Thread: Metric Space Proof TFAE

1. Metric Space Proof TFAE

There is no rush for this. I am reading my analysis book and this is left without a proof... so if anybody wants to offer help, that would be great.

Prop: If A is a subset of a metric space X, then for a point x which is an element of X, TFAE:

(i) x is an accumulation point of A
(ii) For all ε>0, Uε(x)∩A is infinite.
(iii) For all ε>0, (Uε(x)\{x}) ∩A is infinite

2. We define x to be an accumulation point of a set A if every neighbourhood of x contains infinitely many points of A. Let $B_r(x)$ denote the open ball of radius r around x: then a set N is a neighbourhood of x if and only if it contains some open ball of positive radius and in particular of course those open balls themselves are neighbourhoods of x.

So if x is an accumulation point of A then every open ball, being a neighbourhood, contains infinitely many points of A: that is, (i) implies (ii).

An infinite set with one point added or deleted is still infinite, so (ii) is equivalent to (iii).

Now assume (ii). Suppose if possible that x is not an accumulation point of A, so that there is a neighbourhood N of x which contains only finitely many points of A. Since N is a neighbourhood, there is a positive r such that $N \supseteq B_r(x)$ and so $(A \cap N) \supseteq (A \cap B_r(x))$ and so the latter is finite. This contradicts (ii), and this contradiction shows that (ii) implies (i).