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3 Texified, since problem was on front-page anyway
space X.

• X $X$.

• $X$ is compact.

• Xκ

• $X^\kappa$ is Lindelöf for any cardinal

• Xω1$\kappa$.

• $X^{\omega_1}$ is Lindelöf.

• for 1 implies 2, since if X $X$ is compact, thenXκ$X^\kappa$ is compact and hence Lindelöf.

So suppose that we have a space X $X$ that is not compact, butXω1$X^{\omega_1}$ is Lindelöf. Itfollows that X $X$ is Lindelöf. Thus, there is a countableU0$U_0 \subset U1U_1 \subset ... Un... \dots U_n \dots$with the union U{ Un$\bigcup\lbrace U_n \; | \; n \in ω} \omega \rbrace = XX$.

For each $J \subset ω1\omega_1$ of size n, $n$, let UJ$U_J$ bethe set {$\lbrace s \in Xω1X^{\omega_1} \; | s(α)\; s(\alpha) \in UnU_n$ for each α $\alpha \in J}. J \rbrace$. As the size of J$J$ increases, the set UJ$U_J$ allows more freedom on thecoordinates in J, $J$, but restricts more coordinates. If J $J$ hassize n, $n$, let us call UJ$U_J$ an open n-box, $n$-box, since itrestricts the sequences on n $n$ coordinates. Let F $F$ be thefamily of all such UJ$U_J$ for all finite $J \subset ω1.\omega_1$

This F $F$ is a cover of Xω1. $X^{\omega_1}$. Tosee this, consider any point $s \in Xω1X^{\omega_1}$. For each α $\alpha \inω1, \omega_1$, there is some n $n$ with s(α) $s(\alpha) \inUnU_n$. Since ω1$\omega_1$ is uncountable,there must be some value of n $n$ that is repeated unboundedlyoften, in particular, some n $n$ occurs at least n $n$ times. Let Jbe the coordinates where this n $n$ appears. Thus, s $s$ is inUJ. $U_J$. So F $F$ is a cover.

Since Xω1$X^{\omega_1}$ is Lindelöf,there must be a countable subcover F0. $F_0$. Let J* $J^*$ bethe union of all the finite J $J$ that appear in theUJ$U_J$ in this subcover. So J* $J^*$ is a countable subsetof ω1. $\omega_1$. Note that J* $J^*$ cannot be finite,since then the sizes of the J $J$ appearing in F0$F_0$Xω1. $X^{\omega_1}$. We may rearrange indicesand assume without loss of generality that J*=ω $J^*=\omega$ isthe first ω $\omega$ many coordinates. So F0$F_0$ isreally a cover of Xω, $X^\omega$, by ignoring the

But this is impossible. Define a sequence $s \inXω1X^{\omega_1}$ by choosing s(n) $s(n)$ to beoutside Un+1, $U_{n+1}$, and otherwise arbitrary. Note thats $s$ is in Un$U_n$ in fewer than n $n$ coordinates belowω, $\omega$, and so s $s$ is not in any n-box $n$-box with $J \subset ω, \omega$, since any such box has n $n$ values in Un.Thus, s $s$ is not in any set in F0, $F_0$, so it is not a

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

2 fixed typo.

The answer is Yes.

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

1. X is compact.

2. Xκ is Lindelöf for any cardinal κ.

3. Xω1 is Lindelöf.

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

So suppose that we have a space X that is not compact, but Xω1 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 ... Un ... with the union U{ Un | n in ω} = X.

For each J subset ω1 of size n, let UJ be the set {s in Xω1 | s(α) in Un for each α in J}. As the size of J increases, the set UJ allows more freedom on the coordinates in J, but restricts more coordinates. If J has size n, let us call UJ an open n-box, since it restricts the sequences on n coordinates. Let F be the family of all such UJ for all finite J subset ω1.

This F is a cover of Xω1. To see this, consider any point s in Xω1. For each α in ω1, there is some n with s(α) in Un. 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 UJ. So F is a cover.

Since Xω1 is Lindelöf, there must be a countable subcover F0. Let J* be the union of all the finite J that appear in the UJ 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 Xω1. 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 Xω, by ignoring the other coordinates.

But this is impossible. Define a sequence s in Xω1 by choosing s(n) to be outside Un+1, and otherwise arbitrary. Note that s is in Un 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 Un. 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.

1

The answer is Yes.

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

1. X is compact.

2. Xκ is Lindelöf for any cardinal κ.

3. Xω1 is Lindelöf.

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

So suppose that we have a space X that is not compact, but Xω1 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 ... Un ... with the union U{ Un | n in ω} = X.

For each J subset ω of size n, let UJ be the set {s in Xω1 | s(α) in Un for each α in J}. As the size of J increases, the set UJ allows more freedom on the coordinates in J, but restricts more coordinates. If J has size n, let us call UJ an open n-box, since it restricts the sequences on n coordinates. Let F be the family of all such UJ for all finite J subset ω1.

This F is a cover of Xω1. To see this, consider any point s in Xω1. For each α in ω1, there is some n with s(α) in Un. 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 UJ. So F is a cover.

Since Xω1 is Lindelöf, there must be a countable subcover F0. Let J* be the union of all the finite J that appear in the UJ 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 Xω1. 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 Xω, by ignoring the other coordinates.

But this is impossible. Define a sequence s in Xω1 by choosing s(n) to be outside Un+1, and otherwise arbitrary. Note that s is in Un 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 Un. 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.
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