User nik weaver - MathOverflow

Answer by Nik Weaver for Importance of separability vs. second-countability

2013-05-18T18:13:04Z

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.

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.

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.

Answer by Nik Weaver for Is there any proof that you feel you do not "understand"?

2013-05-17T02:06:13Z

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.

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!

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.

Answer by Nik Weaver for A space parameterizing the choices of orthonormal bases for a Hilbert space

2013-05-14T02:39:16Z

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.

Here is a link that may be helpful.

Answer by Nik Weaver for A characterization of Hilbert spaces?

2013-05-12T04:32:58Z

A related question was asked earlier by Mark Meckes: Self-dual finite-dimensional complex normed spaces. 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.

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.

Answer by Nik Weaver for Decidability of equality of expressions built using 1,+,-,*,/,^

2013-05-03T18:42:55Z

The equational theory of $\langle {\bf N}, 0, 1, +, \times, \uparrow\rangle$ is decidable, but not finitely axiomatizable.

Answer by Nik Weaver for Does the Border (Boundary) Points of a convex body make a concave function?

2013-05-03T03:02:58Z

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.

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 $1$.

Answer by Nik Weaver for Certain bounded linear operators on L^2 of a torus

2013-05-01T00:43:04Z

Yes, this is standard functional analysis. You're starting out the right way.

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 $h \in L^\infty({\bf T}^n)$.

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.

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.

Answer by Nik Weaver for Existence of a projection operator onto a classical set of density matrices

2013-04-26T17:13:00Z

Can we construct a linear projection operator P onto C?

No. The range of any linear operator will be a linear subspace.

If not, is there a nonlinear projection operator and if so how would one construct it?

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.

Answer by Nik Weaver for Uncountable Pre-Image

2013-04-23T00:53:13Z

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.)

Answer by Nik Weaver for Are compact sets in a Banach lattice order bounded?

2013-04-18T05:52:36Z

One way of disproving this conjecture is to construct a norm convergent sequence in E which is not order bounded.

E.g. the sequence $(\frac{1}{n}e_n)$ in $l^1$.

Answer by Nik Weaver for Resolvent of Laplacian

2013-04-04T06:30:19Z

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 $\xi < 0$ then the function $\frac{1}{\xi - f(x)}$ is bounded by $\frac{1}{|\xi|}$ in absolute value, so as $\xi \to -\infty$ we have $\|R(\xi)\| \to 0$.

The point is that $\|R(\xi)\|$ equals one over the distance from $\xi$ to ${\rm spec}(-\Delta)$.

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 $\xi > 0$.

Answer by Nik Weaver for Is the image of discrete set under an open map discrete?

2013-04-02T07:38:25Z

No, that's false. You didn't say that $f$ is a homomorphism, but the answer is still no if we require this.

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 $H$).

Answer by Nik Weaver for Order-isomorphic down-set lattices

2013-04-01T00:35:27Z

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.)

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 ${\bf 2}$.

Fourier transform on the discrete cube

2012-12-31T19:45:03Z

Notation: identify an element of $\{-1,1\}^n$ with the set $S \subseteq \{1, \ldots, n\}$ on which it takes the value $-1$ .

The following is an asymptotic question. "Close to one" means "more than $r_n$ " and "away from $\frac{n}{2}$ " means "outside the interval $[\frac{n}{2} - k_n\sqrt{n}, \frac{n}{2} + k_n\sqrt{n}]$ ", for some $r_n$ and $k_n$ which respectively increase to one and infinity as $n \to \infty$.

Conjecture: for any subset of $\{-1,1\}^n$ there is a complex valued function with $l^2$ norm equal to 1, supported either on that subset or on its complement, whose Fourier transform has $l^2$ norm close to one on $\{S: |S|$ is away from $\frac{n}{2}\}$ .

A positive answer would have very interesting consequences. It would mean that a single subspace of $l^2(\{-1,1\}^n)$ whose dimension is small compared to the whole space is close to every subspace spanned by standard basis vectors or its complement.

I'm not familiar with the literature on Fourier analysis on the discrete cube. Is there anything there that would help settle this question?

faces in the discrete cube

2013-01-02T22:29:37Z

This arose from a question Gil Kalai asked about a problem I posed involving the Fourier transform on the discrete cube. Maybe it is more tractable. I'm afraid I'm not sure how to do this kind of computation.

A $k$-dimensional face of the discrete cube $\{0,1\}^n$ is a set of the form: all vertices which take prescribed values (either $0$ or $1$ ) on some given $n-k$ coordinates and are otherwise arbitrary.

The question is: does a typical subset of $\{0,1\}^n$ approximately contain a face of dimension greater than $.6n$ ?

We are interested in the limit as $n\to\infty$ . So "approximately contains" means "contains all but a fraction which goes to $0$ as $n\to\infty$ ". And "typical subset" means that as $n\to\infty$ the fraction of subsets for which this fails goes to zero. The $.6n$ can be moved a bit closer to $.5n$ but I am assuming this is not crucial.

A positive answer to this question would imply a generically positive answer to the Fourier transform question.

vector balancing problem

2012-05-12T16:20:43Z

This problem may be hard, so I don't expect an off-the-cuff solution. But can anyone suggest possible proof strategies?

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 $S \subset \{1, \ldots, k\}$ such that $$.0001 < \sum_{i \in S} |\langle u,v_i\rangle|^2 < .9999$$ for every unit vector $u$? This will imply the same bounds when summing over the complement of $S$.

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.)

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.

approximate uncertainty principle for finite abelian groups

2012-10-02T19:58:02Z

Can we find, for every $\epsilon > 0$, an example of the following?

$\bullet$ a finite abelian group $G$

$\bullet$ a small subset $T$ of the dual group $\hat{G}$ (meaning $|T| \leq \epsilon |\hat{G}|$)

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 $\|\hat{f} - \hat{f}|_T\|_2 \leq \epsilon \|\hat{f}\|_2$).

It seems unlikely, but I don't know any version of the uncertainty principle that would forbid this.

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 $T$.)

Answer by Nik Weaver for Consistency of the concept of the collection of all collection

2013-03-24T14:36:39Z

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.

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.

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.

Answer by Nik Weaver for Generalization of zero-diagonal square matrices to linear operators

2013-03-21T22:50:15Z

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.

Answer by Nik Weaver for functional analysis

2013-03-21T05:48:29Z

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, Studia Math. 56 (1976), 121–155.

Answer by Nik Weaver for Limit with theorem of dominated convergence

2013-03-13T04:14:50Z

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 $y$.

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.

(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.)

Answer by Nik Weaver for Products of Boolean algebras and probability measures thereon

2013-02-24T15:42:12Z

This is called the "direct sum" or "internal sum" of Boolean algebras. See Introduction to Boolean Algebras 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).

Answer by Nik Weaver for From point-wise to essential supremum of a set of real-valued measurable functions

2013-02-21T18:26:46Z

Okay. Give $[0,1]$ Lebesgue measure. For each $t \in [0,1]$ let $S_t = \{1_{\{t\}}\} \cup \{a\cdot 1_{[0,1]}: a \geq 1\}$ .

Then $\bigcap S_t = \{a\cdot 1_{[0,1]}: a \geq 1\}$ and its inf is the function $1_{[0,1]}$ . For each $t$ the inf of $S_t$ is the function $1_{\{t\}}$ , and the sup of these is also the function $1_{[0,1]}$ . So your premise holds.

The essential inf of $\bigcap S_t$ is also the function $1_{[0,1]}$ but the essential inf of each $S_t$ is the zero function and their essential sup is again the zero function. So the conclusion fails.

Is this really what you meant?

Answer by Nik Weaver for Undecidability and holomorphic functions (Reference request)

2013-02-20T17:09:03Z

Very likely the fact that you are trying to remember is the interpolation problem solved by Erdős.

http://www.renyi.hu/~p_erdos/1964-04.pdf

Answer by Nik Weaver for Is this result of Spain correct?

2013-02-18T15:50:45Z

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.

Answer by Nik Weaver for From point-wise to essential supremum of a set of real-valued measurable functions

2013-02-15T19:59:56Z

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 Lipschitz Algebras for a proof. Let $A$ be a positive measure set that contains no atoms, find $B \subset A$ with $0 < \mu(B) < \mu(A)$, and take $c = \chi_B$ and $d = \chi_{A\setminus B}$.

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?

2013-02-11T21:32:32Z

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.

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.

$0$ 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.

Answer by Nik Weaver for fourier analytic proofs

2013-02-01T18:19:56Z

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.

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?

2013-01-23T12:22:49Z

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.]

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 $\alpha < \aleph_1$. Observe that each $C_\alpha$ is uncountable. Now we construct disjoint countable sets $T_{\alpha,\beta}$ for $\beta < \alpha < \aleph_1$ by recursion on $\alpha$ as follows. At step $\alpha$ let $A = \bigcup_{\beta' < \alpha' <\alpha} T_{\alpha',\beta'}$ (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 $\beta < \alpha$. There are only countably many $\beta < \alpha$ so we can do this, and we can relabel the resulting sets $T_n$ as $T_{\alpha,\beta}$ for $\beta < \alpha$.

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.

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 $\alpha < 2^{\aleph_0}$ since each $T_{\alpha,\beta}$ will have cardinality $< 2^{\aleph_0}$.

Answer by Nik Weaver for Example of a topological space.....

2013-01-18T12:27:20Z

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.

Comment by Nik Weaver

2013-05-24T17:22:49Z

Incidentally, the "probability zero never happens" language came not from the highly upvoted answer he refers to, but from OP's own response to it ...

Comment by Nik Weaver

2013-05-24T17:21:33Z

There are several grounds on which this question could be closed.

Comment by Nik Weaver

2013-05-24T03:16:59Z

Look in Ken Goodearl's book on von Neumann regular rings.

Comment by Nik Weaver

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.

Comment by Nik Weaver

2013-05-18T20:50:17Z

@Martin: do you wish to tell me that this simple fact is not important?

Comment by Nik Weaver

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.

Comment by Nik Weaver

2013-05-12T21:04:12Z

I don't think this question is appropriate for mathoverflow, but ... think about whether the connecting maps are injective.

Comment by Nik Weaver

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 ...

Comment by Nik Weaver

2013-05-03T19:11:01Z

That's right, I agree.

Comment by Nik Weaver

2013-05-03T07:51:27Z

@Wlodzimierz: yes. I've edited my answer to clarify this.

Comment by Nik Weaver

2013-05-02T22:45:56Z

This is a site for research mathematics, not homework.

Comment by Nik Weaver

2013-05-02T03:18:47Z

It isn't a one-parameter semigroup for $\alpha \neq 1$. The product law fails.

Comment by Nik Weaver

2013-05-01T18:43:35Z

No problem, you're welcome.

Comment by Nik Weaver

2013-04-30T17:59:34Z

The question doesn't belong here. This site is for research level math questions.

Comment by Nik Weaver

2013-04-30T17:07:13Z

Oh, I didn't notice the minus sign ... yes, that seems to work. Nice!