(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

If X is a metric space such that every infinite subset has a limit point,

then prove that X is compact.

2. Relevant 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.

3. 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!!

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# Homework Help: Very badly stuck (Prove that X is compact)

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