User martin - MathOverflow

Answer by Martin for Product of Baire sigma-algebras

2013-03-20T22:54:57Z

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, analytic sets 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, Measurable rectangles, Theorem 1 for a proof of this. In my answer to a related question on math.SE there are more explanations and further references.

Answer by Martin for Is the closed unit ball of the Hilbert space homeomorphic to the unit sphere ?

2013-02-13T10:28:01Z

The answer to the question in the title is yes.

In Bessaga and Pelczynski, Selected topics in infinite-dimensional topology, Chapter VI, §2 there is a proof of the following:

Theorem. Each of the following sets is homeomorphic to the countable product $\mathbb{R^N}$ of the real line:

The separable Hilbert space $\ell_2$.

The closed unit ball in $\ell_2$.

The unit sphere in $\ell_2$.

The "upper half space" in $\ell_2$: those vectors with non-negative first entry.

Answer by Martin for Old books still used

2012-12-29T09:53:38Z

How about:

G. H. Hardy, J. E. Littlewood, G. Pólya, Inequalities (1934, second edition 1952).

G. Pólya, G. Szegő, Problems and Theorems in Analysis (first German edition in 1925)

G. Szegő, Orthogonal Polynomials (1939)

Answer by Martin for Degeneracies for semi-simplicial Kan complexes

2012-12-03T02:27:18Z

Kan wrote a very short note while the paper by Rourke and Sanderson (mentioned in the other answers) was in the publishing process:

Daniel M. Kan, Is an ss complex a css complex? Advances in Mathematics Volume 4, Issue 2, April 1970, Pages 170–171.

(as explained in the comments "ss" refers to semi-simplicial set -- without degeneracies -- and "css" to what is known as simplicial set nowadays).

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."

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 On $\Delta$-sets, I).

Answer by Martin for Product of Borel sigma algebras

2012-11-30T21:10:38Z

This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies 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.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.

Comment by Martin

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 "almost orthonormal" sequence of vectors. On the face of it this seems to require an application of the axiom of countable dependent choice: en.wikipedia.org/wiki/Axiom_of_dependent_choice

Comment by Martin

2013-06-02T08:13:22Z

You can try if robjohn's (a moderator on math.stackexchange.com) "render MathJax" bookmarklet helps math.ucla.edu/~robjohn/math/mathjax.html On any page containing LaTeX (for example arXiv abstracts) it will render the formulas with a single click.

Comment by Martin

2013-05-28T02:17:28Z

@Jesse "unknown (google)" uses the terminology of section 4.3 of Zimmer's Ergodic theory and semi-simple groups, Monographs in Mathematics 81, Birkhäuser (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: dx.doi.org/10.1090/S0002-9904-1977-14392-9 (for the purposes of this question one can ignore the ergodicity assumption made there).

Comment by Martin

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*-topology on $V$ induced by $U$. (If it is continuous with respect to the topology induced by the weak*-topology on the subspace, it extends uniquely to the weak*-closure, then Hahn-Banach for the weak*-topology applies).

Comment by Martin

2013-05-24T00:23:01Z

@Wlodimierz Holsztynski: A related discussion can be read in Armand Borel's Twenty-five years with Nicolas Bourbaki ams.org/notices/199803/borel.pdf on page 378, where a short account of the story of the congrès du foncteur inflexible is given. It discusses Grothendieck's proposal how they should treat sheaf theory and why that route wasn't chosen.

Comment by Martin

2013-05-21T12:40:50Z

As André pointed out, separability of a $W^\ast$-algebra means separability of the pre-dual and that is equivalent to the conditions on $m$ and $n$ you give. It's not separability in the ultraweak topology, as his example of $\ell^1[0,1]$ shows. Moreover, you use "standard Borel space" 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.

Comment by Martin

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.

Comment by Martin

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.

Comment by Martin

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?

Comment by Martin

2013-05-18T19:31:45Z

What makes the simple fact that images of separable spaces are separable an important theorem?

Comment by Martin

2013-05-02T16:20:11Z

I think it is due to using * which confuses the renderer because it wants to set things between * as italics. If you replace * by \ast everywhere, then the preview looks fine.

Comment by Martin

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 $\beta (D_1 \times D_2) \neq \beta D_1 \times \beta D_2$.

Comment by Martin

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 $\psi_n$.

Comment by Martin

2013-04-24T10:52:30Z

"There exists a Cauchy sequence ..." seems to be a typo (since Cauchy sequences are bounded, the condition is obviously nonsensical). Delete "Cauchy" 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.

Comment by Martin

2013-03-29T18:01:04Z

No worries, I removed my comment :-)