User nik weaver - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:30:56Z http://mathoverflow.net/feeds/user/23141 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131062/importance-of-separability-vs-second-countability/131074#131074 Answer by Nik Weaver for Importance of separability vs. second-countability Nik Weaver 2013-05-18T18:13:04Z 2013-05-18T18:13:04Z <p>I second the idea that second countability is more fundamental than separability --- topologies are defined in terms of open sets, not points, and second countability is the natural "countability" condition on the family of open sets. It just says that the topology is countably generated.</p> <p>Important theorems where separability is crucial: the basic example is that a continuous image of any separable space is separable. Even a quotient of a second countable space need not be second countable.</p> <p>Is the popularity of the word/concept of "separability" just due to the special case of metric spaces? Yes, I think so. But that's a pretty important special case! I just finished writing a book on measure theory and functional analysis, and I found that by restricting attention to separable Banach spaces and their duals I was able to get by just fine without mentioning generalized convergence (nets/filters). For instance, the weak* topology is metrizable on the unit ball of the dual of a separable Banach space, which is good enough for most purposes by the Krein-Smulian theorem.</p> http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130892#130892 Answer by Nik Weaver for Is there any proof that you feel you do not "understand"? Nik Weaver 2013-05-17T02:06:13Z 2013-05-17T02:06:13Z <p>I remember not understanding the proof of the fundamental theorem of calculus. My teacher, who was otherwise very good, didn't cover the proof; she told us we could look at it ourselves if we were curious. (This was a high school class.) I did take a look, but I couldn't make heads nor tails of it.</p> <p>It so happens that I didn't encounter this proof again until I was a postdoc teaching second quarter calculus. I was relieved to learn that it was now quite trivial!</p> <p>If I were teaching calculus at the high school level I wouldn't leave the students entirely on their own, but I also don't think I would want to tell them anything like a real proof of the fundamental theorem. I would probably be satisfied with trying to get across the general idea in an intuitive way.</p> http://mathoverflow.net/questions/130535/a-space-parameterizing-the-choices-of-orthonormal-bases-for-a-hilbert-space/130538#130538 Answer by Nik Weaver for A space parameterizing the choices of orthonormal bases for a Hilbert space Nik Weaver 2013-05-14T02:39:16Z 2013-05-14T02:39:16Z <p>You could look at the set of operators in $B(H)$ which are diagonalized by a given basis. For each choice of orthonormal basis, this gives you an atomic maximal abelian von Neumann algebra (atomic masa). Now there is a natural topology on the space of all von Neumann algebras sitting inside $B(H)$ called the Effros-Maréchal topology, so you could just restrict this topology to the set of atomic masas.</p> <p><a href="http://www.sciencedirect.com/science/article/pii/S0022123699935383" rel="nofollow">Here</a> is a link that may be helpful.</p> http://mathoverflow.net/questions/130310/a-characterization-of-hilbert-spaces/130390#130390 Answer by Nik Weaver for A characterization of Hilbert spaces? Nik Weaver 2013-05-12T04:32:58Z 2013-05-12T04:32:58Z <p>A related question was asked earlier by Mark Meckes: <a href="http://mathoverflow.net/questions/123124/self-dual-finite-dimensional-complex-normed-spaces" rel="nofollow">Self-dual finite-dimensional complex normed spaces</a>. He pointed out the $l^1$-$l^\infty$ example and noted that it generalizes to unit balls that are regular polygons in the real two-dimensional case. He also told us the $X \oplus_2 X^*$ construction.</p> <p>I believe the specific question asked by Wlodzimierz has a positive answer, based on a comment I heard Giles Pisier make many years ago --- he said something very similar to this, though I don't remember exactly what. I don't have a reference though.</p> http://mathoverflow.net/questions/129563/decidability-of-equality-of-expressions-built-using-1/129568#129568 Answer by Nik Weaver for Decidability of equality of expressions built using 1,+,-,*,/,^ Nik Weaver 2013-05-03T18:42:55Z 2013-05-03T18:42:55Z <p><a href="http://www.pps.univ-paris-diderot.fr/~dufour/zeronfa.pdf" rel="nofollow">The equational theory of $\langle {\bf N}, 0, 1, +, \times, \uparrow\rangle$ is decidable, but not finitely axiomatizable.</a></p> http://mathoverflow.net/questions/129478/does-the-border-boundary-points-of-a-convex-body-make-a-concave-function/129484#129484 Answer by Nik Weaver for Does the Border (Boundary) Points of a convex body make a concave function? Nik Weaver 2013-05-03T03:02:58Z 2013-05-03T07:50:26Z <p>As Wlodzimierz points out, the answer is trivially yes. Maybe it would help to recall that $S$ is compact, so that for each $x$ there exists $y$ such that $(x,y) \in S$ and $f(x) = y$. So if $f(x_1) = y_1$ and $f(x_2) = y_2$ then the point $\frac{1}{2}((x_1, y_1) + (x_2, y_2)) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ belongs to $S$, and hence $f(\frac{x_1 + x_2}{2}) \geq \frac{y_1 + y_2}{2}$. Also it's easy to see that $f$ is continuous.</p> <p>But it's interesting to note that continuity fails in ${\bf R}^3$. Let $S$ be the convex hull of the circle in the $xy$-plane with center $(1,0,0)$ and radius $1$, and the point $(0,0,1)$. So $S$ is a cone. Now if we set $f(x,y) = {\rm max}_{(x,y,z) \in S} z$ for $(x,y)$ belonging to the disc with center $(1,0)$ and radius $1$, we find that $f(x,y) = 0$ on the boundary circle except at the point $(0,0)$, where it takes the value $1$.</p> http://mathoverflow.net/questions/129254/certain-bounded-linear-operators-on-l2-of-a-torus/129268#129268 Answer by Nik Weaver for Certain bounded linear operators on L^2 of a torus Nik Weaver 2013-05-01T00:43:04Z 2013-05-01T00:43:04Z <p>Yes, this is standard functional analysis. You're starting out the right way.</p> <p>I'll sketch the argument. First observe that $\phi$ is invariant for the ${\bf Z}^n$ action if and only if it commutes with the multiplication operators $M_h$ for $h(z_1, \ldots, z_n) = z_1^{l_1}\cdots z_n^{l_n}$. By taking linear combinations (and using the fact that the $l_i$ can be negative) this implies that it commutes with $M_h$ for $h$ any polynomial in the $z_i$'s and $\bar{z}_i$'s. Now these polynomials are uniformly dense in the continuous functions on the torus, $C({\bf T}^n)$, so by taking limits we can conclude that $\phi$ commutes with $M_h$ for any continuous function $h$ on the torus. Since $C({\bf T}^n)$ is weak* dense in $L^\infty({\bf T}^n)$, we can now take weak* limits and deduce that $\phi$ commutes with $M_h$ for any $h \in L^\infty({\bf T}^n)$.</p> <p>So we now know that $\phi(1_A) = \phi(1_A\cdot 1) = 1_A\cdot\phi(1)$ for any measurable set $A$, where $1_A$ is the characteristic function of $A$. If $|\phi(1)| > \|\phi\|$ on a positive measure set we can take $A$ to be this set and get a contradiction. That's how you know $\phi(1)$ is bounded. The rest is easy: we have $\phi(h) = \phi(h\cdot 1) = h\cdot \phi(1)$ for any $h \in L^\infty({\bf T}^n)$, and these functions are dense in $L^2({\bf T}^n)$, so we conclude that $\phi = M_{\phi(1)}$ by continuity.</p> <p>There's some work you have to do to check the approximation arguments if you're doing this from scratch, but that's the proof in outline.</p> http://mathoverflow.net/questions/128701/existence-of-a-projection-operator-onto-a-classical-set-of-density-matrices/128839#128839 Answer by Nik Weaver for Existence of a projection operator onto a classical set of density matrices Nik Weaver 2013-04-26T17:13:00Z 2013-04-26T17:13:00Z <blockquote> <p>Can we construct a linear projection operator P onto C?</p> </blockquote> <p>No. The range of any linear operator will be a linear subspace.</p> <blockquote> <p>If not, is there a nonlinear projection operator and if so how would one construct it?</p> </blockquote> <p>Yes, if $K$ is a closed convex subset of a Hilbert space $H$ there is a standard "projection" map $P: H \to K$ defined by letting $Pv$ be the closest element of $K$ to $v$. I guess the basic properties of this map are that $Pv = v$ for any $v \in K$ and $\|Pv - Pw\| \leq \|v - w\|$, i.e., $P$ is a contraction.</p> http://mathoverflow.net/questions/128416/uncountable-pre-image/128421#128421 Answer by Nik Weaver for Uncountable Pre-Image Nik Weaver 2013-04-23T00:53:13Z 2013-04-23T00:53:13Z <p>I think so, yes. Let $x$ be in the range of $f$ and define $U = f^{-1}((-\infty, x))$, $V = f^{-1}((x,\infty))$. Since $f$ is open, $U$ and $V$ are both nonempty. So they are disjoint nonempty open sets, which means that the complement of $f^{-1}(x)$ is disconnected. But the complement of any countable subset of ${\bf R}^2$ is connected. (Even path connected, by a simple cardinality argument: for any distinct $p$ and $q$ one can find uncountably many paths from $p$ to $q$, any two of which are disjoint except for the points $p$ and $q$. So a countable set can obstruct only countably many of them.)</p> http://mathoverflow.net/questions/127918/are-compact-sets-in-a-banach-lattice-order-bounded/127921#127921 Answer by Nik Weaver for Are compact sets in a Banach lattice order bounded? Nik Weaver 2013-04-18T05:52:36Z 2013-04-18T05:52:36Z <blockquote> <p>One way of disproving this conjecture is to construct a norm convergent sequence in E which is not order bounded.</p> </blockquote> <p>E.g. the sequence $(\frac{1}{n}e_n)$ in $l^1$.</p> http://mathoverflow.net/questions/126477/resolvent-of-laplacian/126482#126482 Answer by Nik Weaver for Resolvent of Laplacian Nik Weaver 2013-04-04T06:30:19Z 2013-04-04T16:53:31Z <p>Sure, since $-\Delta$ is a positive operator, by the spectral theorem it can be realized as multiplication by a positive function $f(x)$ on some $L^2(X)$ space. Then $R(\xi)$ is multiplication by $\frac{1}{\xi - f(x)}$ and its operator norm is the sup norm of this function. If $\xi &lt; 0$ then the function $\frac{1}{\xi - f(x)}$ is bounded by $\frac{1}{|\xi|}$ in absolute value, so as $\xi \to -\infty$ we have $\|R(\xi)\| \to 0$.</p> <p>The point is that $\|R(\xi)\|$ equals one over the distance from $\xi$ to ${\rm spec}(-\Delta)$.</p> <p>If you restrict $\xi$ to be positive the question is more interesting. On the unit circle ${\bf T}^1$ the eigenvalues of $-\Delta$ are the square integers, and for any $\epsilon > 0$ we can find $\xi > 0$ such that $|\xi - n^2| > \frac{1}{\epsilon}$ for every $n \in {\bf Z}$, so we can still ensure that $\|R(\xi)\| \to 0$. In other words, the eigenvalues are spaced farther and farther apart so $\xi > 0$ can be chosen arbitrarily far from the spectrum of $-\Delta$. But on ${\bf T}^4$ the eigenvalues of $-\Delta$ are of the form $a^2 + b^2 + c^2 + d^2$ for $a, b, c, d \in {\bf Z}$, which means that every positive integer is an eigenvalue. Thus any $\xi > 0$ is at most $\frac{1}{2}$ units away from an eigenvalue, and therefore $\|R(\xi)\| \geq 2$ for every $\xi > 0$.</p> http://mathoverflow.net/questions/126236/is-the-image-of-discrete-set-under-an-open-map-discrete/126240#126240 Answer by Nik Weaver for Is the image of discrete set under an open map discrete? Nik Weaver 2013-04-02T07:38:25Z 2013-04-02T07:38:25Z <p>No, that's false. You didn't say that $f$ is a homomorphism, but the answer is still no if we require this.</p> <p>Let $G = {\bf Z} \times C_2 \times C_3 \times C_5 \times \cdots$ be the product of the infinite cyclic group and the cyclic groups of all prime orders. Let $H = C_2 \times C_3 \times C_5 \times \cdots$ and let $f: G \to H$ be the obvious homomorphism with kernel ${\bf Z}$. Let $Y$ be the subgroup of $G$ generated by the element $(1, 1, 1, \ldots)$. It is a discrete infinite cyclic subgroup, but its image in $H$ is not discrete (in fact, it is dense in $H$).</p> http://mathoverflow.net/questions/126138/order-isomorphic-down-set-lattices/126142#126142 Answer by Nik Weaver for Order-isomorphic down-set lattices Nik Weaver 2013-04-01T00:35:27Z 2013-04-01T00:35:27Z <p>I wasn't aware of the previous work which called them "superalgebraic lattices". In my book on Lipschitz algebras I called them "Stone lattices" because they exhibit some analogies to Stone spaces. (In this analogy, completely distributive complete lattices correspond to compact Hausdorff spaces.)</p> <p>To add to Joseph's answer, you can identify ${\mathcal O}(P)$ with the set of order-preserving maps from $P$ into the two-element lattice ${\bf 2}$. And you recover $P$ as the set of complete $0,1$-lattice homomorphisms from ${\mathcal O}(P)$ into ${\bf 2}$.</p> http://mathoverflow.net/questions/117754/fourier-transform-on-the-discrete-cube Fourier transform on the discrete cube Nik Weaver 2012-12-31T19:45:03Z 2013-03-29T09:56:13Z <p>Notation: identify an element of <code>$\{-1,1\}^n$</code> with the set <code>$S \subseteq \{1, \ldots, n\}$</code> on which it takes the value <code>$-1$</code>.</p> <p>The following is an asymptotic question. "Close to one" means "more than <code>$r_n$</code>" and "away from <code>$\frac{n}{2}$</code>" means "outside the interval <code>$[\frac{n}{2} - k_n\sqrt{n}, \frac{n}{2} + k_n\sqrt{n}]$</code>", for some <code>$r_n$</code> and <code>$k_n$</code> which respectively increase to one and infinity as $n \to \infty$.</p> <p>Conjecture: for any subset of <code>$\{-1,1\}^n$</code> there is a complex valued function with <code>$l^2$</code> norm equal to 1, supported either on that subset or on its complement, whose Fourier transform has <code>$l^2$</code> norm close to one on <code>$\{S: |S|$</code> is away from <code>$\frac{n}{2}\}$</code>.</p> <p>A positive answer would have very interesting consequences. It would mean that a single subspace of <code>$l^2(\{-1,1\}^n)$</code> whose dimension is small compared to the whole space is close to every subspace spanned by standard basis vectors or its complement.</p> <p>I'm not familiar with the literature on Fourier analysis on the discrete cube. Is there anything there that would help settle this question?</p> http://mathoverflow.net/questions/117912/faces-in-the-discrete-cube faces in the discrete cube Nik Weaver 2013-01-02T22:29:37Z 2013-03-29T09:44:32Z <p>This arose from a question Gil Kalai asked about a problem I posed involving <a href="http://mathoverflow.net/questions/117754/fourier-transform-on-the-discrete-cube" rel="nofollow">the Fourier transform on the discrete cube</a>. Maybe it is more tractable. I'm afraid I'm not sure how to do this kind of computation.</p> <p>A <em>$k$-dimensional face</em> of the discrete cube <code>$\{0,1\}^n$</code> is a set of the form: all vertices which take prescribed values (either <code>$0$</code> or <code>$1$</code>) on some given <code>$n-k$</code> coordinates and are otherwise arbitrary.</p> <p>The question is: does a typical subset of <code>$\{0,1\}^n$</code> approximately contain a face of dimension greater than <code>$.6n$</code>?</p> <p>We are interested in the limit as <code>$n\to\infty$</code>. So "approximately contains" means "contains all but a fraction which goes to <code>$0$</code> as <code>$n\to\infty$</code>". And "typical subset" means that as <code>$n\to\infty$</code> the fraction of subsets for which this fails goes to zero. The <code>$.6n$</code> can be moved a bit closer to <code>$.5n$</code> but I am assuming this is not crucial.</p> <p>A positive answer to this question would imply a generically positive answer to the Fourier transform question.</p> http://mathoverflow.net/questions/96782/vector-balancing-problem vector balancing problem Nik Weaver 2012-05-12T16:20:43Z 2013-03-29T09:42:22Z <p>This problem may be hard, so I don't expect an off-the-cuff solution. But can anyone suggest possible proof strategies?</p> <p>I have vectors $v_1, \ldots, v_k$ in ${\bf R}^n$. Each of them has euclidean length at most $.01$, and for every unit vector $u \in {\bf R}^n$ they satisfy $$\sum_{i=1}^k |\langle u,v_i\rangle|^2 = 1.$$ Is it possible to find a set of indices <code>$S \subset \{1, \ldots, k\}$</code> such that $$.0001 &lt; \sum_{i \in S} |\langle u,v_i\rangle|^2 &lt; .9999$$ for every unit vector $u$? This will imply the same bounds when summing over the complement of $S$.</p> <p>The $.01$ and $.0001$ aren't important; I just need the result for some positive $\delta$ and $\epsilon$. But they have to be independent of $k$ and $n$. (This may seem unlikely, until you try to construct a counterexample.)</p> <p>The motivation is that this is a (very slightly simplified) equivalent version of the famous Kadison-Singer problem. A solution would have important consequences in operator theory, harmonic analysis, and C*-algebra. Many people have worked on this problem, but perhaps not in the above form, which I feel exposes the combinatorial difficulty which is the real root of the problem.</p> http://mathoverflow.net/questions/108661/approximate-uncertainty-principle-for-finite-abelian-groups approximate uncertainty principle for finite abelian groups Nik Weaver 2012-10-02T19:58:02Z 2013-03-29T07:40:15Z <p>Can we find, for every $\epsilon > 0$, an example of the following?</p> <p>$\bullet$ a finite abelian group $G$</p> <p>$\bullet$ a small subset $T$ of the dual group $\hat{G}$ (meaning $|T| \leq \epsilon |\hat{G}|$)</p> <p>such that for any subset $S$ of $G$, there is a nonzero complex valued function $f$ supported either on $S$ or $S^c$ whose Fourier transform is approximately supported on $T$ (meaning $\|\hat{f} - \hat{f}|_T\|_2 \leq \epsilon \|\hat{f}\|_2$).</p> <p>It seems unlikely, but I don't know any version of the uncertainty principle that would forbid this.</p> <hr> <p>Two comments: (1) This arose out of discussions I had with Chuck Akemann about the Kadison-Singer problem. It wouldn't, as it stands, imply a negative solution to Kadison-Singer, but it would be close and (I believe) could probably be easily converted into a full negative solution. (2) Since either $S$ or $S^c$ has cardinality at least $|G|/2$, one could consider assuming $|S| \geq |G|/2$ and demanding that $f$ be supported on $S$. But I know that this makes the problem too hard. (Given $T$, I can always find an $S$ with $|S^c|/|G| = O(\sqrt{\epsilon})$ such that no nonzero function supported on $S$ has Fourier transform approximately supported on $T$.)</p> http://mathoverflow.net/questions/125446/consistency-of-the-concept-of-the-collection-of-all-collection/125458#125458 Answer by Nik Weaver for Consistency of the concept of the collection of all collection Nik Weaver 2013-03-24T14:36:39Z 2013-03-24T14:47:48Z <p>I presented a system for reasoning about arbitrary concepts, including the concept "is a concept". It's located at arXiv:1112.6129. The system has full comprehension; there is no restriction on "subconcept" formation. The paper includes a consistency proof.</p> <p>The key idea of my paper is that we do not have a global notion of an object falling under a concept, in exactly the same way that we do not have a global notion of an assertion being true. The best we can do is talk about provably falling under, referring (as constructivists do) to the general semantic notion of provability, not to provability within some formal system. My system deals with concepts like "is a concept that does not provably fall under itself" by restricting the axioms relating to provability. One of the basic axioms that you expect to hold for provability turns out not to have a clear justification.</p> <p>Maybe no one can read this paper because you'd have to be familiar with both Fregean concepts and intuitionistic logic, as well as comfortable with substantial technical content.</p> http://mathoverflow.net/questions/125206/generalization-of-zero-diagonal-square-matrices-to-linear-operators/125209#125209 Answer by Nik Weaver for Generalization of zero-diagonal square matrices to linear operators Nik Weaver 2013-03-21T22:50:15Z 2013-03-21T22:50:15Z <p>The obvious answer is those operators $A$ on $l^2$ which satisfy $\langle Ae_k, e_k\rangle = 0$ for all $k$. But you'd really have to tell us more about your motivation.</p> http://mathoverflow.net/questions/125126/functional-analysis/125130#125130 Answer by Nik Weaver for functional analysis Nik Weaver 2013-03-21T05:48:29Z 2013-03-21T05:48:29Z <p>Yes. $\delta(\epsilon)/\epsilon$ is nondecreasing on $[0,2]$, so once $\delta(\epsilon)$ is nonzero it is strictly increasing. See T. Figiel, On the moduli of convexity and smoothness, <em>Studia Math.</em> <strong>56</strong> (1976), 121–155.</p> http://mathoverflow.net/questions/124345/limit-with-theorem-of-dominated-convergence/124377#124377 Answer by Nik Weaver for Limit with theorem of dominated convergence Nik Weaver 2013-03-13T04:14:50Z 2013-03-13T04:14:50Z <p>I think the limit of the integrals does equal the integral of the limit. First of all, outside of a small ball about $y$ we have dominated convergence. So the issue is whether $$\lim_{|x-y| \to 0} \int_{ball(y,\epsilon)} \frac{e^{i\lambda|x-x'|}}{|x-x'|} f(x')dx'$$ equals the same integral with $y$ in place of $x$. Thus we can ignore the exponential factor --- when $x$ gets close to $y$ it's essentially constant on this ball --- and the weight $(1 + |x|^2)^s$ which governs behavior at infinity but is irrelevant near $y$.</p> <p>Fix $\epsilon > 0$ and let $h_x(x') = \frac{1}{|x-x'|}\cdot \chi_{ball(y,\epsilon)}$. I claim that $h_x \to h_y$ weakly in $L^2$ as $x \to y$. This will imply that $\int h_xf \to \int h_yf$, which is all you need for the reasons I just explained. Well, the functions $h_x$ are uniformly bounded in $L^2$, so weak convergence will follow from weak convergence when integrated against a dense subset of $L^2$. But $\int h_xg \to \int h_yg$ for any $L^2$ function $g$ which is $0$ on a neighborhood of $y$, by dominated convergence, and such functions $g$ are dense, so we are done.</p> <p>(In fact weak convergence plus the fact that $\|h_x\| \to \|h_y\|$ implies that $h_x \to h_y$ in norm, which presumably you could show by direct calculation.)</p> http://mathoverflow.net/questions/122800/products-of-boolean-algebras-and-probability-measures-thereon/122802#122802 Answer by Nik Weaver for Products of Boolean algebras and probability measures thereon Nik Weaver 2013-02-24T15:42:12Z 2013-02-24T15:42:12Z <p>This is called the "direct sum" or "internal sum" of Boolean algebras. See <em>Introduction to Boolean Algebras</em> by Givant and Halmos, p. 427 for the abstract definition and p. 432 for the concrete description you want (the elements of the internal sum are finite joins of finite meets of elements and complements of elements from the disjoint union of the summands).</p> http://mathoverflow.net/questions/121926/from-point-wise-to-essential-supremum-of-a-set-of-real-valued-measurable-function/122561#122561 Answer by Nik Weaver for From point-wise to essential supremum of a set of real-valued measurable functions Nik Weaver 2013-02-21T18:26:46Z 2013-02-21T18:26:46Z <p>Okay. Give <code>$[0,1]$</code> Lebesgue measure. For each <code>$t \in [0,1]$</code> let <code>$S_t = \{1_{\{t\}}\} \cup \{a\cdot 1_{[0,1]}: a \geq 1\}$</code>.</p> <p>Then <code>$\bigcap S_t = \{a\cdot 1_{[0,1]}: a \geq 1\}$</code> and its inf is the function <code>$1_{[0,1]}$</code>. For each <code>$t$</code> the inf of <code>$S_t$</code> is the function <code>$1_{\{t\}}$</code>, and the sup of these is also the function <code>$1_{[0,1]}$</code>. So your premise holds.</p> <p>The essential inf of <code>$\bigcap S_t$</code> is also the function <code>$1_{[0,1]}$</code> but the essential inf of each <code>$S_t$</code> is the zero function and their essential sup is again the zero function. So the conclusion fails.</p> <p>Is this really what you meant?</p> http://mathoverflow.net/questions/122406/undecidability-and-holomorphic-functions-reference-request/122426#122426 Answer by Nik Weaver for Undecidability and holomorphic functions (Reference request) Nik Weaver 2013-02-20T17:09:03Z 2013-02-20T18:06:01Z <p>Very likely the fact that you are trying to remember is the interpolation problem solved by Erdős.</p> <p><a href="http://www.renyi.hu/~p_erdos/1964-04.pdf" rel="nofollow">http://www.renyi.hu/~p_erdos/1964-04.pdf</a></p> http://mathoverflow.net/questions/122159/is-this-result-of-spain-correct/122185#122185 Answer by Nik Weaver for Is this result of Spain correct? Nik Weaver 2013-02-18T15:50:45Z 2013-02-18T15:50:45Z <p>I haven't looked at the paper but your counterexample is mistaken. The basis projections generate not $B(l^p)$ but the algebra of multiplication operators, which is isometrically isomorphic to $l^\infty$ and hence is a von Neumann algebra.</p> http://mathoverflow.net/questions/121926/from-point-wise-to-essential-supremum-of-a-set-of-real-valued-measurable-function/121937#121937 Answer by Nik Weaver for From point-wise to essential supremum of a set of real-valued measurable functions Nik Weaver 2013-02-15T19:59:56Z 2013-02-15T19:59:56Z <p>The question is not clear. Are you asking whether the complete distributive law holds in (probably the unit ball of) $L^\infty(X,\mu)$? The answer is no: complete distributivity is characteristic of atomic measure spaces. An easy way to see this is to use the fact that a complete lattice is completely distributive iff it has the property that for all $c$ and $d$ with $c \not\geq d$ there exist $c' \not\leq c$ and $d' \not\geq d$ such that every element of the lattice lies above $c'$ or below $d'$. I refer you to Theorem 5.3.5 of my book <em>Lipschitz Algebras</em> for a proof. Let $A$ be a positive measure set that contains no atoms, find $B \subset A$ with $0 &lt; \mu(B) &lt; \mu(A)$, and take $c = \chi_B$ and $d = \chi_{A\setminus B}$.</p> http://mathoverflow.net/questions/121526/are-all-points-x-of-the-boundary-of-a-convex-set-c-of-a-hilbert-space-h-projectio/121532#121532 Answer by Nik Weaver for Are all points x of the boundary of a convex set C of a Hilbert space H projections onto C of a point different than x? Nik Weaver 2013-02-11T21:32:32Z 2013-02-12T23:02:18Z <p>I'm a little concerned that my original incorrect answer was accepted and the only feedback on the edit was a comment that it didn't make sense ... here's a slightly simpler counterexample.</p> <p>Let $H = l^2$ and let $C$ be the set of sequences $(a_n)$ satisfying $|a_n| \leq \frac{1}{n}$ for all $n$. $C$ is clearly closed and convex.</p> <p>$0$ is a boundary point of $C$ (in fact $C$ has no interior), but it is not the nearest element of $C$ to any point outside of $C$. If $(a_n)$ is any nonzero sequence in $l^2$ then some entry $a_{n_0}$ must be nonzero, and then we can find a point in $C$ that is closer to $(a_n)$ than $0$ is. For sufficiently small $\epsilon$, the point $\epsilon a_{n_0}e_{n_0}$ (where $e_n$ is the standard basis) works. Another way to say this is that $0$ has no support hyperplane.</p> http://mathoverflow.net/questions/120485/fourier-analytic-proofs/120531#120531 Answer by Nik Weaver for fourier analytic proofs Nik Weaver 2013-02-01T18:19:56Z 2013-02-01T18:19:56Z <p>Hermann Weyl's delightful proof that for irrational $\alpha$ the sequence of values $k\alpha$ mod $1$, $k \in {\bf N}$, is uniformly distributed in $[0,1]$ deserves a mention. It's so simple I can summarize it here. First we check that for any nonzero $n \in {\bf Z}$ we have $$\frac{1 + e^{2\pi i n\alpha} + \cdots + e^{2\pi in(k-1)\alpha}}{k} \to 0$$ as $k \to \infty$. This is just a simple computation since the numerator is a geometric series. For $n = 0$ the displayed fraction reduces to $\frac{k}{k} = 1$. Since $\int_0^1 e^{2\pi i nx} dx = 1$ or $0$ depending on whether $n = 0$ or $n \neq 0$, it follows that $$\frac{1}{k}\sum_{j=0}^{k-1} e^{2\pi i nj\alpha} \to \int_0^1 e^{2\pi inx} dx$$ for all $n \in {\bf Z}$. Setting $x_j = j\alpha$ mod $1$ and taking linear combinations then yields $$\frac{1}{k}\sum_{j=0}^{k=1} f(x_j) \to \int_0^1 f(x) dx$$ for any trigonometric polynomial $f$, and by straightforward approximation arguments we get the same conclusion, first for any continuous function $f$ on $[0,1]$ and then for $f = \chi_{[a,b]}$. But with this $f$ the left side becomes the fraction of values $j\alpha$ mod $1$ for $0 \leq j \leq k-1$ which lie in $[a,b]$ and the right side becomes $b-a$, so this is just the statement of uniform distribution.</p> http://mathoverflow.net/questions/119646/is-there-a-maximum-to-the-amount-of-disjoint-non-measurable-subsets-of-the-unit-i/119650#119650 Answer by Nik Weaver for Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure? Nik Weaver 2013-01-23T12:22:49Z 2013-01-23T12:32:37Z <p>I can solve this assuming the continuum hypothesis. (Edit: CH isn't needed, see below.) Lemma: if $A$ is a countable set and $(S_m)$ is a sequence of uncountable sets then we can find a sequence of disjoint countable sets $(T_n)$ such that $A \cap T_n = \emptyset$ for all $n$ and $S_m \cap T_n \neq \emptyset$ for all $m$ and $n$. [Proof: Choose a countable subset $S_1'$ of $S_1 \setminus A$, enumerate it, and put the $n$th element in $T_n$. Then choose a countable subset $S_2'$ of $S_2 \setminus (A \cup S_1')$, enumerate it, and put the $n$th element in $T_n$. Proceed in this way. Each $S_k'$ is countable so the difference $S_{k+1} \setminus (A \cup S_1' \cup \cdots \cup S_k')$ is always uncountable.]</p> <p>Now there are $2^{\aleph_0}$ open subsets of $[0,1]$ because any open subset is a union of rational intervals and there are only countably many rational intervals. So there are only $2^{\aleph_0}$ closed subsets of $[0,1]$. Assume CH and enumerate the closed subsets of $[0,1]$ of positive measure as $C_\alpha$ for $\alpha &lt; \aleph_1$. Observe that each $C_\alpha$ is uncountable. Now we construct disjoint countable sets $T_{\alpha,\beta}$ for $\beta &lt; \alpha &lt; \aleph_1$ by recursion on $\alpha$ as follows. At step $\alpha$ let $A = \bigcup_{\beta' &lt; \alpha' &lt;\alpha} T_{\alpha',\beta'}$ (all the $T$s constructed so far, a countable union of countable sets) and apply the lemma to this $A$ and the sets $C_\beta$ for $\beta &lt; \alpha$. There are only countably many $\beta &lt; \alpha$ so we can do this, and we can relabel the resulting sets $T_n$ as $T_{\alpha,\beta}$ for $\beta &lt; \alpha$.</p> <p>After this process is complete, for each $\beta$ let $T_\beta = \bigcup_{\alpha > \beta} T_{\alpha,\beta}$. Then the sets $T_\beta$ are disjoint, there are $\aleph_1$ of them, and each one intersects every closed subset of $[0,1]$ of positive measure, so each of them has full outer measure.</p> <p>Edit: actually this doesn't need CH. Every closed set is the union of a countable set and a perfect set, so if it has positive measure then it contains $2^{\aleph_0}$ elements. That's enough to keep the induction going for $\alpha &lt; 2^{\aleph_0}$ since each $T_{\alpha,\beta}$ will have cardinality $&lt; 2^{\aleph_0}$.</p> http://mathoverflow.net/questions/119262/example-of-a-topological-space/119265#119265 Answer by Nik Weaver for Example of a topological space..... Nik Weaver 2013-01-18T12:27:20Z 2013-01-18T12:27:20Z <p>Not possible, in a completely regular space every open set is a union of cozero sets. Just collect the clopen sets that you use for these subordinate cozero sets.</p> http://mathoverflow.net/questions/131748/can-an-uniformly-picked-real-number-be-an-integer Comment by Nik Weaver Nik Weaver 2013-05-24T17:22:49Z 2013-05-24T17:22:49Z Incidentally, the &quot;probability zero never happens&quot; language came not from the highly upvoted answer he refers to, but from OP's own response to it ... http://mathoverflow.net/questions/131748/can-an-uniformly-picked-real-number-be-an-integer Comment by Nik Weaver Nik Weaver 2013-05-24T17:21:33Z 2013-05-24T17:21:33Z There are several grounds on which this question could be closed. http://mathoverflow.net/questions/131673/proof-that-a-finitely-generated-projective-module-over-a-von-neumann-regular-ring Comment by Nik Weaver Nik Weaver 2013-05-24T03:16:59Z 2013-05-24T03:16:59Z Look in Ken Goodearl's book on von Neumann regular rings. http://mathoverflow.net/questions/131062/importance-of-separability-vs-second-countability/131074#131074 Comment by Nik Weaver Nik Weaver 2013-05-19T06:08:55Z 2013-05-19T06:08:55Z It sounds like you are asking why separability is important. Speaking from my area of expertise, I could tell you any number of basic theorems about separable C*-algebras which fail in the nonseparable case. http://mathoverflow.net/questions/131062/importance-of-separability-vs-second-countability/131074#131074 Comment by Nik Weaver Nik Weaver 2013-05-18T20:50:17Z 2013-05-18T20:50:17Z @Martin: do you wish to tell me that this simple fact is <i>not important</i>? http://mathoverflow.net/questions/130750/calculus-of-variations-and-quantum-information Comment by Nik Weaver Nik Weaver 2013-05-15T19:21:48Z 2013-05-15T19:21:48Z The question seems quite arbitrary. Depending on how hard you want to strain, I suppose you could find a connection between any two areas of mathematics. You don't give any motivation for why you picked these two, other than that you like both of them ... voting to close. Yes, please try to ask something more focused next time. http://mathoverflow.net/questions/130433/direct-limit-union Comment by Nik Weaver Nik Weaver 2013-05-12T21:04:12Z 2013-05-12T21:04:12Z I don't think this question is appropriate for mathoverflow, but ... think about whether the connecting maps are injective. http://mathoverflow.net/questions/130045/is-bv2-space-closed-in-l2-space Comment by Nik Weaver Nik Weaver 2013-05-08T04:27:56Z 2013-05-08T04:27:56Z Probably $S$ contains a dense subspace of $L^2$ like, say, the $C^\infty$ functions with compact support, but easy examples show that it is not all of $L^2$, so ... http://mathoverflow.net/questions/129563/decidability-of-equality-of-expressions-built-using-1/129568#129568 Comment by Nik Weaver Nik Weaver 2013-05-03T19:11:01Z 2013-05-03T19:11:01Z That's right, I agree. http://mathoverflow.net/questions/129478/does-the-border-boundary-points-of-a-convex-body-make-a-concave-function/129484#129484 Comment by Nik Weaver Nik Weaver 2013-05-03T07:51:27Z 2013-05-03T07:51:27Z @Wlodzimierz: yes. I've edited my answer to clarify this. http://mathoverflow.net/questions/129465/number-of-linear-maps-is-less-or-equal-than-the-dimension-of-the-vector-space Comment by Nik Weaver Nik Weaver 2013-05-02T22:45:56Z 2013-05-02T22:45:56Z This is a site for research mathematics, not homework. http://mathoverflow.net/questions/129366/generalization-of-et-a-to-et-alphaa Comment by Nik Weaver Nik Weaver 2013-05-02T03:18:47Z 2013-05-02T03:18:47Z It isn't a one-parameter semigroup for $\alpha \neq 1$. The product law fails. http://mathoverflow.net/questions/129254/certain-bounded-linear-operators-on-l2-of-a-torus/129268#129268 Comment by Nik Weaver Nik Weaver 2013-05-01T18:43:35Z 2013-05-01T18:43:35Z No problem, you're welcome. http://mathoverflow.net/questions/129238/would-matrices-not-exist-if-it-were-not-for-geometry Comment by Nik Weaver Nik Weaver 2013-04-30T17:59:34Z 2013-04-30T17:59:34Z The question doesn't belong here. This site is for research level math questions. http://mathoverflow.net/questions/129160/existence-of-a-continuous-and-unbounded-map-f-with-ffxx/129181#129181 Comment by Nik Weaver Nik Weaver 2013-04-30T17:07:13Z 2013-04-30T17:07:13Z Oh, I didn't notice the minus sign ... yes, that seems to work. Nice!