# Proof

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

### "$\Rightarrow$"

• 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, $x\not\in A$.

### "$\Leftarrow$"

• 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.

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### References

#### Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984