

You are very sharp. You'll do well in this course. Thanks for pointing that out. I forgot about that. That gets rid of the repeated point problem. Yes, your reasoning before is correct. Duh. So what do you think about b)? Read ahead to c) and think of letting some things being unbounded.

Last edited: Feb 24, 2012