Let \(K\subset\mathbb R^n\). Suppose every continuous real-valued function on \(K\) is bounded. Prove that \(K\) is compact.
Solution
1. The set \(K\) is bounded.
The function \[ f:K\to\RR, \qquad f(x)=\norm{x}, \] is continuous. By hypothesis it is bounded, so there is an \(M\geq 0\) such that \[ \norm{x}\leq M \] for every \(x\in K\). Hence \(K\) is bounded.
2. The set \(K\) is closed in \(\RR^n\).
Suppose instead that \(K\) is not closed. Then there is a point \[ p\in\overline K\setminus K. \] Because \(p\notin K\), the function \[ g:K\to\RR, \qquad g(x)=\frac{1}{\norm{x-p}}, \] is well-defined and continuous.
For every positive integer \(m\), the condition \(p\in\overline K\) gives a point \(x_m\in K\) with \[ \norm{x_m-p}<\frac1m. \] Therefore \[ g(x_m)>m. \] Thus \(g\) is unbounded on \(K\), contradicting the hypothesis. Hence \(K\) is closed.
3. The set \(K\) is compact.
By steps 1 and 2, \(K\) is a closed and bounded subset of \(\RR^n\). The Heine–Borel theorem therefore implies that \(K\) is compact.
4. Q.E.D.
Step 3 is the required conclusion.