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.



How far is  ﻿Lindelöf  from compactness?
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 $\kappa$. 


