◀ ▲ ▶Branches / Topology / Proof

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