# Proof

(related to Theorem: Theorem of Bolzano-Weierstrass)

• Let $X$ be a metric space.
• Let $$A\subset X$$ be a compact subset.
• Let $(x_n)_{n\in\mathbb N}$ be a sequence of points $$x_n\in A$$.
• We will prove by contradiction that $(x_n)_{n\in\mathbb N}$ contains a subsequence $(x_{n_k})_{k\in\mathbb N}$, which converges against some point $$a\in A$$.
• Assume, there is no such subsequence.
• Then for every $x\in A$ we can choose arbitrary small $$\epsilon_x > 0$$ such that the open ball $B(x,\epsilon_x)$ contains only finitely many sequence members. (Otherwise we could construct a subsequence convergent against $x$, since we can choose $$\epsilon_x$$ arbitrarily small).
• Note that $(B(x,\epsilon_x))_{x\in A}$ is an open cover of $A$, i.e. $$A\subset \bigcup_{x\in A} B(x,\epsilon_x).$$
• Since $A$ is compact, there are finitely many points $$x_1,\ldots,x_n\in A$$ forming with their open balls a finite open subcover of $A$, formally $$A\subset \bigcup_{k=1}^n B(x_k,\epsilon_{x_k}).$$
• But then, the sequence $(x_n)_{n\in\mathbb N}$ would contain only finitely many points, which is a contradiction.
• Therefore, the assumption that there is no convergent subsequence $(x_{n_k})_{k\in\mathbb N}$ of $(x_n)_{n\in\mathbb N}$ is wrong.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

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