User martin - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:47:12Z http://mathoverflow.net/feeds/user/29555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125100/product-of-baire-sigma-algebras/125103#125103 Answer by Martin for Product of Baire sigma-algebras Martin 2013-03-20T22:54:57Z 2013-03-20T22:54:57Z <p>Assuming that $X$ is an uncountable Polish space, the desired conclusion that $\mathcal{E} \otimes \mathcal{E}$ contains all sets with the property of Baire is not true. In fact, <a href="http://en.wikipedia.org/wiki/Analytic_set" rel="nofollow">analytic sets</a> have the property of Baire and it is a variant of a result due to Mansfield and Rao that no universal analytic set belongs to $\mathcal{E} \otimes \mathcal{E}$. See Miller, <a href="http://arxiv.org/abs/math/9211206" rel="nofollow">Measurable rectangles</a>, Theorem 1 for a proof of this. In my answer to a <a href="http://math.stackexchange.com/q/177416" rel="nofollow">related question</a> on math.SE there are more explanations and further references.</p> http://mathoverflow.net/questions/121685/is-the-closed-unit-ball-of-the-hilbert-space-homeomorphic-to-the-unit-sphere/121698#121698 Answer by Martin for Is the closed unit ball of the Hilbert space homeomorphic to the unit sphere ? Martin 2013-02-13T10:28:01Z 2013-02-13T10:28:01Z <p>The answer to the question in the title is yes. </p> <p>In Bessaga and Pelczynski, <em>Selected topics in infinite-dimensional topology</em>, Chapter VI, §2 there is a proof of the following:</p> <blockquote> <p><strong>Theorem.</strong> Each of the following sets is homeomorphic to the countable product $\mathbb{R^N}$ of the real line:</p> <ol> <li>The separable Hilbert space $\ell_2$.</li> <li>The closed unit ball in $\ell_2$.</li> <li>The unit sphere in $\ell_2$.</li> <li>The "upper half space" in $\ell_2$: those vectors with non-negative first entry.</li> </ol> </blockquote> http://mathoverflow.net/questions/117415/old-books-still-used/117493#117493 Answer by Martin for Old books still used Martin 2012-12-29T09:53:38Z 2012-12-29T10:01:04Z <p>How about:</p> <p>G. H. Hardy, J. E. Littlewood, G. Pólya, <a href="http://books.google.com/books?id=t1RCSP8YKt8C" rel="nofollow">Inequalities</a> (1934, second edition 1952).</p> <p>G. Pólya, G. Szegő, <a href="http://books.google.com/books?id=b9l2NqGEFzgC" rel="nofollow">Problems and Theorems in Analysis</a> (first German edition in 1925)</p> <p>G. Szegő, <a href="http://books.google.com/books?id=3hcW8HBh7gsC" rel="nofollow">Orthogonal Polynomials</a> (1939)</p> http://mathoverflow.net/questions/57653/degeneracies-for-semi-simplicial-kan-complexes/115232#115232 Answer by Martin for Degeneracies for semi-simplicial Kan complexes Martin 2012-12-03T02:27:18Z 2012-12-03T02:27:18Z <p>Kan wrote a very short note while the paper by Rourke and Sanderson (mentioned in the other answers) was in the publishing process:</p> <p>Daniel M. Kan, <em><a href="http://dx.doi.org/10.1016/0001-8708%2870%2990021-6" rel="nofollow">Is an ss complex a css complex?</a></em> Advances in Mathematics Volume 4, Issue 2, April 1970, Pages 170–171.</p> <p>(as explained in the comments "ss" refers to semi-simplicial set -- without degeneracies -- and "css" to what is known as simplicial set nowadays).</p> <p>The proposition is: "An ss complex $X$ which satisfies the extension condition can be completed (although, in general, in many diferent ways)." and he states afterwards that "It is clear that any two such completions will have the same homotopy type."</p> <p>The proof proceeds by an inductive construction of a simplicial set $WX$ such that there is an isomorphism $FWX \to X$ where $F$ is the forgetful functor from simplicial sets to semi-simplicial sets. As a crucial ingredient he uses the geometric realization from Rourke and Sanderson (he gives a reference to [1,1.3], which should probably be Proposition 2.1. on p.325 in the published version of Rourke and Sanderson's <a href="http://www.maths.ed.ac.uk/~aar/papers/deltars.pdf" rel="nofollow">On $\Delta$-sets, I</a>).</p> http://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras/115024#115024 Answer by Martin for Product of Borel sigma algebras Martin 2012-11-30T21:10:38Z 2012-11-30T21:15:52Z <p>This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.</p> <p>Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does <em>not</em> assume the continuum hypothesis.</p> <p>Arnold W. Miller showed in section 4 of <a href="http://www.math.wisc.edu/~miller/res/hier.pdf" rel="nofollow">On the length of Borel hierarchies</a> that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.</p> <p>See my answer to <a href="http://math.stackexchange.com/q/177416" rel="nofollow">Universally measurable sets of $\mathbb{R}^2$</a> on math.stackexchange.com for related results and more details and references.</p> http://mathoverflow.net/questions/77750/is-there-an-infinite-dimensional-banach-space-with-a-compact-unit-ball/133790#133790 Comment by Martin Martin 2013-06-14T21:05:16Z 2013-06-14T21:05:16Z The Riesz lemma itself holds true without the axiom of choice. But you will apply it countably many times to choose an &quot;almost orthonormal&quot; sequence of vectors. On the face of it this seems to require an application of the axiom of countable dependent choice: <a href="http://en.wikipedia.org/wiki/Axiom_of_dependent_choice" rel="nofollow">en.wikipedia.org/wiki/Axiom_of_dependent_choice</a> http://mathoverflow.net/questions/132546/is-there-an-rss-reader-for-mathematicians Comment by Martin Martin 2013-06-02T08:13:22Z 2013-06-02T08:13:22Z You can try if robjohn's (a moderator on math.stackexchange.com) &quot;render MathJax&quot; bookmarklet helps <a href="http://www.math.ucla.edu/~robjohn/math/mathjax.html" rel="nofollow">math.ucla.edu/~robjohn/math/mathjax.html</a> On any page containing LaTeX (for example arXiv abstracts) it will render the formulas with a single click. http://mathoverflow.net/questions/131598/characterization-of-amenable-actions Comment by Martin Martin 2013-05-28T02:17:28Z 2013-05-28T02:17:28Z @Jesse &quot;unknown (google)&quot; uses the terminology of section 4.3 of Zimmer's <i>Ergodic theory and semi-simple groups</i>, Monographs in Mathematics 81, Birkh&#228;user (1984), where all the relevant terminology and references can be found. The question asks whether the property proved in Zimmer's proposition 4.3.9 characterizes amenable actions. The definitions are also available in the preliminaries of this article by Zimmer: <a href="http://dx.doi.org/10.1090/S0002-9904-1977-14392-9" rel="nofollow">dx.doi.org/10.1090/S0002-9904-1977-14392-9</a> (for the purposes of this question one can ignore the ergodicity assumption made there). http://mathoverflow.net/questions/131871/hahn-banach-restricted-to-a-pre-dual Comment by Martin Martin 2013-05-25T22:10:57Z 2013-05-25T22:10:57Z I'm not sure this is the kind of answer you want: it is necessary and sufficient for the functional to be continuous with respect to the weak&#42;-topology on $V$ induced by $U$. (If it is continuous with respect to the topology induced by the weak&#42;-topology on the subspace, it extends uniquely to the weak&#42;-closure, then Hahn-Banach for the weak&#42;-topology applies). http://mathoverflow.net/questions/131657/why-did-bourbaki-ignore-the-theory-of-categories Comment by Martin Martin 2013-05-24T00:23:01Z 2013-05-24T00:23:01Z @Wlodimierz Holsztynski: A related discussion can be read in Armand Borel's <i>Twenty-five years with Nicolas Bourbaki</i> <a href="http://www.ams.org/notices/199803/borel.pdf" rel="nofollow">ams.org/notices/199803/borel.pdf</a> on page 378, where a short account of the story of the <i>congr&#232;s du foncteur inflexible</i> is given. It discusses Grothendieck's proposal how they should treat sheaf theory and why that route wasn't chosen. http://mathoverflow.net/questions/129531/when-does-a-w-algebra-have-a-standard-borel-spectrum/129625#129625 Comment by Martin Martin 2013-05-21T12:40:50Z 2013-05-21T12:40:50Z As Andr&#233; pointed out, separability of a $W^\ast$-algebra means separability of the pre-dual and <i>that</i> is equivalent to the conditions on $m$ and $n$ you give. It's <i>not</i> separability in the ultraweak topology, as his example of $\ell^1[0,1]$ shows. Moreover, you use &quot;standard Borel space&quot; differently from the OP (whose usage is is the usual one): you can allow countably many atoms, so the conclusion is actually yes with those two fixes. http://mathoverflow.net/questions/131206/whats-the-definition-of-continuous-of-set-valued-functions/131209#131209 Comment by Martin Martin 2013-05-20T13:06:23Z 2013-05-20T13:06:23Z @MarkGrant: If $X$ and $Y$ are Hausdorff, a closed-valued map $\phi\colon X \to \operatorname{Cl}(Y)$ is upper- and lower semicontinuous iff $\phi$ is continuous with respect to the Vietoris topology. A sub-basis for the Vietoris topology is given by $U^- = \lbrace F \in \operatorname{Cl}(Y) \mid F \cap U \neq \emptyset\rbrace$ and $U^+ = \lbrace F \in \operatorname{Cl}(Y) \mid F \subseteq U\rbrace$ where $U$ runs through the open sets of $Y$. [Specialize to $Y$ discrete to get $\mathcal{P}(Y)$]. Gerald points out that upper and lower continuity are more related to order than to a topology. http://mathoverflow.net/questions/131029/drect-limit-of-sequences Comment by Martin Martin 2013-05-19T04:08:20Z 2013-05-19T04:08:20Z @David White: There is no reason that one should be able to choose the splittings to be compatible with the direct system so as to yield a splitting in the limit. For example, the submodule $\bigoplus_{\mathbb{N}} \mathbb{Z}$ of $\prod_{\mathbb{N}} \mathbb{Z}$ is pure, but it is not a direct summand. http://mathoverflow.net/questions/131062/importance-of-separability-vs-second-countability/131074#131074 Comment by Martin Martin 2013-05-19T03:53:04Z 2013-05-19T03:53:04Z No, I was asking an honest question. I agree that it is a useful fact, but it seems more related to continuity than separability per se: It is a characteristic property of continuous functions that the image of the closure is contained in the closure of the image. Thus, dense sets are mapped to dense sets. What makes the instantiation of this observation to countable dense sets particularly noteworthy and important? http://mathoverflow.net/questions/131062/importance-of-separability-vs-second-countability/131074#131074 Comment by Martin Martin 2013-05-18T19:31:45Z 2013-05-18T19:31:45Z What makes the simple fact that images of separable spaces are separable an important theorem? http://mathoverflow.net/questions/129350/reason-for-studying-coherent-sheaves-on-complex-manifolds/129390#129390 Comment by Martin Martin 2013-05-02T16:20:11Z 2013-05-02T16:20:11Z I think it is due to using &#42; which confuses the renderer because it wants to set things between &#42; as italics. If you replace &#42; by \ast everywhere, then the preview looks fine. http://mathoverflow.net/questions/128901/tensor-product-of-c-algebras-of-bounded-uniformly-continuous-functions-on-metri/128924#128924 Comment by Martin Martin 2013-04-27T17:52:35Z 2013-04-27T17:52:35Z Does the following work? If $X_1$ and $X_2$ are not totally bounded, they contain $\varepsilon$-discrete infinite subsets $D_1$ and $D_2$ for some $\varepsilon$. Their closures in the Samuel compactifications should be $\beta D_1$ and $\beta D_2$. Moreover, $D_1 \times D_2$ is $\varepsilon$-discrete, so its closure in the Samuel compactification of $X_1 \times X_2$ is $\beta (D_1 \times D_2)$. But $\beta (D_1 \times D_2) \neq \beta D_1 \times \beta D_2$. http://mathoverflow.net/questions/128595/separable-spaces-qm-vs-functional-analysis Comment by Martin Martin 2013-04-25T10:51:22Z 2013-04-25T10:51:22Z Since $\psi_n$ is supposed to be a Cauchy sequence, it is contained in some ball of radius $R$ around zero. Assuming $R \gt \varepsilon$, no vector $\psi$ of norm $\geq 2R$ can be near any $\psi_n$. http://mathoverflow.net/questions/128595/separable-spaces-qm-vs-functional-analysis Comment by Martin Martin 2013-04-24T10:52:30Z 2013-04-24T10:52:30Z &quot;There exists a Cauchy sequence ...&quot; seems to be a typo (since Cauchy sequences are bounded, the condition is obviously nonsensical). Delete &quot;Cauchy&quot; and you get a dense sequence $\psi_n$, hence a countable dense set. For the other direction enumerate the countable dense set to get a sequence satisfying Zettili's condition. http://mathoverflow.net/questions/125929/if-s-times-re-is-diffeomorphic-to-t-times-re-then-are-s-and-t-diffeomorph/125930#125930 Comment by Martin Martin 2013-03-29T18:01:04Z 2013-03-29T18:01:04Z No worries, I removed my comment :-)