Question

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.

Answer 1

The answer is Yes.

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

1. X is compact.

2. X^κ is Lindelöf for any cardinal κ.

3. X^ω₁ is Lindelöf.

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

So suppose that we have a space X that is not compact, but X^ω₁ 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 U₀ subset U₁ subset ... U_n ... with the union U{ U_n | n in ω} = X.

For each J subset ω₁ of size n, let U_J be the set {s in X^ω₁ | s(α) in U_n for each α in J}. 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 ω₁.

This F is a cover of X^ω₁. To see this, consider any point s in X^ω₁. For each α in ω₁, there is some n with s(α) in U_n. Since ω₁ 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^ω₁ is Lindelöf, there must be a countable subcover F₀. 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 ω₁. Note that J* cannot be finite, since then the sizes of the J appearing in F₀ would be bounded and it could not cover X^ω₁. We may rearrange indices and assume without loss of generality that J*=ω is the first ω many coordinates. So F₀ is really a cover of X^ω, by ignoring the other coordinates.

But this is impossible. Define a sequence s in X^ω₁ 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 ω, and so s is not in any n-box with J subset ω, since any such box has n values in U_n. Thus, s is not in any set in F₀, so it is not a cover. QED

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

Answer 2

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.

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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.

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I wonder if there is any actual connection here?