Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 1 / 1F (Analysis II)

We prove the equivalence by a cycle of implications. First assume closedness and boundedness. Given a sequence $(x_j)$ in $E$, boundedness puts each coordinate in a fixed closed bounded real interval. Apply the one-dimensional Bolzano-Weierstrass theorem to its first coordinate to extract a convergent subsequence, then apply it to the second coordinate of that subsequence, and continue through the finitely many coordinates. The final subsequence converges coordinatewise to some $x\in {\mathbb{R}}^n$. It also converges in the Euclidean norm, since the sum of the squares of its finitely many coordinate differences tends to zero. Since $E$ is a closed set, this limit belongs to $E$. Thus closedness and boundedness imply the required subsequence property. No higher-dimensional compactness theorem has been assumed.

Assume the subsequence property, and let $f:E\to \mathbb{R}$ be a continuous function. If $f$ were unbounded, for each positive integer $j$ we could choose $x_j\in E$ with $|f(x_j)|>j$. There is a subsequence $x_{j_k}\to x$ with $x\in E$. By continuity, $f(x_{j_k})\to f(x)$, so that real sequence is bounded. But $|f(x_{j_k})|>j_k\to \infty$, a contradiction. Hence every continuous real-valued function on $E$ is bounded. This is the bounded-function consequence of sequential compactness.

Assume every continuous function $E\to \mathbb{R}$ is bounded. The Euclidean norm function $x\mapsto \Vert x\Vert$ is continuous, so its boundedness proves that $E$ is a bounded set.

If $E$ were not a closed set, there would be $a\in \overline{E}\setminus E$. Choose $x_j\in E$ with $\Vert x_j-a\Vert <1/j$. The function $f(x)=1/\Vert x-a\Vert$ is continuous everywhere on $E$, because its denominator is nonzero there. Yet $f(x_j)>j$, contradicting the hypothesis. Thus $E$ is closed. Together with the two previous implications this proves all three conditions are equivalent. The empty set satisfies all three conditions trivially. This is the Euclidean case of the bounded-continuous-function characterization of compact metric spaces.