Sequential Compactness of Sets: A Definition

[Eclair_de_XII]

TL;DR

non-trivial intersection. Denote the sequence as $K_n$ with the property that for any $n$, $K_{n+1}\subset K_n$. Show that there is a point in $\cap_{n\in \mathbb{N}} K_n$.

Definition: A set is sequentially compact if all sequences contained in the set contain a subsequence that converges to a point in the set.

Let $N\in \mathbb{N}$ and suppose that $m\geq N$. Let $x\in K_m$. Since $K_m\subset K_{m-1}\subset \ldots \subset K_N$, it follows that $x$ is an element of all $K_n$ for $n\leq N$ and all $n\in [N,m]$.

I'm very sure I'm misunderstanding something very crucial here, since I didn't need to invoke the fact that these sets are sequentially compact.

 

[Office_Shredder]

Haven't you only shown that
$$x\in \bigcap_{i=1}^{m} K_i?$$ What about $K_{m+1}$?

 

[WWGD]

Doesn't this need assumptions on the host space, like it being complete, etc?

 

[Eclair_de_XII]

I cannot say without giving more information about what $x$ is, I think. If $x$ is for example, the point that a subsequence contained in $K_m$ converges to, then $x$ would belong to $K_{m+1}$ but only if $K_m$ did not contain any other sequences. Otherwise, you could choose $K_{m+1}$ that included an entirely different sequence in $K_m$ that converged to some point not equal to $x$.

You would have to give me more time to think about the case if $x$ belonged to a sequence in $K_m$. But as it stands, $x$ would not necessarily belong to $K_{m+1}$ because of the reasons stated.

host space

Sorry about that. That would be the real numbers; I'm working with a real analysis/elementary calculus hybrid book.

 

[Office_Shredder]

Ok, so it doesn't sound like you've proven the statement if you think $x$ might not be in $K_{m+1}$. Do you have any ideas about how to fix it?

 

[Eclair_de_XII]

Do you have any ideas about how to fix it?

I was thinking of proceeding algorithmically:

1.
Let $l$ be a positive integer and let $a_n$ be a sequence in $K_l$.
Then there must exist a subsequence $a_{n_k}\rightarrow a\in K_l$.

2.
Now consider $K_{l+1}$.
Let $b_n$ be a sequence in $K_{l+1}$ whose subsequence converges to some point $b\in K_{l+1}$.

3.
If $b=a$, then we have found points, namely $a$ and the members of the sequence that converges to it, belonging to $K_m$ for $m\leq l+1$.
If $b\neq a$, then $b\in K_{l+1}\subset K_l$ and the members of the sequence converging to $b$ are also in $K_l$.

4.
Repeat steps two and three for as many times as you wish.
No matter how many times you execute these steps, there will always be a sequence of points and a limit point contained in the intersection of all $K_l$ in the sequence of sets thus described.

 

[Office_Shredder]

In general you don't expect the sequence you construct to work for any of the deeper sets, so this process doesn't stabilize ever.

You have a sequence of sets and want to construct a sequence of points. I think the natural thing to do is to pick one point in each set to be in your sequence.

 

[Eclair_de_XII]

In general you don't expect the sequence you construct to work for any of the deeper sets, so this process doesn't stabilize ever.

Is your meaning that no matter how many times the algorithm is performed (let's call the number of times $N$), it will never be the case in step three where $b$ has to be equal to $a$ for all $n\geq N$?

I think the natural thing to do is to pick one point in each set to be in your sequence.

Oh, so by creating a sequence of points by selecting a point from each set in the sequence of sets, you can find a subsequence that converges to a point that I'm guessing is contained in every single element of the sequence of sets?

 

[Office_Shredder]

Is your meaning that no matter how many times the algorithm is performed (let's call the number of times $N$), it will never be the case in step three where $b$ has to be equal to $a$ for all $n\geq N$?

Exactly. At the very least you will need to use sequential compactness to prove that it works, since $K_n=(-1/n,0)$ will not let your algorithm succeed.

Oh, so by creating a sequence of points by selecting a point from each set in the sequence of sets, you can find a subsequence that converges to a point that I'm guessing is contained in every single element of the sequence of sets?

Why don't you give it a shot and see what happens?

 

[Eclair_de_XII]

Okay. For each $K_n$, choose $x_n$ to be appended to a sequence of points.

Doing so, we have a sequence $\{x_n\}$ contained in $K_1$.
There must exist a subsequence $\{x_{n_k}\}_{k\in\mathbb{N}}\subset K_{n_1}\subset K_1$ that converges to some $x\in K_1$ as a result.
Moreover, consider the subsequence $\{x_{n_k}\}_{k>1}$. This is a sequence contained in $K_{n_2}\subset K_{2}$. Since this sequence was originally the subsequence of a sequence converging to $x$, this sequence must also converge to $x$. Every subsequence of $K_{2}$ must converge to a point contained within $K_{2}$. Hence, $x\in K_{2}$.
Continuing in this fashion, we conclude that $x$ must be an element of $K_l$ for $l\geq 1$.

 

[Office_Shredder]

Looks good to me. I guess it's worth emphasizing that for $K_2$, we are only guaranteed thata subsequence of $x_{n_k}$, say $x_{n_{k_j}}$
, converges to some point, which must be $x$, so $x\in K_2$ (and obviously at that point, the whole sequence $x_{n_k}$ converges to $x$)