Compactness of Nested Subsets in Metric Spaces

[HumbleHarry]

Homework Statement

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.

Homework Equations

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

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.

 

[Dick]

Homework Statement

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.

Homework Equations

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

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.

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?

 

[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 $X_1$?

 

[Dick]

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 $X_1$?

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.

 

[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.

 

[Dick]

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.

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)?

 

[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?

 

[Dick]

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?

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, $X_{100}$?

 

[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 $X_(n \geq k) \subseteq X_100$?

 

[Dick]

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 $X_(n \geq k) \subseteq X_100$?

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.

 

[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.

 

[Dick]

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.

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.

 

[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.