Thread: Concepts in metric spaces

Hey,

I just moved from euclidean spaces to generic metric spaces, and I'm trying to avoid missing some basic concepts that links these spaces.

For example, in a metric space $\displaystyle (S,\rho) $, how can I define that some set is bounded? Can I say that $\displaystyle A \subseteq S$ is bounded if $\displaystyle \exists M \in \mathbb{R}$ s.t. $\displaystyle \rho (x,y)<M, \forall x,y \in A$ ? Assuming that S doesn't necessarily has a zero.

What I wanna show is that if $\displaystyle K \in S $ is compact, then every sequence in K has at least one convergent subsequence.

I can prove it in Rn using bolzano-weierstrass, but i'm unsure about what I can use in such a generic metric space.

Thanks in advance,
JP

Originally Posted by javeiro

Hey,

I just moved from euclidean spaces to generic metric spaces, and I'm trying to avoid missing some basic concepts that links these spaces.

For example, in a metric space $\displaystyle (S,\rho) $, how can I define that some set is bounded? Can I say that $\displaystyle A \subseteq S$ is bounded if $\displaystyle \exists M \in \mathbb{R}$ s.t. $\displaystyle \rho (x,y)<M, \forall x,y \in A$ ? Assuming that S doesn't necessarily has a zero.

What I wanna show is that if $\displaystyle K \in S $ is compact, then every sequence in K has at least one convergent subsequence.

I can prove it in Rn using bolzano-weierstrass, but i'm unsure about what I can use in such a generic metric space.

Thanks in advance,
JP

A set is bounded if there exists an real number r > 0 and $\displaystyle s \in S $ such that $\displaystyle S \subset B_r(s) $, which is the open ball of radius r around s. in other words, your definition is too strong, you just have to fix one of x or y.

As for your second question, what definition of compactness are you using? You cannot use the closed and bounded definition.

Your boundness definition looks good. Thanks!

I realize that the closed and bounded definition (heine-borel) is just for the Rn. Here, I'm considering that K is compact if for any arbitrary open cover of K there is a finite subcover.

I did not know there were multiple definitions of compactness, I'm checking that right now, but I'm pretty sure that this definition is the one my professor is using.

Originally Posted by javeiro

Your boundness definition looks good. Thanks!

I'm considering that K is compact if for any arbitrary open cover of K there is a finite subcover.

I did not know there were multiple definitions of compactness, I'm checking that right now, but I'm pretty sure that this definition is the one my professor is using.

There are a million different equivalent definitions of compactness in metric spaces. You are trying to prove that what I would consider to be the normal definition implies sequential compactness.

Try this implication

$\displaystyle \text{Compactness }\implies\text{ B.W.P.}$- Let $\displaystyle X$ be a compact metric space and $\displaystyle E\subseteq X$ infinite. Suppose that $\displaystyle E$ did not have a limit point. Then, around each point of $\displaystyle x\in X$ we may form a ball $\displaystyle B_{\delta}(x)$ such that $\displaystyle B_{\delta}(x)\cap E\subseteq\{x\}$. Clearly, the set of all these balls covers $\displaystyle X$ and so by definition it must have a finite subcover $\displaystyle \left\{B_{\delta_1}(x_1),\cdots,B_{\delta_n}(x_n)\ right\}$ and so $\displaystyle E=E\cap X=E\cap\left(B_{\delta_1}(x_1)\cup\cdots\cup B_{\delta_n}(x_n)\right)=\left(E\cap B_{\delta_1}(x_1)\right)\cup\cdots$$\displaystyle \cup\left(E\cap B_{\delta_n}(x_n)\right)\subseteq\{x_1,\cdots,x_n\ }$. But, this contradicts the assumption that $\displaystyle E$ is infinite. The conclusion follows. $\displaystyle \blacksquare$

$\displaystyle \text{B.W.P }\implies\text{ Sequential Compactness}$: Let $\displaystyle \left\{x_n\right\}$ be a sequence in $\displaystyle X$. If $\displaystyle \left\{x_n:n\in\mathbb{N}\right\}$ has only finitely man values it's easy to see there is a convergent subsequence, so assume not. Then, by assumption $\displaystyle \left\{x_n:n\in\mathbb{N}\right\}$ must have a limit point $\displaystyle \xi$. Then, merely take a subsequence $\displaystyle \{y_n\}$ of $\displaystyle \{x_n\}$ by choosing any point in $\displaystyle B_{\frac{1}{n}}(\xi)$. Either way the conclusion follows. $\displaystyle \blacksquare$

Suppose that $\displaystyle \mathcal{A}$ is an infinite subset of $\displaystyle \mathcal{K}$.
If $\displaystyle \mathcal{A}$ has no cluster point in $\displaystyle \mathcal{K}$ then $\displaystyle \left( {\forall x \in \mathcal{K}} \right)\left( {\exists \delta _x > 0} \right)\left[ {B(x;\delta _x )} \right]$ that contains at most one point in $\displaystyle \mathcal{A}$.
Use compactness to get a contradiction.

Thank you all.

What i had before was that every bounded and infinite set on Rn has a limit point, and that every bounded sequence on Rn has a convergent subsequence.

quick question: on (S,p) is not enough that the sequence is bounded to guarantee that it has a convergent subsequence cause we don't know if the limit is in S, right? if (S,p) is complete, I guess it is enough that the sequence is bounded, just like on Rn, right?

Thanks again!

Originally Posted by javeiro

Thank you all.

What i had before was that every bounded and infinite set on Rn has a limit point, and that every bounded sequence on Rn has a convergent subsequence.

quick question: on (S,p) is not enough that the sequence is bounded to guarantee that it has a convergent subsequence cause we don't know if the limit is in S, right? if (S,p) is complete, I guess it is enough that the sequence is bounded, just like on Rn, right?

Thanks again!

Well, yes and no. Another equivalent definition of compactness in metric spaces is that the space is totally bounded and complete. So, every compact metric space is complete.

Originally Posted by javeiro

on (S,p) is not enough that the sequence is bounded to guarantee that it has a convergent subsequence cause we don't know if the limit is in S, right? if (S,p) is complete, I guess it is enough that the sequence is bounded, just like on Rn, right?

I think that you are very confussed about these concepts.
Please take time to review all the replies and all definitions in your notes.
It appears to me that you have changed the focus of the post.

Hum, new stuff but very interesting stuff

Yeah, I'm kind of confused...but about the completeness of (S,p) I was just trying to guess...Hehe...

Thank you all very much!

Originally Posted by javeiro

Hum, new stuff
Yeah, I'm kind of confused...but about the completeness of (S,p) I was just trying to guess...Hehe...

Do not make fun of this.
I understand that you are at a very basic level.
It is unfortunate that we have responders who what to show how much they know rather than helping you.
Your question has to do with very basic concepts about compact sets in a metric space.
You need to understand the basic idea of compactness.
You get that down, then move on to ‘sequential compactness’.
This problem requires not more than basic definitions,