# If $S \subseteq T \subseteq M$. Then show that $S$ is compact in metric space $(M,d) \iff S$ is compact in the metric subspace $(T,d)$.

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?

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!

• Hint: Write the open sets of $T$ as $T \cap O$, $O$ a open set of $M$ – Jonas Gomes Oct 2 '14 at 21:57
• 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? – MathMan Oct 2 '14 at 21:57
• 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 $\mathbb{R}$. – Jonas Gomes Oct 2 '14 at 22:00
• 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 $M$ – MathMan Oct 2 '14 at 22:06
• 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 $M$. – Jonas Gomes Oct 2 '14 at 22:24

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 $K\subset N$.

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 $X$..