Very badly stuck (Prove that X is compact)

[rumjum]

Homework Statement

If X is a metric space such that every infinite subset has a limit point,
then prove that X is compact.

Homework Equations

Hint from Rudin: X is separable and has a countable base. So, it has
countable subcover {Gn} , n=1,2,3... Now, assume that no finite sub
collection of {Gn} covers X and take complement Fn of Union of {Gn}
n=1,2...n. Now, Fn is nonempty as it shall have some element of X not covered
by the finite subcover of {Gn}. Now, if E is a set that contains one
element from each of Fn, then consider a limit point of E & obtain a
contradiction.

The Attempt at a Solution

My attempt:

I can't understand the last part of the hint as how to obtain a contradiction.
Here is my attempt anyways.Please do help.

Given that X has a countable base , say, {Vn} , such that for all x belongs
to an open space G, x belongs to some Vi. In other words,
Gn = Union of {Vn}, say from i = 1,2...n. (finitely many).
Hence, every open cover of X shall have countable subcover

Now,let there be an open cover G1 U G2 ... Gn U..
such that X is a subset of this cover. And
let there be no sub-cover such that X is a subset of {Gi} i=1,2...n.
(finitely many).

Let Fn = complement of (G1 U G2 ...Gn).i.e, complement of any combinations
of Gi's. Now, Fn is non empty as any finite collection of Gi's do not cover
X. Let us chose a point from each Fn, and form a set E.

Now, E shall be an infinite subset of X as otherwise, there shall be finite
number of Fn's. Hence, E should have a limit point. Let the limit point be
"p".

[I am stuck as to what the contradiction should be]. Pls. help!

 

[Dick]

X is the union of all of the G's and p is in X. So p is in one of the Gn's and the Gn's are open. Do you see it now?

 

[rumjum]

I think I do see what you are saying. If I understand this correctly, "p" is in one of the Gn's as you said and Gn is open. Hence, all points of Gn are internal points. In other words, one can find a neighborhood Nr(p) such that it is a subset of a Gn. But E contains points of Fn that are not in Gn, and so the intersection of Nr(p) with E shall give finitely many points of X or E. [The only other way one can obtain infinitely many points of intersection is such that for m >n, the {Gm} covers X entirely.

Am I on the right track?

 

[Dick]

Right. For m>n, where Gn is the set that contains p, the sequence point m was chosen so that it was in Fm, so it's not in Gn. So you know it can't be very close to p. Contradiction. So the sequence doesn't have a limit point.

 

[rumjum]

Thank you very much!