How do i prove this about cluster points?

Question

x is a cluster point of A contained in R iff every neighborhood of x has infinitely many points of A?

Answer 1

1) if

If every neighborhood of x contains infinitely many points of A, then every neighborhood of x contains an element of A distinct from x. Hence, x is a cluster point of A.


2) Only if

Suppose some neighborhood V of x contains finitely many elements of A. Then, the set V intersection A is finite and, therefore, closed (every finite subset of R is closed). It follows the set K = (V intersection A) - {x} is also finite and closed. Since V is an open set containing x, the set V - K is an open set containing x that, with possible exception of x, contains no element of A. So, V - K is a neighborhood of x that contains no element of A distinct from x, which shows x is not a cluster point of A. By contraposition, the conclusion follows.


I like that proof for the only if part because it's not restricted to metric spaces. It's valid in every topological spaces where sets composed of only one element are closed.

Answer 2

Let B(x, r) denote the open ball of radius r around the point x; that is, B(x, r) = {y in R such that |x - y| < r}.

Suppose that x is a cluster point of A. I will show that every neighborhood of x contains infinitely many points of A.

Let U be a neighborhood of x. I will construct a sequence {p_n} of distinct points contained in U.

Since U is a neighborhood of x, then U contains some ball B(x, r_1). Since x is a cluster point, then B(x, r_1) intersects A in some point; call it p_1. Set r_2 = |x - p_1|; note r_2 < r_1. Then B(x, r_2) is another neighborhood of x; since x is a cluster point, then B(x, r_2) intersects A. It can't intersect A at p_1, since p_1 lies at a distance r_2 from x, so it must intersect A at some other point p_2. Set r_3 = |x - p_2|; note r_3 < r_2 < r_1. Then B(x, r_3) is another neighborhood of x; since x is a cluster point, then B(x, r_3) intersects A. It can't intersect A at p_1 or p_2, since p_1 and p_2 lie a distance of at least r_3 from x, so it must intersect A at some other point p_3. By continuing in this manner, we may select an infinite sequence {p_n} of distinct points contained in our neighborhood U of x.

So U contains infinitely many points of A. Since U was an arbitrary neighborhood of x, then every neighborhood of x contains infinitely many points of A.

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Now suppose every neighborhood of x has infinitely many points of A. Then certainly every neighborhood of x intersects A in a point other than x itself. Hence x is a cluster point of A.