Let $X$ be a compact metric space, show that a continuous function $f:X\rightarrow\mathbb{R}$ attains a maximum and a minimum value on $X$.

Attempt: So the important thing is that I have previously shown that such a function is bounded and that for compact $X$, $f(X)$ is compact given $f$ continuous. In $\mathbb{R}$, compact $\implies$ closed and bounded. So $f(X)$ is closed and contains its accumulation points, and it is bounded so $\exists \sup(A),\inf(A)$ and since closed $\implies \sup(A)\in A, \inf(A)\in A$.

Did I miss anything/make an unwarranted leap of logic?


2 Answers 2


Here’s an example of how the same argument could be written up nicely.

Since $X$ is compact and $f$ is continuous, $f[X]$ is a compact subset of $\mathbb{R}$ and therefore closed and bounded. Since $f[X]$ is bounded, it has both a supremum and an infimum, and since it is closed, $\sup f[X]\in f[X]$ and $\inf f[X]\in f[X]$. Thus, there are $x_0,x_1\in X$ such that $f(x_0)=\inf f[X]$ and $f(x_1)=\sup f[X]$; that is, $f$ attains its minimum and maximum values at $x_0$ and $x_1$, respectively.

  • $\begingroup$ Nice proof, how does this generalize to X is an arbitrary topological space? will the same proof hold? $\endgroup$
    – grayQuant
    Oct 19, 2015 at 3:35
  • 6
    $\begingroup$ @grayQuant: The proof actually used only the compactness of $X$; the hypothesis that $X$ is metric isn’t needed. If $X$ is not compact, however, there may be continuous real-valued functions on $X$ that do not attain at least one extremum. $\endgroup$ Oct 19, 2015 at 3:39
  • $\begingroup$ This has been bothering me for a while, to see that $f[X]$ is bounded implies $f[X]$ has both a supremum and infimum, we need the fact that $f[X]$ is nonempty, right? but if we take $X$ to be the empty set, then it is compact and $f$ is then the empty set, so it is continuous, so that $f[X]$ is also an empty set. Thanks. $\endgroup$
    – hteica
    Jul 20, 2020 at 13:06
  • $\begingroup$ @hteica: The default assumption is that spaces are not empty. $\endgroup$ Jul 20, 2020 at 17:13
  • 1
    $\begingroup$ @CrashBandicoot: This is a very belated answer, since I only just realized that you never did ask the question separately. The question was in fact asked and answered somewhat earlier: we cannot always construct such a function. Indeed, there are even non-compact spaces on which every continuous real-valued function is constant. $\endgroup$ Jan 15, 2023 at 3:50

Your argument is fundamentally sound, but you have to assume that your metric space is nonempty. Here is a direct proof that requires no other results (the proof generalizes, like yours, to arbitrary topological spaces):

Let $X$ be a nonempty metric space and $f:X\to\mathbb{R}$ a continuous function without a maximum. Then $X$ has an open cover without a finite subcover.


  1. Suppose $f(X)$ is unbounded. Then $\big\{f^{-1}\big((-\infty, n)\big):n\in\mathbb{N}\big\}$ is an open cover without finite subcover.

  2. Suppose $f(X)$ is bounded, with supremum $s$. Since $f$ has no maximum, $s\notin f(X)$ and $\big\{f^{-1}\big((-\infty, s-1/n)\big):n\in\mathbb{N}\big\}$ is an open cover without finite subcover.


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