Discussion
Let \(K \le \mathbf{R}^{n}\) be with the property that any real valued continuous function on K is bounded. Then K is :
Answer: B
K must be compact.
In \(\mathbf{R}^{n}\) a set is compact exactly when it is closed and bounded. If K were not bounded, \(f(x) = \|x\|\) would be a continuous unbounded function on K. If K were bounded but not closed, there would be a limit point \(a\) of K outside K, and \(f(x) = 1/\|x - a\|\) would be continuous and unbounded on K. So the property forces K to be closed and bounded, that is, compact; conversely every continuous function on a compact set is bounded.
'Bounded' alone is not enough (the open interval (0, 1) carries the unbounded \(1/x\)), and connectedness or density has nothing to do with it.
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