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Today I learnt in class that if $X$ is compact then any continuous map $f:X\to\mathbb{R}$ attains max and min. I was thinking if the converse is true:

If every continuous map $f:X\to\mathbb{R}$ attains max and min, then $X$ is compact.

And I use open cover compactness definition or sequential compactness definition. I could not prove it and I suspect there might be counter-examples but I'm yet to find one.

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For metric spaces, see here: math.stackexchange.com/questions/114123/… –  Seirios Apr 2 '13 at 12:35
    
If a counterexample existed, $X$ would have to not be a subset of $\mathbb{R}^n$ for any finite $n$ because if $X$ is a subset of $\mathbb{R}^n$ and isn't compact, then either $X$ is unbounded or not closed. Both of these instances lead to real valued functions on $X$ that can grow without bound. I'm guessing the proposition is true but don't know a proof yet. If it's not true, I imagine you can put some rather weak constraints on $X$ to make it true (connectedness constraints I'd imagine). –  Daniel Rust Apr 2 '13 at 12:35
    
It's easy to come up with counterexamples if you're just mapping to subsets of $\Bbb R$, but this is a much more interesting question! +$1$ –  Clayton Apr 2 '13 at 12:38

2 Answers 2

up vote 10 down vote accepted

Spaces in which all continuous real-valued functions achieve their extrema are called pseudocompact. (Note that if $X$ is pseudocompact and $f : X \to \mathbb{R}$ is continuous with $\alpha = \inf f [ X ]$ and $\beta = \sup f [ X ]$, then if, say, $\alpha \notin f[X]$ we may compose $f$ with a homeomorphism between $( \alpha , \beta + 1 )$ and $\mathbb{R}$ to obtain an unbounded continuous real-valued function.)

The general theory shows that compact $\Rightarrow$ countably compact $\Rightarrow$ pseudocompact, and neither arrow reverses.

Example 1: The ordinal space $\omega_1 = [ 0 , \omega_1 )$ is countably compact but not compact.

details.

  • If $\langle \alpha _n \rangle_{n \in \mathbb{N}}$ is a one-to-one sequence in $\omega_1$, then by moving to a subsequence if necessary we may assume that $\langle \alpha _n \rangle_{n \in \mathbb{N}}$ is strictly increasing. Then there is a least $\beta \in \omega$ such that $\alpha_n < \beta$ for all $n \in \mathbb{N}$, and it isn't difficult to show that $\beta = \lim_n \alpha_n$. Therefore $\omega_1$ is countably compact.
  • The cover $\{ [ 0 , \alpha ] : \alpha \in \omega \}$ is an (uncountable) open cover without a finite subcover. (In fact, it has no countable subcover!)


Example 2: The deleted Tychonoff plank is pseudocompact but not countably compact.

details.

  • The Tychonoff plank is the product $\hat{X} = ( \omega_1 + 1) \times ( \omega + 1 ) = [ 0 , \omega_1 ] \times [ 0 , \omega ]$, and the deleted Tychonoff plank is the subspace $X = \hat{X} \setminus \{ \langle \omega_1 , \omega \rangle \}$
  • It can be shown that every continuous $f : X \to \mathbb{R}$ has a continuous extension $\hat{f} : \hat{X} \to \mathbb{R}$, and since $\hat{X}$ is compact then $\hat{f}$ is bounded, and thus so, too, is $f$. Therefore $X$ is pseudocompact.
  • The countably infinite subset $\{ \langle \omega_1 , n \rangle : n \in \omega \}$ has no accumulation point in $X$, and so $X$ is not countably compact.
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Another nice example of a pseudocompact space that is not countably compact is a Mrówka $\Psi$-space; the proof of pseudocompactness is very easy. –  Brian M. Scott Apr 2 '13 at 21:06
    
@Brian: Just go ahead and improve all of my answers, why don't you! :-) –  Arthur Fischer Apr 2 '13 at 22:27
    
Hey, I like your answers. –  Brian M. Scott Apr 2 '13 at 22:28

I'm looking for more references, but I recall the statement isn't true in general. There exist Hausdorff spaces whose only continuous maps are constant. An example of a countably compact space(which is weaker than compact) can be found here. Another example can be found at this link.

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Is it true that if $X$ is metrizable then the statement is true? –  mez Apr 2 '13 at 13:03

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