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A while ago I heard of a nice characterization of compactness but I have never seen a written source of it, so I'm starting to doubt it. I'm looking for a reference, or counterexample, for the following . Let $X$ be a Hausdorff topological space. Then, $X$ is compact if and only if $X^{\kappa}$ is Lindelöf for any cardinal $\kappa$.

If the above is indeed a fact, can one restrict the class of $\kappa$'s for which the characterization is still valid?

Note: Here I'm thinking under ZFC.

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3  
You should edit your title. Every separable metric space is Lindel\" of and in that setting being Lindel\" of is hereditary (compactness is almost never hereditary). So I would argue that the answer to the question in your title is "rather far." But you are asking something rather different. –  Justin Moore Mar 12 '11 at 1:47

4 Answers 4

up vote 33 down vote accepted

The answer is Yes.

Theorem. The following are equivalent for any Hausdorff space $X$.

  1. $X$ is compact.

  2. $X^\kappa$ is Lindelöf for any cardinal $\kappa$.

  3. $X^{\omega_1}$ is Lindelöf.

Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if $X$ is compact, then $X^\kappa$ is compact and hence Lindelöf.

So suppose that we have a space $X$ that is not compact, but $X^{\omega_1}$ is Lindelöf. It follows that $X$ is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets $U_0 \subset U_1 \subset \dots U_n \dots$ with the union $\bigcup\lbrace U_n \; | \; n \in \omega \rbrace = X$.

For each $J \subset \omega_1$ of size $n$, let $U_J$ be the set $\lbrace s \in X^{\omega_1} \; | \; s(\alpha) \in U_n$ for each $\alpha \in J \rbrace$. As the size of $J$ increases, the set $U_J$ allows more freedom on the coordinates in $J$, but restricts more coordinates. If $J$ has size $n$, let us call $U_J$ an open $n$-box, since it restricts the sequences on $n$ coordinates. Let $F$ be the family of all such $U_J$ for all finite $J \subset \omega_1$

This $F$ is a cover of $X^{\omega_1}$. To see this, consider any point $s \in X^{\omega_1}$. For each $\alpha \in \omega_1$, there is some $n$ with $s(\alpha) \in U_n$. Since $\omega_1$ is uncountable, there must be some value of $n$ that is repeated unboundedly often, in particular, some $n$ occurs at least $n$ times. Let $J$ be the coordinates where this $n$ appears. Thus, $s$ is in $U_J$. So $F$ is a cover.

Since $X^{\omega_1}$ is Lindelöf, there must be a countable subcover $F_0$. Let $J^*$ be the union of all the finite $J$ that appear in the $U_J$ in this subcover. So $J^*$ is a countable subset of $\omega_1$. Note that $J^*$ cannot be finite, since then the sizes of the $J$ appearing in $F_0$ would be bounded and it could not cover $X^{\omega_1}$. We may rearrange indices and assume without loss of generality that $J^*=\omega$ is the first $\omega$ many coordinates. So $F_0$ is really a cover of $X^\omega$, by ignoring the other coordinates.

But this is impossible. Define a sequence $s \in X^{\omega_1}$ by choosing $s(n)$ to be outside $U_{n+1}$, and otherwise arbitrary. Note that $s$ is in $U_n$ in fewer than $n$ coordinates below $\omega$, and so $s$ is not in any $n$-box with $J \subset \omega$, since any such box has $n$ values in $U_n$. Thus, $s$ is not in any set in $F_0$, so it is not a cover. QED

In particular, to answer the question at the end, it suffices to take any uncountable $\kappa$.

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Oh, I see you wanted either a reference or a counterexample, whereas I gave a proof. I'm sorry that I don't know a reference, but since it appears to be true, it must have been known classically, so surely there is a reference. What a fun problem! –  Joel David Hamkins Dec 24 '09 at 4:44
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what you wrote is more than perfect,and actually I wonder if your proof can be slightly modify to prove a more general result. If a Hausdorff space $X$ has the property that $X^{\aleph_{\alpha+1}}$ is $\aleph_{\alpha+1}$-compact, does it follows that $X$ is $\aleph_{\alpha}$-compact? –  Guillermo Mantilla Dec 25 '09 at 8:32
    
I'm not sure what the right generalization is. There was something special about omega in my argument, since with the product topology the basic open sets have finite support, and this was what allowed for the diagonalization argument at the end. Perhaps for larger cardinals, there might be a clever workaround... –  Joel David Hamkins Dec 26 '09 at 3:15

This is a complement to Joel´s answer and some further generalizations.

In "Paracompactness and product spaces" (1948), Stone proved that if a product space is Lindelof and regular then all but countably many factors are compact.

In "Compact factors in finally compact products of topological spaces" (2005), Lipparini removed the regularity condition and generalized the result to weaker forms of compactness. For instance, it follows that if $X^{\aleph_{\alpha+n+1}}$ is finally $\aleph_{\alpha+n+1}$-compact then $X$ is finally $\aleph_\alpha$-compact (a space is finally $\kappa$-compact if any open cover admits a subcover of size less than $\kappa$). A corollary of this is that if $X^{\aleph_n}$ is finally $\aleph_n$-compact then $X$ is compact. In particular if $X^{\aleph_1}$ is Lindelof then $X$ is compact.

In a different direction (generalizing Tychonoff), in "Products of initially m-compact spaces" (1974), Stephenson and Vaughan proved that if $\kappa$ is a singular strong limit cardinal, then any product of initially $\kappa$-compact spaces is initially $\kappa$-compact (a space is initially $\kappa$-compact if any open cover of size $\kappa$ admits a finite subcover). Note that initially $\aleph_0$-compact is just countably compact, and that there are spaces $X$ such that $X$ is countably compact but $X^2$ is not (see Novák´s "On the cartesian product of two compact spaces", 1953).

All the information above was taken from: http://biblioteca.uniandes.edu.co/Tesis_2006_primer_semestre/00006522.pdf

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@Ramiro: +++1 por esa referencia. Ojalá tuviera tiempo para poder leer esa tesis, de reojo se ve bien. –  Guillermo Mantilla Apr 12 '12 at 9:38

I've never heard of that result (which is not to say that I doubt its truth -- I have no opinion either way), but it reminds me of the following

Theorem (N. Noble): If each power of a $T_1$-space is normal, then the space is compact.


See

MR0283749 (44 #979) Noble, N. Products with closed projections. II. Trans. Amer. Math. Soc. 160 1971 169--183

and for a simpler proof,

MR0415571 (54 #3656) Franklin, S. P.; Walker, R. C. Normality of powers implies compactness. Proc. Amer. Math. Soc. 36 (1972), 295--296.


I wonder if there is any actual connection here?

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I haven't had time to see your references yet but I like a lot the theorem you mention, and I also find it similar to the one in my question. About connections I'd ask, how far is T_2 and Lindelöf form normal? –  Guillermo Mantilla Dec 25 '09 at 8:24
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Every T_2 Lindelöf is normal. Even more, every Lindelöf is strongly paracompact hence paracompact hence normal. –  Ostap Chervak Apr 15 '11 at 17:10
    
@Ostap: Nice! I suspected this to be the case, so an alternative proof --to the one provided by Joel-- follows from your comment and Noble's thm in Pete's answer. –  Guillermo Mantilla Jun 12 '11 at 7:32

Here is a very surprising fact I was completely unaware of until yesterday, when I found it in Herrlich´s book "Axiom of Choice". It shows that the answer to the question in the title could be "very very close":

There are models of $ZF$ in which for every $T_1$ space $X$, $X$ is Lindelöf if and only if $X$ is compact.

For instance this holds in what is called Cohen first model. Also in this model, Tychonoff´s theorem holds for Hausdorff spaces, so arbitrary products of Lindelöf Hausdorff spaces are Lindelöf.

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I wish I knew this result while Arhangelskii was still around our department. I could've answered lots of his questions! –  Todd Eisworth May 15 '13 at 14:46

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