Let X and Y be sets satisfying Y X (- , ). Suppose that X is compact and that for every x X, there is a neighborhood of x which contains only a finite number of points of Y. This number may vary with x. Show that Y is a finite set.
What I have so far is that since X is compact it is closed and bounded and since it is closed there is an accumulation point in each neighborhood of X. then the neighborhood contains the point (x- ,x+ . And the Ix's are open sets that contain X.
I'm not sure what to do next and/or how to actually do the proof of it, hope someone can help. thanks
B-W says that every bounded sequence in R^n has a convergent subsequence. So suppose Y is infinite. Then Y has a countably infinite subset. Enumerate it as . Then there's a convergent subsequence . But since X is closed and every element of the sequence is in X, the limit is in X. Let L be the limit. Then every neighborhood of L contains infinitely many elements of the sequence, which is a contradiction.
Thats what i meant, contradiction! lol So if i were to show this with B-W-T do I still need to state that since X is closed there is an accumulation point in each neighborhood of X and that the neighborhood (x-epsilon, x+epsilon)? or can am I allowed to just go straight into the B-W?
Rather than say there's an accumulation point in some neighborhood, you want to first show that (at least) one accumulation point exists in X (L in my post above), then say that any neighborhood of this point must contain infinitely many elements of Y, contrary to the hypothesis. You don't need any epsilonics here. (Or maybe you do...if this is all analysis and no topology...in that case, yeah, say that contains a point of Y distinct from L, for any epsilon, so in fact it contains infinitely many for any epsilon.)