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It is well known that compactness implies pseudocompactness; this follows from the Heine–Borel theorem. I know that the converse does not hold, but what is a counterexample?

(A pseudocompact space is a topological space $S = \langle X,{\mathfrak I}\rangle$ such that every continuous function $f:S\to\Bbb R$ has bounded range.)

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You are aware of the limited applicability of the Heine-Borel Theorem (i.e. it is valid basically for finite dimensional manifolds)? –  user20266 Aug 11 '12 at 17:51
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6 Answers

up vote 7 down vote accepted

A favorite example (and counterexample) to may things is the first uncountable ordinal $\omega_1$ in its order topology: $[0,\omega_1)$. It is pseudo-compact but not compact.

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In fact, for every infinite ordinal $\alpha$, we have $[0,\alpha)$ is countably compact, and hence pseudocompact, however it is not compact. Am I right? –  Paul Aug 12 '12 at 1:22
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@Paul: For any ordinal $\alpha$, $[0,\alpha+\omega)$ is not countably compact. Similarly, $\omega_\omega$ is not countably compact. However, $[0,\alpha)$ is countably compact if $\operatorname{cf}\alpha>\omega$. –  Brian M. Scott Aug 12 '12 at 7:34
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Spacebook, a searchable version of Steen and Seebach's Counterexamples in Topology, gives the following examples of pseudocompact spaces that are not compact. You should be able to find more about these spaces online, but let me know if any of the names are mysterious.

An Altered Long Line
Countable Complement Topology
Countable Particular Point Topology
Deleted Tychonoff Plank
Divisor Topology
Double Pointed Countable Complement Topology
Gustin’s Sequence Space
Hewitt’s Condensed Corkscrew
Interlocking Interval Topology
Irrational Slope Topology
Minimal Hausdorff Topology
Nested Interval Topology
Novak Space
Open Uncountable Ordinal Crossed with Uncountable Cartesian Product of Unit Interval
Open Uncountable Ordinal Space
Prime Integer Topology
Relatively Prime Integer Topology
Right Order Topology on the Reals
Roy’s Lattice Space
Strong Ultrafilter Topology
The Long Line
Tychonoff Corkscrew
Uncountable Particular Point Topology

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I like the idea, but the definitions or classifications used in your site seem to be a little off for me. For example, this is listed as $T_4$ (even $T_5$!), but not $T_1$ through $T_{3\frac{1}{2}}$. –  tomasz Aug 11 '12 at 18:01
    
@tomasz At the time Counterexamples was written, there were two competing sets of separation axioms. The one used in Counterexamples seems to be the less popular of the two nowadays. I have been meaning to make a note of this on the site. Thank you for reminding me. In the meantime, see this: en.wikipedia.org/wiki/… –  Austin Mohr Aug 11 '12 at 18:03
    
I like the idea of this as a resource, but I think it needs better navigation. For example, I should be able to click on the name of a space and see its definition and a list of properties it satisfies. –  MJD Aug 11 '12 at 19:51
    
@MarkDominus Clicking on the name of a space does list the properties it satisfies. As for definitions, most of the spaces are discussed on Wikipedia (see en.wikipedia.org/wiki/…). I agree it would be better to incorporate some of this to Spacebook, but I do not have the time to devote to it at present. –  Austin Mohr Aug 12 '12 at 0:56
    
Austin, have you seen this database? –  Brian M. Scott Aug 12 '12 at 7:37
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I couldn't think of an obvious counterexample, so I looked in Wikipedia and it suggested the particular point topology on an infinite set.

$\def\p{{\bf x}}$In the particular point topology, we have a distinguished point, $\p\in X$, and the topology is that a set is open if and only if it is either empty, or includes $\p$.

Let $S=\langle X,{\mathfrak I}\rangle$ be an infinite particular-point space with distinguished point $\p$. It is clear that $S$ is not compact: the open cover consisting of $\{p, \p\}$ for each $p\in X$ other than $\p$ is an infinite open cover of $S$ with no proper, and therefore no finite subcover.

$\def\R{{\Bbb R}}$However, the space is pseudocompact. Let $f:S\to\R$ be a continuous function. Then $f^{-1}[\Bbb R\setminus \{f(\bf x)\}]$ is an open set not containing $\bf x$, so it must be empty, hence $f$ is constant.

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Note that sequential compactness also implies pseudocompactness, so any sequentially compact space which is not compact will work as well (the particular point topology is not sequentially compact, either, so it's different kind).

For example, the Corson space $\Sigma([0,1]^\kappa)$ of sequences of length $\kappa$ with countable support is not compact for uncountable $\kappa$ (which is easy to see), but is sequentially compact (which is a bit harder to see). It is also completely regular Hausdorff, which makes it sort of a "stronger" example than particular point topology. $\Sigma(2^\kappa)$ should work fine for this, too.

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Is the topology for the Corson space just the subspace topology of the product space $[0,1]^\kappa$? –  MJD Aug 18 '12 at 5:20
    
@MarkDominus: Yes. –  tomasz Aug 18 '12 at 11:49
    
@MarkDominus: I think there's even a specific kind of compactness the space satifsies, “Corson compactness”, but I don't really know – I've only encountered it as an exercise in one of the courses I've had last year, and encountered the name while googling information on it (but ultimately had to do the exercise myself ;) ). –  tomasz Aug 18 '12 at 12:00
    
@tomasz Corson compact is just being compact and embeddable in a $\Sigma$-product like that. So it's a special class of (ordinarily) compact Hausdorff spaces. The whole $\Sigma$-product is not compact, so also not Corson compact. –  Henno Brandsma Feb 23 '13 at 11:34
    
@HennoBrandsma: Thanks for clarifying, I was wondering about that. :) –  tomasz Feb 23 '13 at 13:04
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An example that does not depend on countable compactness is Mrówka’s space $\Psi$. Subsets of $\omega$ are said to be almost disjoint if their intersection is finite. Let $\mathscr{A}$ be a maximal almost disjoint family of subsets of $\omega$, and let $\Psi=\omega\cup\mathscr{A}$. Points of $\omega$ are isolated. Basic open nbhds of $A\in\mathscr{A}$ are sets of the form $\{A\}\cup(A\setminus F)$, where $F$ is any finite subset of $A$. $\Psi$ is not even countably compact, since $\mathscr{A}$ is an infinite (indeed uncountable) closed, discrete set in $\Psi$. (In fact it’s not hard to ensure that $|\mathscr{A}|=2^\omega$.)

To see that $\Psi$ is pseudocompact, suppose that $f:\Psi\to\Bbb R$ is continuous. Since $\omega$ is dense in $\Psi$, it suffices to show that $f[\omega]$ is bounded. If not, we can choose $S=\{n_k:k\in\omega\}\subseteq\omega$ such that $f(n_{k+1})\ge f(n_k)+1$ for each $k\in\omega$. The maximality of $\mathscr{A}$ ensures that there is an $A\in\mathscr{A}$ such that $A\cap S$ is infinite. Let $A_0=\{k\in\omega:n_k\in A\cap S\}$. Then $\langle f(n_k):k\in A_0\rangle\to f(A)$, which contradicts the choice of $S$.

$\Psi$ clearly is $T_2$ and has a clopen base, so it’s Tikhonov. It’s not normal, however, since in $T_4$ spaces pseudocompactness is equivalent to countable compactness.

Added: This example is somewhat akin to what Steen & Seebach call the strong ultrafilter topology.

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Your explanations are allways so helpfull! Thank you! –  Shir Sivroni Mar 28 at 9:23
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Some exotic examples of pseudocompact and non-compact spaces are constructed in my paper “Pseudocompact paratopological groups that are topological”:

Example 1 (p.6). A $T_1$ space having each power countably pracompact (and, hence pseudocompact). (See p.1 for the definition of a countably pracompact space).

Example 2 (p.6). [Under $MA_{countable}$] A functionally Hausdorff countably compact space $X$. (See p.6 for axiomatic assumptions which I use for the example construction).

Example 3 (p.8) A functionally Hausdorff second countable space having each power countably pracompact (and, hence pseudocompact).

Example 5 (p. 14) A $T_0$ sequentially compact, not totally countably compact space and a $T_1$ pseudocompact, not countably pracompact space. (See p.1 for the definition of a totally countably compact space).

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