User bojan kwitek - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-24T12:05:01Zhttp://mathoverflow.net/feeds/user/29433http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/118575/kadison-singer-problem-in-exotic-hilbert-spacesKadison-Singer problem in exotic Hilbert spacesBojan Kwitek2013-01-10T21:23:12Z2013-01-10T23:39:13Z
<p>The Kadison-Singer problem is considered in relation to the separable Hilbert space:</p>
<p>KS: Does every pure state on the diagonal (atomic) masa of $B(\ell_2)$ has a unique extension to $B(\ell_2)$? </p>
<p>What is the status of this problem for non-separable Hilbert spaces? Most of operator-algebraits are uninterested in (starting a day with) non-separable Hilbert spaces, so probably they don't care. But if the (classical) KS problem has a negative solution, this solution <em>must</em> depend on the underlying ultrafilter $p\in \beta \mathbb{N}\setminus \mathbb{N}$ (it cannot be, for example, rare).</p>
<p>Sometimes weird ultrafilter over uncountable sets are (consistently) easier to grasp. This indicates that perhaps one should look at $\ell_2(\kappa)$ for some $\kappa$ big enough.</p>
http://mathoverflow.net/questions/118080/non-super-reflexive-spaceNon-super reflexive spaceBojan Kwitek2013-01-04T19:40:57Z2013-01-04T20:49:29Z
<p>Suppose $X$ is a reflexive space (possibly non-separable) which is not super-reflexive. Then (by definition) there exists a non-reflexive Banach space $Y$ which is non-reflexive but is finitely representable in $X$, meaning that for each $\lambda >1$, every finite dimensional subspace of $Y$ is $\lambda$-isomorphic to a subspace of $X$. Can we always find such $Y$ (i.e. non-reflexive) which is separable? In this spirit, what are examples of reflexive but not super-reflexive spaces in which neither $\ell_1$ nor $c_0$ is finitely representable?</p>
http://mathoverflow.net/questions/117527/reflexive-saturated-banach-spacesReflexive-saturated Banach spacesBojan Kwitek2012-12-29T15:36:03Z2012-12-29T15:36:03Z
<p>Say that a Banach space $X$ is strongly saturated by reflexive subspaces if every closed subspace $Y\subset X$ contains a further reflexive subspace $Z\subset Y$ with $\mbox{dens }Y=\mbox{dens Z}$. If I recall correctly, the long James space has this property (it is strongly saturated by Hilbert spaces).</p>
<p>I would like to know how far from a reflexive space can be a space with this property. To quantify "the distance to a reflexive space" I ask the following question:</p>
<p>Is there a Banach space $X$ strongly saturated by reflexive subspaces such that $\ell_\infty(\omega_1)$ (or $\ell_\infty / c_0$) embeds into $X^{**}$?</p>
<p>(in my vague understanding, containment of $\ell_\infty$, say, is still not very far...)</p>
http://mathoverflow.net/questions/117038/automatic-continuity-of-the-inverse-mapAutomatic continuity of the inverse mapBojan Kwitek2012-12-22T17:59:47Z2012-12-22T20:52:26Z
<p>All topological spaces considered here are Hausdorff.</p>
<p>It is a well-known consequence of the minimality of a compact topology that an injective continuous map</p>
<p>$f\colon X\to Y$</p>
<p>where $X$ is compact, must be automatically a homeomorphism onto its range. I am interested in possibly non-compact spaces which share this property. I would like to kindly ask whether there is a characterisation of this class of spaces.</p>
http://mathoverflow.net/questions/115402/quotients-of-cantor-cubes-onto-spacesQuotients of Cantor cubes onto spaces Bojan Kwitek2012-12-04T14:42:07Z2012-12-04T19:24:48Z
<p>Let $\lambda$ be an infinite cardinal. Consider the Cantor cube $\Delta_\lambda = \{0,1\}^\lambda$. It is a standard fact in topology that the topological weight (= minimal cardinality for a basis) of $\Delta_\lambda$ is $\lambda$. Let $S$ be a zero-dimensional compact space of weight $\lambda$ and suppose $s\colon \Delta_\lambda\to S$ is a continuous suriection. Does there exists a closed subspace $D$ of $\Delta_\lambda$, which is homeomorphic to $S$ such that $p|_D$ is a homeomorphism?</p>
http://mathoverflow.net/questions/114922/embedding-of-ell-p-into-infinite-direct-sumsEmbedding of $\ell_p$ into infinite direct sumsBojan Kwitek2012-11-29T20:08:12Z2012-12-01T14:33:10Z
<p>Let $p\in (1,\infty)$ and let $q$ be conjugate to $p$. Is there a subspace of $\ell_1(\ell_p)$ isomorphic to $\ell_q$? Of course, I am uninterested in the case $p=2$.</p>
http://mathoverflow.net/questions/114925/number-of-ii-1-factorsNumber of II${}_1$ factorsBojan Kwitek2012-11-29T20:16:52Z2012-11-30T15:47:58Z
<p>McDuff proved that there exist continuum many non-isomorphic (separable) II${}_1$ factors. I would like to politely ask whether it is known/open if one can find $2^{\mathfrak{c}}$ (or at least $\mathfrak{c}^+$) many such factors.</p>
<p>My feeling is that this is not possible to construct more than $\mathfrak{c}$ separable von Neumann algebras by a simple cardinality argument. The ball of $B(H)$ for $H$ separable, is metrisable under the ultraweak topology, so it has at most $\mathfrak{c}$ ultraweakly closed subsets. So we cannot have more than $\mathfrak{c}$ different balls, and consequently, have more than $\mathfrak{c}$ non-isomorphic algebras. Is this correct?</p>
http://mathoverflow.net/questions/114938/algebras-with-countable-chains-onlyAlgebras with countable chains onlyBojan Kwitek2012-11-29T23:08:54Z2012-11-30T01:19:49Z
<p>Is there an example of an uncountable Boolean algebra $B$ in which every chain is countable and such that $\ell_\infty$ embeds into the Banach space $C(\mbox{Stone }B)$? The latter requirement is not very important, I just want to exclude some trivial cases, like the algebra of finite/cofinite subsets on some uncountable set.</p>
http://mathoverflow.net/questions/114542/independent-families-and-chainsIndependent families and chainsBojan Kwitek2012-11-26T15:51:02Z2012-11-26T19:22:48Z
<p>My question will be very short. </p>
<p><em>Suppose we have a Boolean algebra $B$ which admits an uncountable independent family. Does it follow that there is an uncountable chain of elements in $B$?</em></p>
<p>Manifestly, this is the case for (infinite) complete Boolean algebras, although the proofs of existence of uncountable independent families/chains in these algebras seem to have nothing in common. </p>
http://mathoverflow.net/questions/120147/realcompact-spaceComment by Bojan KwitekBojan Kwitek2013-01-28T21:56:50Z2013-01-28T21:56:50ZTry Engelking's "General topology" (they are named "Hewitt spaces" therein).http://mathoverflow.net/questions/119985/showing-a-banach-space-is-reflexive/120004#120004Comment by Bojan KwitekBojan Kwitek2013-01-27T12:37:29Z2013-01-27T12:37:29ZEvery reflexive space is weakly sequentially complete, $C(K)$ spaces contain copies of $c_0$ which is not WSC (and this property is hereditary). http://mathoverflow.net/questions/118579/question-about-getting-review-servicesComment by Bojan KwitekBojan Kwitek2013-01-10T22:36:32Z2013-01-10T22:36:32ZHisanobu Shinya, since you presumably claim your result is correct, why didn't you choose any of the leading journals? http://mathoverflow.net/questions/118080/non-super-reflexive-space/118087#118087Comment by Bojan KwitekBojan Kwitek2013-01-06T22:56:36Z2013-01-06T22:56:36ZOK, thank you. I haven't spotted this paper. By the way, can we deduce from the fact $\ell_1$ is finitely representable in $X$ that $c_0$ is finitely representable in $X^*$?http://mathoverflow.net/questions/118080/non-super-reflexive-space/118087#118087Comment by Bojan KwitekBojan Kwitek2013-01-04T22:34:42Z2013-01-04T22:34:42ZDear Prof. Johnson. Thank you. I've been trying to find the papers with no success yet, but I'll try again. Let me ask then whether the answer to the second question is positive or negative. :)http://mathoverflow.net/questions/112468/ultrapowers-of-operators/112476#112476Comment by Bojan KwitekBojan Kwitek2012-12-29T13:20:09Z2012-12-29T13:20:09ZJust out of curiosity, does the following hold for countably complete ultrafilters: $(X\oplus Y)_U \isom X_U \oplus Y_U$, $X,Y$ Banach spaces?http://mathoverflow.net/questions/115402/quotients-of-cantor-cubes-onto-spaces/115436#115436Comment by Bojan KwitekBojan Kwitek2012-12-05T12:15:22Z2012-12-05T12:15:22ZThis is very clever, thank you. By the way, do you think is there any name for the following property (?) of a compact space: $X$ has (?) if for every surjection $s\colon X\to X$ there is a copy $Y$ of $X$ such that $s|_Y$ is injective (a homeomorphism onto its image).http://mathoverflow.net/questions/115402/quotients-of-cantor-cubes-onto-spaces/115406#115406Comment by Bojan KwitekBojan Kwitek2012-12-04T15:24:24Z2012-12-04T15:24:24ZOf course, it was a typo. Sorry for this. I want obviously $D$ homeomorphic to $S$.http://mathoverflow.net/questions/114542/independent-families-and-chains/114552#114552Comment by Bojan KwitekBojan Kwitek2012-11-26T17:23:19Z2012-11-26T17:23:19Z$\beta N$ has uncountable chains of clopen sets while it is separable (and hence c.c.c.), so I think there is no obvious relation.http://mathoverflow.net/questions/114542/independent-families-and-chains/114552#114552Comment by Bojan KwitekBojan Kwitek2012-11-26T16:37:03Z2012-11-26T16:37:03ZMuch appreciated.