Let $X$ be a closed subset of $\Bbb R^2$. Prove that $X$ is compact if, and only if, every continuous function $f:X \to \Bbb R$ is bounded.
So far I have:
Assume X is compact and X is a closed subset of $ \Bbb R^2$.
The continuous image of a compact set is compact (Theorem 26.5)
$f[X] \subset \Bbb R$
All compact sets in $\Bbb R$ are bounded, because $\Bbb R$ is metrizable.
Therefore, $f$ is bounded (By definition).
Conversely, I don't believe it is true. Am I right that every continuous function is bounded does not imply that $X$ is compact?
Best Answer
Here a proof of the missing "arrow"
Suppose $X$ is a closed subset of $\mathbb R^2$, and suppose that any contiunuos function on $X$ is bounded.
Let $p\in\mathbb R^2$ and set $f(x)=d(x,p):X\to\mathbb R$. $f$ is clearly continuous, whence it is bounded.
Hence $X$ is a closed and bounded subset of $\mathbb R^2$ and therefore is compact.