1
$\begingroup$

Let $(M,d)$ be an arbitrary metric space and $S,T$ be subsets of $M$. Assume $S \subseteq T \subseteq M$. Then show that $S$ is compact in $(M,d) \iff S$ is compact in the metric subspace $(T,d)$.

Attempt: Suppose $S $ is compact in the metric space $(T,d)$, then for every open cover of $S$ in $T$, we can find a finite subcover that also covers $S$.

Let this collection be $\{A_1,A_2,\cdots A_p\}$ .Then, since each $A_i$ is open in $T$, it is also open in $M$.

Hence, if $S$ is compact in $T \implies S$ is compact in $M$ as well.

To prove the other way around : we need to prove that if $S$ is compact in $M$, then it's compact in $T$ as well

$ S$ is compact in $M \implies$ for every open cover of $S$ in $M$, we can find a finite subcover that also covers $S$.

Let this collection be $\{C_1,C_2,\cdots C_p\}$ .Then, since each $C_i$ is open in $M$, should it be open in $T$ as well?

Can it be a possibility that $C_j \cap T = \{ \phi \}$ for some $j$?

Did I attempt the first part of this problem correctly and how do I proceed to prove th other direction of the problem?

Thank you for your help.

$\endgroup$
2
$\begingroup$

There are a couple of problems here:

"Since each $A_i$ is open in $T$, it is also open in $M$"

This is false.

"Since each $C_i$ is open in $M$, should it be open in $T$"

This is false too.

"Can it be a possibility that $C_j \cap T = \emptyset$?"

Yes, but that's not an issue, because, as $S \subset T$ then $C_j \cap S = \emptyset$ as well.

Think about open sets in a subspace and you can right the two statements you have written here!

$\endgroup$
  • 1
    $\begingroup$ Hint: Write the open sets of $T$ as $T \cap O$, $O$ a open set of $M$ $\endgroup$ – Jonas Gomes Oct 2 '14 at 21:57
  • $\begingroup$ Each $A_i$ is open in $T \implies \exists~$ an open Ball $B(x,r) \subseteq T$ . Since, $T \subseteq M$, hence , $B(x,r) \subseteq M$. Hence, $A_i$ is open in $M$. Did I go somewhere wrong in this inference? $\endgroup$ – MathMan Oct 2 '14 at 21:57
  • 1
    $\begingroup$ What if $M = \mathbb{R}$, $T = [0,1]$ and $A_i = [0,\frac{1}{2}[$? $A_i$ is a open set of $T$ but not of $\mathbb{R}$. $\endgroup$ – Jonas Gomes Oct 2 '14 at 22:00
  • $\begingroup$ Oh right. However, this result would have been true in $\mathbb R^k$ right? So, I will attempt to write the open sets of $T$ as $T \cap O$ where $O$ is an open set of $M$. I am thinking on the lines of this theorem : if $X \subseteq S , X$ is open in $S \iff X = A \cap S$ for some $A$ open in $M$ $\endgroup$ – MathMan Oct 2 '14 at 22:06
  • $\begingroup$ If you mean the result would have been true with $T = \mathbb{R}^k$? Or with $M$? The crucial point is that $T$ must be open in order for every open subset of $T$ also be a open subset of $M$. $\endgroup$ – Jonas Gomes Oct 2 '14 at 22:24
1
$\begingroup$

If you are allowed to use the equivalent definition:

$K$ is compact if and only ifevery sequence $\{x_n\}_{n\in\mathbb N}\subset K$ possesses a converging subsequence with its limit in $K$.

Then, in this definition it does not matter whether $K\subset M$ or $K\subset N$.

$\endgroup$
1
$\begingroup$

You're wrong, you are using what you need to prove.

you can easily prove that $A$ is open in a subset $Y$ of a metric space $X$ iff $A=Y\cap B$ with $B$ an open set in $X$..

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.