Proof
(related to Proposition: Compact Subset of Real Numbers Contains its Maximum and its Minimum)
- Let $A$ be a non-empty compact subset of the real numbers $\mathbb R$.
- We have to show that $\max(A)\in A$ and $\min(A)\in A$.
- According to the Heine-Borel theorem, \(A\) is bounded and closed.
- Because $A$ is a non-empty bounded subset of real numbers, it has an upper bound and a lower bound.
- Because real numbers have the supremum and the infimum property, every non-empty subset, which has a upper and an lower bound, has also a supremum and an infimum.
- Thus, also $A$ has a supremum and an infimum.
- Therefore there exist convergent real sequences $(x_k)_{k\in\mathbb N}$ and $(y_k)_{k\in\mathbb N}$ with \(x_k\in A\) and $y_k\in A$ for all $k\in\mathbb N$ and \(\lim x_k=\sup(A)\), $\lim y_k=\inf(A)$.
- In any metric space, a closed set (like $A$ is) can be characterized by the limit of its sequences as follows: if there is a convergent sequence of points $(x_k)_{k\in\mathbb N}$ contained in $A$, $A$ will also contain its limit.
- Real numbers are a metric space.
- It follows that $\sup(A)\in A$ and $\inf(A)\in A$.
- By definition of maximum and minimum, it follows that $\max(A)\in A$ and $\min(A)\in A$.
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References
Bibliography
- Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984
