Question

Prove that (x₀) is a limit point of (S) iff there is a sequence of distinct points (xₙ) in (S) such that (xₙ) converges to (x₀).

a) True
b) False

Answer 1

True

A limit point of a set S is a point x₀ such that every neighborhood of x₀ contains infinitely many points of S. The given statement is true: (x₀) is a limit point of (S) if and only if there is a sequence of distinct points (xₙ) in (S) such that (xₙ) converges to (x₀).

Step-by-step explanation:

A limit point of a set S in a metric space is a point x₀ such that every neighborhood of x₀ contains infinitely many points of S.

To prove that (x₀) is a limit point of (S) if and only if there is a sequence of distinct points (xₙ) in (S) such that (xₙ) converges to (x₀):

1. If (x₀) is a limit point of (S), then for every positive integer n, there exists a point xₙ in the intersection of the neighborhood Nₙ(x₀) and S, where Nₙ(x₀) is a neighborhood of x₀. The sequence (xₙ) is then a sequence of distinct points in S that converges to x₀.
2. If there is a sequence of distinct points (xₙ) in (S) such that (xₙ) converges to (x₀), then for every positive integer n, (xₙ) is contained in the neighborhood Nₙ(x₀). Since each neighborhood Nₙ(x₀) contains infinitely many points of S, (x₀) must be a limit point of S.

Therefore, the statement is true.