(related to Lemma: Characterization of Closed Sets by Limits of Sequences)

Let $X$ be a metric space and let \(A \) be its subset.

- Let $A$ be closed.
- Take any convergent sequence $(x_k)_{k\in\mathbb N}$ with $x_k\in A$ for all $k\in\mathbb N$ with $\lim_{k\to\infty} x_k=x$.
- Assume, $x\not\in A$.
- Because \(X\setminus A\) is open, \(X\setminus A\) is a neighborhood of \(x\).
- By the definition of convergence there is a \(N\in\mathbb N\) such that \(x_k\in X\setminus A\) for all \(k\ge N\).
- This contradicts $x_k\in A$ for all $k\in\mathbb N$.

- Let $x\in A$ for any convergent sequence $(x_k)_{k\in\mathbb N}$ with $x_k\in A$ for all $k\in\mathbb N$ with $\lim_{k\to\infty} x_k=x$. Further let \(x\in A\).
- We have to show that $A$ is closed.
- Assume, $A$ is open.
- Then \(A\) is a neighborhood of \(x\).
- Therefore there is an \(\epsilon > 0\) such that the open ball \(B(x,\epsilon)\) is contained in \(A\).
- Since $(x_k)_{k\in\mathbb N}$ was an arbitrary sequence, we can choose it in a way that \(x\) is contained in the boundary \(\delta A\).
- But then, for all \(\epsilon > 0\), the open ball \(B(x,\epsilon)\) will have points in common with \(A\) and with \(X\setminus A\).
- This contradicts $A$ being open.∎

**Forster Otto**: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984