# Proof

• Let $$(X,d)$$ be a metric space.
• Let the points $$x_n$$ of a convergent sequence $$(x_n)_{n\in\mathbb N}$$ be all contained in $X$, i.e. $$x_n\in X$$, for all $$n\in\mathbb N$$.
• Let the limit $x=\lim_{n\to\infty} x_n$ be also contained in $$X$$.
• We have to show that $$U:=\{x_n:~n\in\mathbb N\}\cup \{x\}$$ is a compact subset of $$X$$.
• For let an open cover $(U_i)_{i\in I}$ of $U$ be given.
• Since $$x\in U$$ by definition, there is an index $$i^*\in I$$ with $x\in U_{i^*}$.
• Because $U_{i^*}$ is open, $U_{i^*}$ is a neighborhood of $$x$$.
• Because $$(x_n)_{n\in\mathbb N}$$ converges against $x$, there is an index $$N\in\mathbb N$$ such that $x_n\in U_{i^*}$ for all $$n > N$$.
• Moreover, every sequence member $$x_k$$ is covered by some open set $$U_{i_k}$$ for the finitely many indices $$k=0,1,\ldots,N$$.
• By construction, it follows that $U\subset U_{i_0}\cup U_{i_1}\cup\ldots\cup U_{i_N}\cup U_{i^*}.$
• Thus, we have found a finite subcover of $$U$$ in the given open cover $(U_i)_{i\in I}$.
• Therefore, $$U$$ is compact.

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### References

#### Bibliography

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