How far is ﻿Lindelöf from compactness? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:12:46Z http://mathoverflow.net/feeds/question/9641 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness How far is ﻿Lindelöf from compactness? Guillermo Mantilla 2009-12-23T21:38:41Z 2013-05-15T13:56:32Z <p>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 $\kappa$. </p> <p>If the above is indeed a fact, can one restrict the class of $\kappa$'s for which the characterization is still valid? </p> <p>Note: Here I'm thinking under ZFC.</p> http://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness/9642#9642 Answer by Pete L. Clark for How far is ﻿Lindelöf from compactness? Pete L. Clark 2009-12-23T21:47:38Z 2012-04-11T01:12:48Z <p>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</p> <p>Theorem (N. Noble): If each power of a $T_1$-space is normal, then the space is compact. </p> <hr> <p>See</p> <p>MR0283749 (44 #979) Noble, N. Products with closed projections. II. Trans. Amer. Math. Soc. 160 1971 169--183</p> <p>and for a simpler proof,</p> <p>MR0415571 (54 #3656) Franklin, S. P.; Walker, R. C. Normality of powers implies compactness. Proc. Amer. Math. Soc. 36 (1972), 295--296. </p> <hr> <p>I wonder if there is any actual connection here?</p> http://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness/9651#9651 Answer by Joel David Hamkins for How far is ﻿Lindelöf from compactness? Joel David Hamkins 2009-12-24T03:02:12Z 2012-04-11T01:27:11Z <p>The answer is <b>Yes</b>.</p> <p>Theorem. The following are equivalent for any Hausdorff space $X$.</p> <ol> <li><p>$X$ is compact.</p></li> <li><p>$X^\kappa$ is Lindel&ouml;f for any cardinal $\kappa$.</p></li> <li><p>$X^{\omega_1}$ is Lindel&ouml;f.</p></li> </ol> <p>Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if $X$ is compact, then $X^\kappa$ is compact and hence Lindel&ouml;f.</p> <p>So suppose that we have a space $X$ that is not compact, but $X^{\omega_1}$ is Lindel&ouml;f. It follows that $X$ is Lindel&ouml;f. 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 $\bigcup\lbrace U_n \; | \; n \in \omega \rbrace = X$.</p> <p>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 $J \subset \omega_1$</p> <p>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.</p> <p>Since $X^{\omega_1}$ is Lindel&ouml;f, 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.</p> <p>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</p> <p>In particular, to answer the question at the end, it suffices to take any uncountable $\kappa$. </p> http://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness/93707#93707 Answer by Ramiro de la Vega for How far is ﻿Lindelöf from compactness? Ramiro de la Vega 2012-04-10T20:55:07Z 2012-04-10T21:07:29Z <p>This is a complement to Joel´s answer and some further generalizations.</p> <p>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. </p> <p>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 <em>finally $\kappa$-compact</em> 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.</p> <p>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 <em>initially $\kappa$-compact</em> 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).</p> <p>All the information above was taken from: <a href="http://biblioteca.uniandes.edu.co/Tesis_2006_primer_semestre/00006522.pdf" rel="nofollow">http://biblioteca.uniandes.edu.co/Tesis_2006_primer_semestre/00006522.pdf</a></p> http://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness/130719#130719 Answer by Ramiro de la Vega for How far is ﻿Lindelöf from compactness? Ramiro de la Vega 2013-05-15T13:56:32Z 2013-05-15T13:56:32Z <p>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":</p> <blockquote> <p>There are models of $ZF$ in which for every $T_1$ space $X$, $X$ is Lindelöf if and only if $X$ is compact.</p> </blockquote> <p>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.</p>