Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
The ideas are all there, and they’re connected properly, but it’s not really a well-written proof as it stands. –  Brian M. Scott Feb 15 '12 at 7:24
By the way, $\sup$ and $\inf$ are predefined: use \sup, \inf. –  Brian M. Scott Feb 15 '12 at 7:25
It is better to avoid $\exists$ and $\implies$ symbols when writing math. Just write "there exist" and "then". After all, you write math in english not in other strange symbolic language. –  leo Jun 9 '12 at 3:15

2 Answers 2

up vote 11 down vote accepted

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.

share|cite|improve this answer
+1 Textbook proof. –  user38268 Feb 15 '12 at 9:44

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.