Let $(X,d)$ be a compact metric space, and let $f:X\rightarrow\textbf{R}$ be a continuous function. Then $f$ is bounded. Furthermore, $f$ attains its maximum at some point $x_{max}\in X$ and also attains its minimum at some point $x_{min}\in X$.


Since $f$ is continuous, it maps compact sets onto compact sets.

Once compact sets are closed and bounded, we conclude that $f(X)$ is closed and $f(X)\subseteq [-L,L]\subset\textbf{R}$.

Given that $f(X)$ is bounded, it admits a supremum $M = \sup f(X)$ and an infimum $m = \inf f(X)$.

But both $m$ and $M$ are adherent points of $f(X)$. Thus $f(X)\ni m$ and $f(X)\ni M$.

In other words, $m = f(x_{min})$ for some $x_{min}\in X$ and $M = f(x_{max})$ for some $x_{max}\in X$, as previously mentioned.

Any comments or contributions to my solution?

  • $\begingroup$ looks good. I've never seen "adherent points." They are limit points, and hence contained in the image of $f$ by closedness. Assuming thats what you mean, it makes sense to me $\endgroup$ – Andres Mejia May 11 '20 at 22:52
  • $\begingroup$ The book which I am reading makes a distinction between adherent points and limit points. It says that $x\in\textbf{R}$ is an adherent point of $E$ iff for every $\varepsilon > 0$ there exists a $y\in E$ such that $|x-y|\leq\varepsilon$. It also says that $x\in\textbf{R}$ is a limit point of $E$ iff it is an adherent point of $E\backslash\{x\}$. $\endgroup$ – APCorreia May 11 '20 at 23:24
  • $\begingroup$ I have just included it. Thanks for the comment. $\endgroup$ – APCorreia May 11 '20 at 23:50

Your approach is absolutely correct.


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