# Analysis Compactness Question

1. Oct 20, 2013

### HumbleHarry

1. The problem statement, all variables and given/known data
Let ${X_n}$ be a sequence of nonempty compact subsets of the metric space Ω such that ${X_n+1}$ $\subseteq$ ${X_n}$, with $n: 1→∞$. Prove that their intersection is nonempty.

**By the way, I mean to subscript "n+1", not have ${X_n}$ + 1 as it seems.

2. Relevant equations
One can use sequential compactness to describe the compactness of each subset. Every sequence in ${X_n}$ has a subsequence ${X_k}$ where $x_k$→$x_o$ $\in$ ${X_n}$.

3. The attempt at a solution
I tried using sequential compactness to find some limit point that's found in ${X_n}$ and all of its subsets, but I have no clue on how to do so. I've tried setting up singleton sequences around the points in each nonempty set to say they must have a subsequence that converges to said point, implying said point is in ${X_n}$, but I don't know if this implies that these points lie in the other sets, as well.
On another line of thought, since ${X_n+1}$ is a subset of ${X_n}$, there's some sequence in ${X_n}$ that's also in ${X_n+1}$, both of which should converge to a point in ${X_1}$.
Help would be appreciated.

2. Oct 20, 2013

### Dick

Start by picking any point $x_n$ in each $X_n$. The sequence $x_n$ has a convergent subsequence, right? If L is the limit of the convergent subsequence, what can you say about what sets it must belong to?

3. Oct 20, 2013

### HumbleHarry

When you say $x_n$, first as a point and then a sequence, do you mean the single point sequence ${x_n}$?
Shouldn't $L$ be in $X_n$, implying $L$ is in every $X_n$ before it, up until $X_1$?

4. Oct 20, 2013

### Dick

No, I mean a sequence of possibly different points. Each point $x_n$ is a element of the corresponding set $X_n$. So the sequence is $x_1,x_2,x_3,...$ with $x_1$ in $X_1$, $x_2$ in $X_2$, etc etc.

5. Oct 20, 2013

### HumbleHarry

Oh, so what you're doing is creating a subsequence of points from each $X_n$, and saying that this sequence has a limit $L$?
In this case, shouldn't the limit lie in $X_1$, and since it's a limit point for a sequence containing points from each $X_n+1$ it should get arbitrarily close to $x_n$ in the sequence after a certain point? This would imply that some $X_n+k$ will eventually contain points infinitesimally close to $L$, and these points would be in all sets before and after it, implying the intersection of these sets are nonempty.

6. Oct 20, 2013

### Dick

You might have the right idea, but it's hard to tell from your description of it. Could you try to write that down in a clearer way (i.e. more like a proof)?

7. Oct 20, 2013

### HumbleHarry

Sorry:
Create a sequence ${x_n}$ where $x_n$ $\in$ $X_n$.
The sequence ${x_n} \subseteq {X_1}$.
Since the sequence of ${X_n}$ is compact, $\forall ε >0$, $\exists x_k$, $k \in N$ such that $d(x_k, L) < ε$ for $k \geq n$, and where $L$ is the limit point of the subsequence.
$\forall ε$ create a ball $B(L,ε)$. This ball contains $x_k$ for $k \geq n$.
This implies that the ball contains points from $X_n$ whose $n \leq k$, and since $X_n+1 \subseteq X_n$, $B(L,ε) \subseteq X_n$ whose $n<k$.
Thus any element from the ball is an intersection point of these $X_n$.
Is this right?

8. Oct 20, 2013

### Dick

You are saying both more and less than you have to. Here, I'll get you started. Start with the sequence $x_1,x_2,x_3,...$ that we chose before. Now you can't just assume that is convergent. But because $X_1$ is compact and that is a sequence in $X_1$ (since all of the other sets are subsets of $X_1$), then it has a convergent subsequence, call that $x_{k_1},x_{k_2},x_{k_3},...$ that does converge to some limit L. Now, sure, L must be in $X_1$. Now can you try and tell me clearly why L has to also be in, say for example, $X_{100}$?

Last edited: Oct 20, 2013
9. Oct 20, 2013

### HumbleHarry

Hmm... would it be because if $L \not\in X_100$, then that would imply that the ball $B(L,ε)$ would not be a subset of $X_100$, though we've established that when $n \geq k$, $\exists k \in N$, $x_k \in B(L,ε) \subseteq X_(n \geq k)$ for $\forall ε > 0$, implying $X_(n \geq k) \subseteq X_100$?

10. Oct 20, 2013

### Dick

Sort of. But no talk of balls is needed. Can you think of a way to define a sequence that converges to L in $X_{100}$?? And don't start from the beginning. There's an easy way to find a subsequence of $x_{k_1},x_{k_2},x_{k_3},...$ that does.

11. Oct 20, 2013

### HumbleHarry

Hmm... I am really pulling a blank here.
I know the general definition of a convergent sequence is that $\exists k \in N$, such that $d(x_n,L)< ε$, $\forall ε > 0$, $k \geq n$.
Maybe construct a sequence $x_kn$ where $d(x_kn,L)<ε$, for $ε>0$? Would this just be the terms from $x_n$ whose index is $k \geq n$ in relation to each term being infinitesimally close to L? Such a sequence would obviously converge to L.
I am sorry if this is not it.

Last edited: Oct 20, 2013
12. Oct 20, 2013

### Dick

Don't be sorry. I know you've got exactly the right idea. You just aren't expressing it as clearly as you might. Talking about the balls is just making in confusing. If you know $x_{k_1},x_{k_2},x_{k_3},...$ converges to L then $x_{k_{100}},x_{k_{101}},x_{k_{102}},...$ also converges to L. The neat way to do this is to prove that $x_{k_{100}}$ must be in $X_{100}$ and so must $x_{k_{101}}$ etc. But you don't need to be that neat. If $x_{k_j}$, for ANY j, is in $X_{100}$ then so are $x_{k_{j+1}}$, $x_{k_{j+2}}$ etc. Remember that $x_{k_j}$ is in $X_{k_j}$ and $k_j<k_{j+1}<k_{j+2}<...$ since it is a subsequence.

13. Oct 21, 2013

### HumbleHarry

I see what you mean, actually, that's what I thought concerning $x_kj+c \in X_j$ for $c \in N$, ie $x_k100+c \in X_100$.
I think I can write a coherent proof now that you've helped me resolve my questions. Thank you very much for having the patience to help me! I appreciate it.