analysis (show that E has infinitely many points)?

Question

Let E be a subset of X, and E has a cluster point. Show
that E has infinitely many points.

I need the process, thank you~!

Answer 1

I'll assume X is a metric space. Suppose e is a cluster point of E. The ball B(e,1) contains a point e_1 of E, e_1 ≠ e. The B(e, d(e1, e)/2) contains a point e2 of E,

e2 ≠ e. Note that e2 ≠ e1 since d(e2, e1) ≥ d(e1,e) - d(e2,e) > d(e1,e) - d(e1,e)/2 = d(e1,e)/2 > 0. Continue the process to obtain a sequence S = {e_n} ⊂ E with e1 ≠ e_2 all for all n > 1, e_n is in B(e, d(e(n-1),e)/n) . Hence for all n > 1,



d(e_n, e_(n-1)) ≥ d(e_(n-1), e) - d(e_n, e)



> d(e_(n-1), e) - (1/n)d(e_(n-1), e) = (n - 1)/n d(e_(n - 1), e) > 0.



This implies all the elements of S are distinct. So S is an infinite subset of E, and E has infinitely many points.



Edit: I corrected the mistake in the first in second lines. Note that in the second line of the inequality, I only changed the > into a = in the 2nd to last string. Recall e_n is in B(e, d(e_(n-1),e)/n), so d(e_n, e) < d(e_(n-1), e)/n. Thus d(e_(n-1), e) - d(e_n, e) > d(e_(n-1), e) - (1/n) d(e_(n-1), e) = (n - 1)/n d(e_(n - 1), e) > 0. Note that 1/2 is not present in the second line.



Edit: The proof I gave is constructive, but here is an indirect way to prove the statement. If E has finitely many points, say, E = {e_1,...,e_n}, then since each one-point set {e_i} is closed and the union of finitely many points are closed, each subset of E is closed. By complementation, each subset of E is open. So each {e_i} is open and none of the points are cluster points. This argument works in general when X is a T1 topological space.

Answer 2

Actually, this is true in every topological space where singletons are closed (this includes every Hausdorff space and, therefore, every metrizable space). Recall that, in such spaces, finite sets are closed.



Let a be a cluster point of E. Suppose a has a neighborhood V that contains only finitely many points of E. Then, the set W =E ∩ V - {a} is finite and, therefore, closed. Since V is open and W ⊂ V, then V' = V - W is open, contains a and contains no element of E distinct from a. This means that V' is a neighborhood of a that, with possible exception of a iself (if a ∈ E), contains no point of E. So, a, contrarily to the assumption, is not a cluster point of E.



It follows that every neighborhood of a contains infinitely many elements of E, which, in turn, sows E is infinite.

Answer 3

Hint: Instead show that if E is finite, then it can't have a cluster point.