If \(X\) is a metric space, \(E\subset X\), and \(x\) is a limit point of \(E\), then there exists a sequence \(\{x_n\}\subset E\) which converges to \(x\).
Solution
1. Use the definition of limit point.Proof: \(x\) is a limit point of \(E\) means every neighborhood of \(x\) contains a point of \(E\) other than \(x\).In particular, for each \(n \ge 1\), the ball \(B(x, 1/n)\) contains a point \(x_n \in E\) with \(x_n \ne x\).
2. The sequence \((x_n)\) converges to \(x\).Proof: for every \(\epsilon > 0\) choose \(N\) with \(1/N < \epsilon\); then for \(n \ge N\), \(d(x_n, x) < 1/n \le 1/N < \epsilon\).Hence \(x_n \to x\).
3. Q.E.D.