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Let $A$ be a subset of $\Bbb{R}^2$ with the property that every continuous function $f:A\to\Bbb{R}$ has a maximum in $A$. I have to prove that $A$ is compact.

Is $A$ were a subset of $\Bbb{R}$ then we could take a function that is monotone over the components of $A$ and show that each of the components is closed and bounded hence proving $A$ to be compact. But couldn't generalize this idea to a subset of $\Bbb{R}^2$ that easily.

QED
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    See here: https://math.stackexchange.com/questions/99637/k-subseteq-mathbbrn-is-a-compact-space-iff-every-continuous-function-in-k – Math1000 May 03 '18 at 04:11
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    There's also this (even more general) one: https://math.stackexchange.com/questions/668905/if-every-real-valued-continuous-function-is-bounded-on-x-metric-space-then – Micah May 03 '18 at 04:32

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If $A$ is not compact, it is either not bounded or not closed. If it is not bounded, simply put $f(x) = |x|$, which is then unbounded on $A$. If $A$ is not closed, there is a limit point $y$ of $A$ which is not contained in $A$. Define $f(x) = \frac{1}{|x - y|}$. This is continuous on $A \subset \mathbb{R}^2 \setminus \{y\}$.

Chris
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