Which subsets of a discrete metric space are compact?

Question

anonymous · Answer

@eliassaab @zakron

anonymous · Answer

@ganeshie8

perl · Answer

what is the definition of a compact set

anonymous · Answer

a set A is said to be compact if every opencovering of A has a finite subcovering

anonymous · Answer

@SolomonZelman  @perl

anonymous · Answer

Every finite subset.

anonymous · Answer

I need the proof too

anonymous · Answer

@Alchemista  thanks, can you help me with a few more, i will post in different post.

anonymous · Answer

@Alchemista  i am having a bit of problem .. can you clarify?

anonymous · Answer

Hmm?

anonymous · Answer

in the fourth line of the proof, G is a open cover of U or X?

anonymous · Answer

What you called G is actually a cursive C, but it is an open cover of $U$.

anonymous · Answer

yeah i tried to copy it, but it was showing C so i wrote G

anonymous · Answer

An open cover for some set $U$ is a collection of open sets whose union is $U$. Clearly every set in $\mathscr{C}$ is open, as mentioned in the proof. Furthermore since it contains every point in $U$ the union is clearly $U$. Do you see why it is an open cover for $U$?

anonymous · Answer

Sorry, whose union contains $U$.

anonymous · Answer

yeah i got that part. next from line number 5 you say

anonymous · Answer

Do you see why $\mathscr{C}$ is an open cover with no finite subcover?

anonymous · Answer

In the first part.

anonymous · Answer

yeah i got it.. Since C consists of only isolated points, C cannot have a finite subcover, since U is infinite

anonymous · Answer

If you leave out any set in the collection of sets $\mathscr{C}$ you are missing a point. Every set in the collection only contains a single point.

anonymous · Answer

And a subcover must include every point (the union must contain $U$)

anonymous · Answer

So you must take them all, but there are an infinite number of sets in the collection.

anonymous · Answer

Therefore any subcover cannot be finite.

anonymous · Answer

now any finite subset is compact.. i am looking into that part

anonymous · Answer

Power set is finite?

anonymous · Answer

The power set is the set of all subsets.

anonymous · Answer

yeah..

anonymous · Answer

Think, why must any cover be a subset of the power set.

anonymous · Answer

Ok it isn't quite right let me fix it a bit.

anonymous · Answer

okayy

anonymous · Answer

Consider any discrete metric space $X$. Let $U \subset X$ be an arbitrary subset. Suppose $U$ is not finite. We show it is not compact. Consider the covering $\mathscr{C} = \{\{x\} : x \in U\}$. Since each $\{x\} \subset X$ in a discrete metric space is a clopen set, clearly $\mathscr{C}$ is a open cover. However, $U$ is not finite so there is no finite subcover since any subcover must include all of the elements of $\mathscr{C}$ since it consists of only isolated points. So if $U$ is not finite it is not compact.

Now suppose $U$ is finite. Now consider any open cover $\mathscr{C}$. We can find a finite subcover as follows. For each $x \in U$ simply pick a set in $A \in\mathscr{C}$ containing $x$. In other words our finite subcover is as follows.
$C = \{A \in \mathscr{C} \text{ and } x \in A  : x \in U\}$

anonymous · Answer

Do you see why that works?

anonymous · Answer

did you change anything in the first part??

anonymous · Answer

Only a minor grammatical error/typo, the logic is the same.

anonymous · Answer

For the second part, let me clarify. Since $\mathscr{C}$ covers $U$, for every $x\in U$ there must be some open set in $\mathscr{C}$ that contains $x$. So we can simply pick, for each $x \in U$ an open set from our cover to form a subcover. Since $U$ is finite we made a finite number of selections for our subcover. So clearly we have a finite subcover.

anonymous · Answer

hmm i got it...

anonymous · Answer

thanks..