User justin moore - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T09:27:25Z http://mathoverflow.net/feeds/user/10774 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction What is the effect of adding 1/2 to a continued fraction? Justin Moore 2013-04-25T03:40:04Z 2013-04-29T02:43:46Z <p>Is there a simple answer to the question "what happens to the continued fraction expansion of an irrational number when you add 1/2?" A closely related question is "what happens to such an expansion when you multiply by 2?"</p> <p>Remarks: I'm not sure what qualifies for an answer. The motivation comes from wanting to better understand the equivalence relation on integer sequences generated by tail equivalence (which is generated by adding integers and taking reciprocals) and closure under doubling/halving. It is known that this Borel equivalence relation is not hyperfinite, so the answer cannot be too simple.</p> <p>Edit: The answers are not really what I am asking for. It is clear there is some recursive procedure for doing this, just like there is a recursive procedure for taking a square root of a decimal expansion. I'm looking for something which one might call a "closed form". For instance, if you start with a periodic expansion, adding 1/2 produces a new periodic expansion. Is there a simple transformation on the initial and periodic parts which corresponds to adding 1/2? For instance, can this be done with a finite state automaton? An authoritative "there is no such nice answer" would actually be an acceptable answer.</p> http://mathoverflow.net/questions/118254/usage-of-set-theory-in-undergraduate-studies/118451#118451 Answer by Justin Moore for Usage of set theory in undergraduate studies Justin Moore 2013-01-09T14:50:39Z 2013-01-09T14:50:39Z <p>As a set theorist, I feel some obligation to offer an answer here. First, the difficulties students may have in proving set theoretic containments like the one you mention above or in constructing $\epsilon$-$\delta$ proofs is not a matter of them struggling with set theory but rather of them struggling with something new: constructing a proof. In the case of $\epsilon$-$\delta$ proofs, a large part of this difficulty is in understanding quantifiers and how they work (for instance "for all ... exists ..." is not the same as "exists ... for all ...". This is because math is not a trivial subject to learn and some difficulty is required as the students minds stretch and grow. Surely there is no way around this.</p> <p>That really has nothing to do with set theory so far. In spite of a common misconception, set theorists do no actually care how it is that ordered pairs are defined. Or how exactly one codes the notion of a function. Set theory is not in competition with category theory, in spite of what category theory thinks. A very good analogy is provided by computer science: set theory is machine language (or maybe better: a low level language like C) and category theory is object oriented programming. While object oriented programming may provide a useful way of thinking about how to write a program, there still needs to be a machine language for the computer to run on. Moreover, there are occasionally things for which it is just better (or even necessary) to code in a low level language.</p> <p>Set theory provides an exact standard by which to discuss questions like "is there a subset of the real line which is uncountable but not of cardinality $|\mathbb{R}|$" (Hilbert's First Problem) or, maybe better, "is there an almost free, non free group?" (Whitehead's problem) or "If $h$ is a homomorphism a commutative Banach algebra into $C[0,1]$, is $h$ continuous?" (the negation being Kaplanski's conjecture). With the exception of the first question, these were asked, to my knowledge at least, without any thought that there was a foundational issue involved. Surely these are questions which could reasonably be asked regardless of how one sets up their foundations. To my knowledge category theory has never resolved these questions; set theory has in as satisfactory a manner possible (or at least until we adopt a more complete set of axioms). Now, one can argue at length about whether such questions are asked in poor taste or whether we should allow them to be asked at all. Readers interested in the question of "why care about set theory" should take a look at <a href="http://www.math.cornell.edu/~justin/Ftp/utility_unctble.pdf" rel="nofollow">this</a> (which might have been titled "why care about the uncountable").</p> http://mathoverflow.net/questions/114996/do-distinct-idempotent-measures-on-finite-binary-systems-have-distinct-supports Do distinct idempotent measures on finite binary systems have distinct supports? Justin Moore 2012-11-30T16:16:02Z 2012-11-30T18:40:11Z <p>Suppose that <code>$(S,*)$</code> is a finite set equipped with a binary operation. Extend the binary operation to the vector space <code>$V$</code> with basis <code>$S$</code>. The set of probability measures on <code>$S$</code>, viewed as a compact convex subset of <code>$V$</code> is closed under <code>$*$</code> and, since <code>$*$</code> is continuous, there are idempotent measures on <code>$S$</code>.</p> <p>Must two idempotent measures on $S$ have distinct supports?</p> <p>I am also interested in the more general question where the assumption of finiteness is dropped and one considers the extension (by convolution) of $*$ to the family of all finitely additive measures on $S$ (in that context, define the support of a measure to be all subsets of $S$ with positive measure).</p> http://mathoverflow.net/questions/107391/applications-of-idempotent-ultrafilters/114952#114952 Answer by Justin Moore for Applications of idempotent ultrafilters Justin Moore 2012-11-30T02:32:26Z 2012-11-30T02:32:26Z <p>I think one could make the case that idempotent ultrafilters are so closely related to Ramsey theory that anything which uses them does relate to Ramsey theory in some way (almost by definition). A less obvious example though would be Gowers's result that $c_0$ is oscillation stable --- that every bounded Lipschitz function on the sphere of $c_0$ is $\epsilon$-constant when restricted to an infinite dimensional subspace. That utilizes a system of idempotent ultrafilters on a families of (partial) semigroups known as $\mathrm{FIN}_k$ (where $k$ ranges over the natural numbers).</p> http://mathoverflow.net/questions/57903/idempotent-measures-on-the-free-binary-system Idempotent measures on the free binary system? Justin Moore 2011-03-09T02:06:32Z 2012-11-25T16:22:59Z <p>Let $(S,*)$ be the free (non associative) binary system on one generator (so $S$ is just the set of terms in $*$ and $1$). There is an extension of $*$ to the space $P(S)$ of finitely additive probability measures on $S$ defined as follows: $$\mu * \nu (A) = \int \int \mathbf{1}_{*^{-1}(A)} (x,y)\ d \nu (y)\ d \mu (x)$$ Here is the question: Is there an idempotent measure in $P(S)$ (i.e. a $\mu$ such that $\mu * \mu = \mu$)? Has this question been considered in the literature?</p> <p>Here is the motivation: there is a natural way to identify $(S,*)$ with the positive elements of Thompson's group $F$ (elements of $S$ are "rooted ordered binary trees"). It is not hard to show that an idempotent measure is in fact an invariant measure with respect to the action of $F$. Now, I don't expect someone to produce a positive answer (although I will conjecture the even stronger statement that every compact convex $C \subseteq P(S)$ which is $*$-closed contains an idempotent). I am mostly asking if someone sees how to refute the existence of such a measure or if this question has appeared in the literature.</p> <p>Some further observations: the map $(\mu,\nu) \mapsto \mu * \nu$ is NOT continuous ($\mu \mapsto \mu * \nu$ is, for each $\nu$, but this is about the extent of continuity). Thus the map $\mu \mapsto \mu * \mu$ is not continuous (if it were, we could apply a fixed point theorem...). If one drops the assumption of freeness, then it is possible to find idempotents if any of the following are true:</p> <ul> <li>$S$ is finite (in this case everything is continuous and so fixed point theorems apply).</li> <li>$(S,*)$ is associative (i.e. $S$ is a semigroup) (this is Ellis's Lemma'').</li> <li>$*$ depends only on one argument (again, fixed point theorems apply).</li> </ul> <p>An auxiliary question is to characterize when a monogenic binary system $(S,*)$ satisfies that $P(S)$ contains an idempotent. It should be noted that for associative binary operations, it is possible to find an idempotent ultrafilter (i.e. taking values in $\{0,1\}$) but that this is impossible for the free (non associative) binary system one one generator.</p> http://mathoverflow.net/questions/26821/is-thompsons-group-f-amenable/45891#45891 Answer by Justin Moore for Is Thompson's Group F amenable? Justin Moore 2010-11-13T02:47:53Z 2012-10-04T13:16:33Z <p>While I did not participate in most of the checking of Shavgulidze's argument, I can offer the following partial account of the situation. I am told the paper was correct except for a lemma (or sequence of them) claiming that a sequence of auxiliary measures had certain properties. These were Borel measures on the $n$-simplex (one for each $n$). I believe it was shown that the original proposed auxiliary sequence of measures did <em>not</em> have one of the two properties. Shavgulidze proposed other sequences of measures. The most recent attempt that I am aware of (which was presented during his 2010 trip to the US mentioned by Mark Sapir in the above comment) involved the direct construction of Folner sets for the action of $F$ on the finite subsets of dyadic rationals (see the next paragraphs). The details were somewhat sparse and the definitions involved many unspecified numerical parameters, but it appeared to be the case that these sets could not be Folner in the necessary sense (see below for a clarification of "necessary sense"). This is because they would likely both contradict the iterated exponential lower bound on the Folner function which I have demonstrated and because they appear to violate the qualitative properties which I have demonstrated that Folner sets of trees must have (see the pre-print on my webpage; the qualitative condition appears in lemma 5.7, noting that marginal implies measure 0 with respect to any invariant measure).</p> <p>Meanwhile I was able to provide a direct elementary proof that the existence of such a sequence having these properties implied the amenability of $F$. In fact the proof gives an explicit procedure for constructing (weighted) Folner sets from the sequence of measures satisfying the hypotheses mentioned above. A note containing the details was circulated to a few people around the time of Shavgulidze's visit to Vanderbilt. While I am reluctant to speak for anyone else (including the author), it appears to me that after the dust had settled (which took a considerable amount of time), the problem with the proof seems to have at least some of its roots in the following observation (which I now include for the sake of prosperity). $F$ acts on the finite subsets of the dyadic rations (let's call this set $\mathcal{D}$) by taking the set-wise image (here I am utilizing the piecewise linear function model of $F$). Now let $\mathcal{T}$ denote the finite subset of $[0,1]$ which contain $0$ and $1$ and are such that any consecutive pair is of the form $p/2^q,(p+1)/2^q$ (for natural numbers $p,q$). $F$ only acts <em>partially</em> on $\mathcal{T}$: the action $T \cdot f$ is defined if $f'$ is defined on the complement of $T$ in $[0,1]$ (there may be other cases when $T \cdot f$ is in $\mathcal{T}$, but let's restrict the domain of the action as above). The full action of $F$ on $\mathcal{D}$ <em>is</em> amenable. The point here is that the action of the standard generators on the sets $\{0,1-2^{-n},1\}$ is the same for large enough $n$ and thus we can build Folner sets as in a $\mathbb{Z}$ action. The amenability of partial action of $F$ on $\mathcal{T}$ is, on the other hand, equivalent to the amenability of $F$ (this is well known, but see the preprint above to see this spelled out in the present jargon). </p> <p>Now here is the catch, if we also require that the invariant measure/Folner sets for the action of $F$ on $\mathcal{D}$ to concentrate on sets of mesh less than $1/16$, then one again arrives at an equivalent formulation of the amenability of $F$. The author was aware of the need for the mesh condition, but (in the most recent example) arranged it only in a modification after the fact (which interferes with invariance).</p> <p>Incidentally the hypotheses on the sequence of measures mentioned above are a condition requiring that the measures concentrate on sets of arbitrarily small mesh as $n$ tends to infinity and a condition which is an analog of translation invariance.</p> <p>I apologize if this borders on too much information.''</p> <p>[Added 1/28/2011] Shavgulidze's 1/14/2011 posting to the ArXiv is essentially a more detailed version of what he was saying in notes, seminars, and private communication in January 2010 during his visit to the US mentioned in Mark Sapir's post above. The present note is still sufficiently vague and full of sufficiently many errors (many typographical in nature) that it is hard (or easy, if you like) to say explicitly which line of the proof is incorrect. It is possible, however to point to places where crucial details are missing and where there are certainly going to be errors (specifically the problems will be on page 11, if not elsewhere as well). The comments from my answer above still apply equally well to the present version. It appears that the present version (or any perturbation of it) still would violate the lower bound on the growth of the Folner function which I have established. The present version still totally ignores that the combinatorial statements on page 11 themselves readily imply the amenability of F, without the involvement of any analytical concepts.</p> <p>[Added 2/3/2011] Details on what is incorrect with Shavgulidze's proof of the amenablity of $F$ can be found <a href="http://arxiv.org/abs/1102.0747" rel="nofollow"> here</a>.</p> <p>[Added 10/3/2012] Well, well, well: now <i>I'm</i> in the position of having announced a proof that $F$ is amenable only to have an error be found. The error was finally found by Azer Akhemedov after being overlooked for roughly 4 weeks by myself and 9 or more people who had checked the proof and found no problems. The basic strategy of the proof still may be valid: it began by considering an extension of the free binary system $(\mathbb{T},*)$ on one generator to the finitely additive probability measures on this system: $$\mu * \nu (E) = \int \int \chi_E(s * t) d \nu (t) d \mu (s).$$ It was shown (correctly) that any idempotent measure is $F$-invariant (there is a natural way of identifying $\mathbb{T}$ with the positive elements of $F$). The difficulty came in constructing the idempotent measure. A version of the Kakutani Fixed Point Theorem was used to construct approximations $K_{\mathcal{B},k,n}$ to the set of idempotent measures. The error occurs in attempting to intersect these compact families of measures. In the proof, it was claimed that the parameter $k$ could be stablized along the an ultrafilter (Lemma 4.13 in the most recent version), allowing one to take a directed intersection of nonempty compact sets. This lemma is likely false and at least is not proved as claimed. One may still be able to argue that a relevant intersection of these approximations is nonempty and hence that there is an idempotent. This seems to require new ideas though.</p> http://mathoverflow.net/questions/58007/is-there-an-idempotent-measure-on-the-free-ld-system Is there an idempotent measure on the free LD system? Justin Moore 2011-03-09T22:43:32Z 2012-09-13T18:20:17Z <p>This is follow up question to MO question <a href="http://mathoverflow.net/questions/57903" rel="nofollow">"Idempotent measures on the free binary system?"</a> Let $(A,*)$ be the free binary operation on one generator which satisfies the left self distributive law: $a * (b * c) = (a * b)* (a* c)$. Is there a finitely additive probability $\mu$ on $A$ such that $\mu * \mu = \mu$?</p> <p>Here $\mu * \nu$ is defined by $$\mu * \nu (X) = \int \int \mathbf{1}_{*^{-1}(X)} (x,y)\ d\mu (x) \ d \nu (y).$$ Note: because of the asymmetry of the left self distributive law, the above order of integration is important (it is possible to show that there is an idempotent with the opposite order of integration). I conjecture a positive answer (this would follow from a positive answer for the free binary system on one generator) and this question may be easier to resolve than the other.</p> http://mathoverflow.net/questions/58007/is-there-an-idempotent-measure-on-the-free-ld-system/107112#107112 Answer by Justin Moore for Is there an idempotent measure on the free LD system? Justin Moore 2012-09-13T18:20:17Z 2012-09-13T18:20:17Z <p>The answer is yes and follows from the positive answer to <a href="http://mathoverflow.net/questions/57903/idempotent-measures-on-the-free-binary-system/107111#107111" rel="nofollow">MO question 57903</a>.</p> http://mathoverflow.net/questions/87187/a-question-on-infinite-dimensional-gaussian-measure-and-affine-tranformations/87950#87950 Answer by Justin Moore for A question on infinite dimensional Gaussian measure and affine tranformations. Justin Moore 2012-02-09T01:27:24Z 2012-02-09T01:27:24Z <p>The answer is that $(a,b)$ must satisfy $a^2 = b^2 + 1$. It is possible to verify (see the preprint above) that for any $b$ and Borel $K \subseteq \mathbb{R}^{\mathbb{N}}$, that $$\gamma_\infty (K) = \int \gamma_\infty (\sqrt{1+b^2} K + b y) d \gamma_\infty (y).$$ By replacing $K$ by $(a/\sqrt{1+b^2})K$ in this equation, we obtain $$\gamma_\infty (\frac{a}{\sqrt{1+ b^2}}K) = \int \gamma_\infty (a K + b y) d \gamma_\infty (y).$$ Let $K$ be all $x$ in $\mathbb{R}^{\mathbb{N}}$ such that $$\lim_{n \to \infty} \left(\frac{1}{n} \sum_{i&lt; n} x_n^2 \right)^{\frac{1}{2}} = 1.$$ Then $\gamma_\infty(K) = 1$ and therefore $(a/\sqrt{1+b^2}) K$ has measure $0$ unless $a^2 = b^2 + 1$. Thus if $a^2 \ne b^2 + 1$, then $\int \gamma_\infty (a K + b y) d \gamma_\infty (y) = 0$ and hence the integrand vanishes for almost every $y$.</p> http://mathoverflow.net/questions/87187/a-question-on-infinite-dimensional-gaussian-measure-and-affine-tranformations A question on infinite dimensional Gaussian measure and affine tranformations. Justin Moore 2012-02-01T01:19:51Z 2012-02-09T01:27:24Z <p>Let $\gamma_\infty$ denote the product Gaussian measure on $\mathbb{R}^\mathbb{N}$. Which $a,b \geq 0$ satisfy that for every Borel set $K\subseteq \mathbb{R}^\mathbb{N}$ of positive measure, $a K + b y$ has positive measure for $\gamma_\infty$-a.e. $y$?</p> <p>The knee-jerk answer is only $a = 1$, $b=0$.'' The Cameron-Martin Theorem tells us that $x \mapsto x + y$ is absolutely continuous exactly when $y$ is in $\ell^2$ (a set of $\gamma_\infty$-measure $0$). Similar arguments apply to linear transformations like those above.</p> <p>This does <em>not</em> answer the question, however. In fact Solecki and I have recently proved that if $a \in \mathbb{R}$ and $K \subseteq \mathbb{R}^\mathbb{N}$ is Borel and of positive measure, then $\gamma_\infty(\sqrt{1+b^2} K + b K) > 0$ for $\gamma_\infty$-a.e. $y$ (i.e. a sufficient conditions is that $a^2 = b^2 +1$). See <a href="http://arxiv.org/abs/1201.3947" rel="nofollow">http://arxiv.org/abs/1201.3947</a> for the preprint. It is not difficult to prove that a necessary requirement for a positive answer is that $a^2 + b^2 -2ab \leq 1 \leq a^2 + b^2 + 2ab$. This leaves some discrepancy, however, and that is the question at hand.</p> http://mathoverflow.net/questions/87188/is-the-property-of-not-containing-mathbbf-2-invariant-under-quasi-isometry Is the property of not containing $\mathbb{F}_2$ invariant under quasi-isometry? Justin Moore 2012-02-01T01:22:38Z 2012-02-01T02:29:25Z <p>Is the property of not containing the free group on two generators invariant under quasi-isometry? Amenability is, so if there is a counterexample it is also a solution to the von Neumann-Day problem (which of course already has a solution).</p> http://mathoverflow.net/questions/58276/parts-of-set-theory-immune-to-independence/58285#58285 Answer by Justin Moore for Parts of Set Theory immune to independence Justin Moore 2011-03-12T20:48:38Z 2011-05-25T16:09:58Z <p>It is a theorem of Woodin that if there is a proper class of Woodin cardinals, then the theory of $L(\mathbb{R})$ can not be changed by forcing. Since forcing and large cardinals are essentially our only means for establishing independence results, this can be interpreted as saying that the theory of $L(\mathbb{R})$ is immune to independence phenomena (except for that which G\" odel's theorem imposes). Here $L(\mathbb{R})$ is the smallest model of ZF which contains all of the reals. $L(\mathbb{R})$ does satisfy the Axiom of Dependent Choice under this assumption, as well as the Axiom of Determinacy. Most of not all theorems in real and complex analysis, measure theory (in the setting of standard Borel space), manifolds, geometry, and number theory can be regarded as statements about what is true in $L(\mathbb{R})$. It is the ideal model in which to study descriptive set theory. Uncountable sets and cardinals, however, often behave strangely in this model (largely because of the influence of the Axiom of Determinacy). For instance, in $L(\mathbb{R})$ and under the above assumptions, $\omega_1$ and $\omega_2$ are measurable cardinals, $\omega_n$ is a singular cardinal for each $n > 2$, there is no uncountable well orderable set of reals, and there are no non-principal ultrafilters on $\omega$. </p> <p>Ironically, Woodin's theorem was the culmination of several deep results concerning iterated forcing, the combinatorics of $\omega_1$, the study of large cardinals, and the fine structure of inner models generalizing $L$.</p> http://mathoverflow.net/questions/60247/when-is-non-amenablity-witnessed-by-a-single-non-measurable-set/60319#60319 Answer by Justin Moore for When is non amenablity witnessed by a single non measurable set? Justin Moore 2011-04-01T20:41:29Z 2011-05-25T12:11:35Z <p>The answer is that the above is equivalent to non amenability. Fix a group $(G,*)$. Since $(G,*)$ is non amenable if and only if every finitely generated subgroup is non amenable, we may assume that $G$ is finitely generated.</p> <p>If $\mu$ and $\nu$ are finitely supported probability measures on $G$, define $$\mu * \nu (Z) = \sum_{x * y \in Z} \mu (\{x\}) \nu (\{y\})$$ Observe that $g * \nu (E) = \nu ( g^{-1} * E)$. If $S$ is a subset of $S$, let $P(S)$ denote all probability measures on $S$ (which are identified with probability measures on $G$ which are supported on $S$). I will identify $G$ with the point masses in $P(G)$.</p> <p>If $A$ and $B$ are subsets of $G$ and $A$ is finite, we say that $B$ is $\epsilon$-Ramsey with respect to $A$ if for every $E \subseteq B$, then there is a $\nu$ in $P(B)$ such that $P(A) * \nu \subseteq P(B)$ and $$|\mu * \nu (E) - \nu (E)| &lt; \epsilon$$ for all $\mu$ in $P(A)$. Notice that in some sense $E$ is defining a partition of $P(B)$ and we are postulating the existence of a copy of $P(A)$ in $P(B)$ which is homogeneous for $E$ up to an error of $\epsilon$.</p> <p>It can be shown with an argument similar to the one below that if $B$ is $\epsilon$-Ramsey with respect to $A$, then for every $f:B \to [0,1]$ there is a $\nu$ in $P(B)$ such that $$|f(\mu * \nu) - f(\nu)| &lt; \epsilon$$ where $f$ has been extended linearly to $P(B)$.</p> <p>We say that $(G,*)$ is Ramsey if for every finite subset $A \subseteq G$ and every $\epsilon > 0$, there is a finite subset $B$ of $G$ with is $\epsilon$-Ramsey with respect to $A$. Notice that if $B$ satisfies that for every $E \subseteq B$ there is a $\nu$ in $P(B)$ such that $$|g * \nu (E) - \nu (E)| &lt; \epsilon$$ for all $g$ in $A$, then $B$ is contained in a finite set which is $\epsilon$-Ramsey (we need only to replace $B$ by $A * B \cup B$).</p> <p>To connect this to the question, suppose that $G$ is not Ramsey, as witnessed by a finite $A \subseteq G$ and $\epsilon > 0$. I claim there is a set $E \subseteq G$ such that for every $\mu \in P(G)$, there is a $g \in A$ such that $|\mu(E \cdot g) - \mu (E)| \geq \epsilon/2$. Let $B_n$ $(n &lt; \infty)$ be an increasing sequence of finite sets covering $G$. Let $T_n$ be the set of all subsets $E$ of $B_n$ which witness that $B_n$ is not $\epsilon$-Ramsey with respect to $A$. Observe that if $E$ is in $T_{n+1}$, then $E \cap B_n$ is in $T_n$. Otherwise there would be a $\nu$ in $P(B_n)$ such that $g * \nu$ is in $P(B_n)$ for each $g$ in $A$ and $$|g * \nu (E \cap B_n) - \nu (E \cap B_n)| &lt; \epsilon$$ Such a $\nu$ would also witness that $E$ is not in $T_{n+1}$. Define $T = \bigcup_n T_n$ and order $E \leq_T E'$ if $E = E' \cap B_m$ where $E$ is in $T_n$. This order makes $T$ into an infinite finitely branching tree. By K&ouml;nig's lemma, $T$ has an infinite path whose union is some $E \subseteq G$. If there were a measure $\mu$ which was $\epsilon/2$-invariant for $E$ with respect to translates by elements of $A$, there would be a finitely supported $\nu$ which was $\epsilon$-invariant for $E$ with respect to translates in $A$. But this would be a contradiction since then the support of $\nu$ would be contained in some $B_n$ and $\nu$ would witness that $E \cap B_n$ was not in $T$.</p> <p>Now the claim is that the Ramsey property of a discrete group is equivalent to its amenability. That amenability implies the Ramsey property follows from F&oslash;lner's characterization of amenability. Also observe that $G$ is amenable provided that for every $\epsilon > 0$, every finite list $E_i$ $(i &lt; n)$ of subsets of $G$, and $g_i$ $(i &lt; n)$ in $G$, there is a finitely supported $\mu$ such that $$|\mu (g_i * E_i) - \mu (E_i) | &lt; \epsilon.$$ Set $B_{-1} = \{1_G\} \cup \{g^{-1}_i :i &lt; n\}$ and construct a sequence $B_i$ $(i &lt; n)$ such that $B_{i+1}$ is $\epsilon/2$-Ramsey with respect to $B_i$.</p> <p>Now inductively construct $\nu_i$ $(i &lt; n)$ by downward recursion on $i$. If $\nu_j$ $(i &lt; j)$ has been constructed, let $\nu_i \in P(B_i)$ be such that $$|\mu * \nu_{i} * \ldots * \nu_{n-1} (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| &lt; \epsilon/2$$ for all $\mu$ in $P(B_{i-1})$. Set $\mu = \nu_0 * \ldots * \nu_{n-1}$. If $i &lt; n$, then since $\nu_0 * \ldots * \nu_{i-1}$ and $g_i^{-1} * \nu_0 * \ldots * \nu_{i-1}$ are in $P(B_{i-1})$, $$|g_i^{-1} * \mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| &lt; \epsilon/2$$ $$|\mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| &lt; \epsilon/2$$ and therefore $|\mu (g_i * E_i) - \mu (E_i)| &lt; \epsilon$.</p> http://mathoverflow.net/questions/60247/when-is-non-amenablity-witnessed-by-a-single-non-measurable-set When is non amenablity witnessed by a single non measurable set? Justin Moore 2011-04-01T01:16:53Z 2011-05-25T12:11:35Z <p>Suppose $G$ is a finitely generated discrete group and that there is a subset $E$ of $G$ such that if $\mu$ is a finitely additive probability measure on $G$, then there is a $g$ in $G$ such that $\mu(E \cdot g) \ne \mu(E)$. Certainly $G$ is non amenable. Can more be said about $G$? Must $G$ contain $\mathbb{F}_2$?</p> <p>It should be noted that the above situation can happen: let $E$ be all elements of $\mathbb{F}_2$ which begin'' with $a$ or $a^{-1}$. Then both $E$ and its complement have infinitely many disjoint translates (by powers of $b$ and $a$, respectively).</p> http://mathoverflow.net/questions/60953/a-question-concerning-separate-and-joint-continuity-of-bilinear-maps A question concerning separate and joint continuity of bilinear maps Justin Moore 2011-04-07T15:40:03Z 2011-04-07T16:27:29Z <p>Suppose that $V$ is a locally convex topological vector space and $f:V^2 \to V$ is a bilinear map. Suppose that $C \subseteq V$ is compact and convex, $f$ maps $C^2$ into $C$ and $f \restriction C^2$ is separately continuous. Must $f \restriction C^2$ be jointly continuous?</p> <p>In the particular application I have in mind, $V = \ell_\infty^*$ with the weak* topology. Moreover the function $f$ is injective. I suspect even in this setting that this is false.</p> <p>I am also interested in a good reference for the optimal results of concerning separate and joint continuity of bilinear maps. Ideally this would be written for someone who is not a functional analyst.</p> http://mathoverflow.net/questions/48526/how-additive-is-lebesgue-measure-in-zfad/60388#60388 Answer by Justin Moore for How additive is Lebesgue measure in ZF+AD ? Justin Moore 2011-04-02T21:16:05Z 2011-04-02T23:43:25Z <p>If $\kappa$ is a well orderable cardinal (i.e. an $\aleph_\alpha$) I don't think there is a need to involve determinacy. If all sets of reals are Lebesgue measurable (and DC holds), then Lebesgue measure is $\aleph_\alpha$ additive for every $\alpha$. This is sufficient (at least granting DC) since AD certainly implies all sets of reals are Lebesgue measurable.</p> <p>First argue that if $\kappa$ is a well orderable cardinal (i.e. an $\aleph_\alpha$) and Lebesgue measure is not $\kappa$ additive, then this can be witnessed in such a way that the sets in the union are all null sets (there is a countable subcollection whose union has measure equal to the sup in your equation). Without loss of generality, we may arrange that $$\lambda ( \bigcup_{\xi \in \alpha} f(\xi) ) = 0$$ for all $\alpha \in \kappa$. Define $Y = \bigcup_{\alpha \in \kappa} f(\alpha)$. Now define $$X = \{(x,y) \in Y^2 : \alpha_x &lt; \alpha_y\}$$ where $\alpha_x = \min\{\alpha \in \kappa : x \in f(\alpha)\}$. The set $X$ now violates Fubini's theorem: the set of all horizontal sections are the union of fewer than $\kappa$ many of the sets $f(\xi)$ and the vertical sections have their complement (in $Y$) being the union of fewer than $\kappa$ sets of the form $f(\xi)$. Thus by integrating in one direction, $X$ has measure 0, while in the other direction, its measure is $\lambda(Y)^2$.</p> http://mathoverflow.net/questions/57980/what-is-the-largest-laver-table-which-has-been-computed/59467#59467 Answer by Justin Moore for What is the largest Laver table which has been computed? Justin Moore 2011-03-24T19:52:01Z 2011-03-24T19:52:01Z <p>I've been in contact with Patrick Dehornoy and Ales Drapal and both thought that $A_{28}$ is likely the current record for a Laver table computation.</p> http://mathoverflow.net/questions/57980/what-is-the-largest-laver-table-which-has-been-computed What is the largest Laver table which has been computed? Justin Moore 2011-03-09T18:48:41Z 2011-03-24T19:52:01Z <p>Richard Laver proved that there is a unique binary operation $*$ on $\{1,\ldots,2^n\}$ which satisfies $$a*1 \equiv a+1 \mod 2^n$$ $$a* (b* c) = (a* b) * (a * c).$$ This is the $n$th Laver table $(A_n,*)$.</p> <p>There is an algorithm for computing $a * b$ in $A_n$, but in general (and especially for small values of $a$), this requires one to compute much of the rest of $A_n$. What is the largest value for $n$ for which someone can, in a modest amount of time, compute an arbitrary entry in $A_n$? I am able to compute entries in $A_{27}$.</p> <p>I should note that the map which sends $a$ to $a\ \mathrm{mod}\ 2^m$ defines a homomorphism from $A_n$ to $A_m$ for $m &lt; n$ and hence the problem becomes strictly harder for larger $n$.</p> <p>Edit: I have actually been able to compute $A_{28}$, not just $A_{27}$.</p> http://mathoverflow.net/questions/58223/what-is-the-origin-of-the-metrization-problem-for-compact-convex-sets What is the origin of the metrization problem for compact convex sets? Justin Moore 2011-03-12T01:43:16Z 2011-03-12T03:02:14Z <p>The following is an old question in analysis:'' Is it true that every <i>perfectly normal</i> compact convex subset of a locally convex topological vector space is metrizable? Here <i>perfectly normal</i> means Hausdorff plus all closed subsets are a countable intersection of open sets.</p> <p>Who first asked this question? The oldest reference I can locate is a 1972 paper by B. MacGibbon in the Journal of Functional Analysis but it is clear from what is written there that she is reporting progress on a known problem.</p> <p>Of course I am also interested in an answer to this question, but I'm really asking about reference information. I should note that Lopez-Abad and Todorcevic have recently demonstrated that it is consistent with ZFC that there is a counterexample to this problem. The question is whether a positive answer is consistent.</p> http://mathoverflow.net/questions/53467/what-is-the-consistency-strength-of-the-failure-of-square-in-terms-of-large-card/58198#58198 Answer by Justin Moore for What is the consistency strength of the failure of square, in terms of large cardinals Justin Moore 2011-03-11T20:43:09Z 2011-03-11T20:43:09Z <p>My understanding is that the weaker form of square was initially needed to get a better lower bound on the consistency strength of PFA but that, with improvements in our understanding of the fine structure of the current inner models, the original form of square now suffices to give the best lower bounds.</p> <p>There are two versions of square (each with their associated weak square'' heierarchy): $\square_\kappa$ (formulated by Jensen) and $\square (\kappa)$ (formulated by Todorcevic). It is immediate from their definitions that $\square_\kappa$ implies $\square (\kappa^+)$ ($\square_\kappa$ and $\square (\kappa^+)$ are both statements about sequences of length $\kappa^+$ --- the apparent shift in indexing is purely notational). Todorcevic's theorem is that PFA implies the failure of $\square (\kappa)$ for every regular $\kappa > \omega_1$. It is often cited in the weaker form stated in item 1 of the question.</p> <p>Traditionally, lower bounds on PFA have been obtained through the failure of $\square_\kappa$ at a single singular strong limit cardinal. To my knowledge, there is currently no method for obtaining better lower bounds on PFA. </p> <p>It was somewhat of a breakthrough when it was realized that the failure of $\square (\kappa)$ at successive values of $\kappa$ was apparently more powerful than the failure of successive instances of $\square_\kappa$. That it took so long for this to be noticed may be due to the fact that Todorcevic's result was commonly cited in its weaker form. I don't believe, however, that this gives an improved lower bound on the consistency strength of PFA. It does, however, yield more strength from the failure of any form of square at cardinals below $\aleph_\omega$.</p> <p>It is at least my impression that it <i>should</i> be the case that the failure of square at all cardinals has a much higher strength than and individual failure of $\square (\kappa)$ or $\square_\kappa$. Whether the global failure of $\square_\kappa$, $\square (\kappa)$ or the weak forms of square mentioned above yield different consistency strengths is a matter which is completely open and for which there is only wild speculation.</p> http://mathoverflow.net/questions/44705/cardinalities-larger-than-the-continuum-in-areas-besides-set-theory/58030#58030 Answer by Justin Moore for Cardinalities larger than the continuum in areas besides set theory Justin Moore 2011-03-10T02:36:29Z 2011-03-11T20:17:14Z <p>For each $n$, there is a unique binary operation $*$ on $\{1,\ldots,2^n\}$ which satisfies $$a * 1 \equiv a+1 \mod 2^n$$ $$a * (b * c) = (a * b) * (a * c)$$ This is the $n$th Laver table $A_n$. The Laver tables have a purely number-theoretic definition (although see <a href="http://mathoverflow.net/questions/57980/what-is-the-largest-laver-table-which-has-been-computed" rel="nofollow">my other post</a> concerning computation in them). These naturally project down to the smaller tables in such a way to give rise to an inverse limit. In this inverse limit, there is a copy of the generator and one can use it to generate an algebra $A_\infty$.</p> <p>Here is the interesting part: the freeness of $A_\infty$ has been established from the assumption that, for each $n$, there is an $n$-huge cardinal (these are among the strongest of the large cardinal hypotheses). No proof is known in ZFC. See Laver's articles in Advances in Mathematics or Dehornoy's book "Braids and self distributivity" for more information. In my opinion this is the greatest problem in reverse mathematics.</p> <p>These tables did originate in set theory, but they have natural connections to the study of braids (Dehornoy's book is testimony to this).</p> http://mathoverflow.net/questions/57994/saturated-extensions-of-zfc-models/58123#58123 Answer by Justin Moore for Saturated extensions of ZFC models Justin Moore 2011-03-11T02:03:14Z 2011-03-11T13:38:42Z <p>I'm addressing both the question and the comments, but possibly this question should be closed. First let's be clear what we mean by a model of set theory. A model is a set $M$ (or perhaps a class) with a binary relation $E$. $(M,E)$ satisfies <i>Foundation</i> if for every $x$ in $M$, there is an $E$-minimal $y$ in $M$ such that $y E x$. $M$ is <i>well founded</i> if for every $x \subseteq M$, there is an $E$-minimal $y$ in $M$ such that $y$ is in $x$. If $E$ is $\in$, then $M$ is <i>transitive</i> if every element of $M$ is a subset of $M$. Well foundedness is implicit in transitivity.</p> <p>The problem is that your model need not be well founded. The Axiom of Foundation only implies that the model of set theory you are working with does not have an element witnessing that it is ill founded. But certainly such a witness can exist outside. The Henkin Construction essentially never will result in a well founded model. A simpler example is given by a non standard model of PA. The model satisfies the induction scheme (which is the analog here of Foundation) but any well founded model of PA is isomorphic to the standard model.</p> <p>Furthermore, any well founded model $(M,E)$ of ZFC is isomorphic to a transitive set (or class) with the membership relation. This is the Mostowski collapse at work. So transitivity is not the issue here. The failure of finiteness to be absolute is tied up in the ill foundedness of the model in question.</p> http://mathoverflow.net/questions/58023/is-it-possible-for-pn-to-be-larger-than-aleph-omega/58024#58024 Answer by Justin Moore for Is it possible for P(N) to be larger than Aleph_omega? Justin Moore 2011-03-10T01:45:37Z 2011-03-11T02:58:16Z <p>It is consistent that $2^{\aleph_0} > \aleph_\omega$. This is immediate from Cohen's work; force with all finite partial functions from $\aleph_{\omega}$ to $2$. More generally Easton has shown that the only constraints on exponentiation of regular cardinals is that it be monotonic and satisfy $\mathrm{cof}(2^\kappa) > \kappa$. For singular cardinals, the matter is much more subtle. For instance one of Shelah's celebrated results is that if $\aleph_\omega$ is a strong limit cardinal, then $2^{\aleph_\omega} &lt; \aleph_{\omega_4}$ (and no, that $4$ is not a misprint).</p> http://mathoverflow.net/questions/57656/standard-model-of-particle-physics-for-mathematicians/57897#57897 Answer by Justin Moore for Standard model of particle physics for mathematicians Justin Moore 2011-03-09T01:36:10Z 2011-03-09T18:25:02Z <p>I'm neither a physicist nor a mathematical physicist but I've taken some recreational interest in learning about the subject and about QFT and the standard model in particular. What follows is the recommendations of a complete QFT novice and is admittedly somewhat off topic, but hopefully will be useful to someone. I found both Feynman's "The strange theory of light and matter" and Griffith's "Introduction to elementary particles" very helpful. These are not math books and Feynman's has essentially no details (or even equations). But the last half of Feynman's book (especially the last lecture) is good for giving an intuitive understanding of what the mathematics is trying to formalize (this is something I found maddening about the many mathematical accounts of QFT I've read). It is also appealing that you can read the book in a couple afternoons (probably no other book on this topic can boast this). Griffith's book fills in a number of blanks in Feynman's book. My main reaction to the mathematical treatments I've seen of QFT is that it is hard to gain intuition as to what the definitions and axioms are really intending to model. Both these books helped a lot in remedying this, at least for me. Read them first and then hunt down the mathematical treatments.</p> http://mathoverflow.net/questions/42569/examples-of-zfc-theorems-proved-via-forcing/57895#57895 Answer by Justin Moore for Examples of ZFC theorems proved via forcing Justin Moore 2011-03-09T01:12:17Z 2011-03-09T01:12:17Z <p>My favorite example of this is Stevo Todorcevic's paper "Compact subsets of the first Baire class" (JAMS, 1999). Fix a Polish space $X$ (for us it will be no loss of generality to take $X = \mathbb{N}^\mathbb{N}$). The Baire class 1 functions on $X$ are those functions which are the limit of a pointwise convergent sequence of continuous functions. A compact space which embeddable into the Baire class 1 functions with the pointwise topology is said to be <i>Rosenthal compact</i>. A typical example of a Rosenthal compacta is the set $\mathbb{H}$ of monotone increasing functions from $[0,1]$ to $[0,1]$. Two others are the split interval'' (which consists of those elements of $\mathbb{H}$ whose range is contained in $\{0,1\}$) and the one point compactification of a discrete set of cardinality at most continuum. The class of Rosenthal compacta is closed under countable products and closed subspaces.</p> <p>Todorcevic proved several ZFC results about Rosenthal compacta using forcing. Probably the best example in the paper (in terms of the use of forcing machinery) is the proof that any Rosenthal compacta contains a dense metrizable subspace. Before this it was an open problem whether there was a c.c.c. non-separable Rosenthal compacta. Todorcevic also proves in this paper that a Rosenthal compacta which does not contain an uncountable discrete subspace must map at most two-to-one into a metric space. Furthermore if such a space is non-metrizable, it must contain a homeomorphic copy of the split interval. Finally, he showed that any non $G_\delta$-point in a separable Rosenthal compacta is the unique accumulation point of a discrete subspace of cardinality continuum. One of the key lemmas of the paper is that the property of being a Rosenthal compacta is preserved when one appropriately reinterprets the compacta in any generic extension.</p> http://mathoverflow.net/questions/14338/what-is-the-general-opinion-on-the-generalized-continuum-hypothesis/56075#56075 Answer by Justin Moore for What is the general opinion on the Generalized Continuum Hypothesis? Justin Moore 2011-02-20T16:31:03Z 2011-02-20T16:31:03Z <p>I think it should be pointed out that, while many people working in set theory have a strong opinion about CH (with many feeling it is "false"), they generally do not have strong feelings against GCH above $\aleph_0$. That is, GCH is not such a controversial statement except that it implies CH.</p> http://mathoverflow.net/questions/55214/does-the-amenability-problem-for-thompsons-group-f-predate-1980 Does the amenability problem for Thompson's group $F$ predate 1980? Justin Moore 2011-02-12T16:41:33Z 2011-02-20T16:01:07Z <p>The first place where the amenability problem for Thompson's group $F$ appears in the literature is, I believe, 1980 in a problems article by Ross Geoghegan. I have heard, however, vague comments to the effect that the problem was considered by other people before this. Does anyone have any knowledge about the existence of this problem prior to 1980?</p> <p>Edit: Following Mark's advice offline, I wrote Richard Thompson to verify the details of Mark's answer. He did confirm that he considered the problem. He first observed that his group $F$ did not contain $\mathbb{F}_2$. He then discovered the material on amenability in Hewitt and Ross's text on abstract harmonic analysis. He then observed that $F$ was not elementarily amenable. This occurred sometime prior to his 1973 visit to University of Illinois at Urbana-Champaign to visit Day.</p> <p>He did not, however, attend Greenleaf's series of lectures in 1967. Instead he read Greenleaf's 1969 text "Invariant Means on Topological Groups and Their Applications." It seems that the best date to attach to Thompson's consideration of the question is 1973. Of course Thompson never published his observations and they were not widely circulated. His observations mentioned above were rediscovered by others in the 1980s (the question itself by Geoghegan).</p> <p>I have invited Richard to post an answer, in which case I will delete this edit.</p> http://mathoverflow.net/questions/23829/solutions-to-the-continuum-hypothesis/54380#54380 Answer by Justin Moore for Solutions to the Continuum Hypothesis Justin Moore 2011-02-05T01:42:22Z 2011-02-05T02:37:10Z <p>First I'll say a few words about forcing axioms and then I'll answer your question. Forcing axioms were developed to provide a unified framework to establish the consistency of a number of combinatorial statements, particularly about the first uncountable cardinal. They began with Solovay and Tennenbaum's proof of the consistency of Souslin's Hypothesis (that every linear order in which there are no uncountable families of pairwise disjoint intervals is necessarily separable). Many of the consequences of forcing axioms, particularly the stronger Proper Forcing Axiom and Martin's Maximum, had the form of classification results: Baumgartner's classification of the isomorphism types of $\aleph_1$-dense sets of reals, Abraham and Shelah's classification of the Aronszajn trees up to club isomorphism, Todorcevic's classification of linear gaps in $\mathcal{P}(\mathbb{N})/\mathrm{fin}$, and Todorcevic's classification of transitive relations on $\omega_1$. A survey of these results (plus many references) can be found in both Stevo Todorcevic's ICM article and my own (the later can be found <a href="http://www.math.cornell.edu/~justin/Ftp/ICM.pdf" rel="nofollow">here</a>). These are accessible to a general audience.</p> <p>What does all this have to do with the Continuum Problem? It was noticed early on that forcing axioms imply that the continuum is $\aleph_2$. The first proof of this is, I believe, in Foreman, Magidor, and Shelah's seminal paper in Martin's Maximum. Other quite different proofs were given by Caicedo, Todorcevic, Velickovic, and myself. An elementary proof which is purely Ramsey-theoretic in nature is in my article "<a href="http://www.math.cornell.edu/~justin/Ftp/OCA_c.pdf" rel="nofollow"> Open colorings, the continuum, and the second uncountable cardinal</a>" (PAMS, 2002).</p> <p>Since it is often the case that the combinatorial essence of the proofs of these classification results and that the continuum is $\aleph_2$ are similar, one is left to speculate that perhaps there may some day be a classification result concerning structures of cardinality $\aleph_1$ which already implies that the continuum is $\aleph_2$. There is a candidate for such a classification: the assertion that there are five uncountable linear orders such that every other contains an isomorphic copy of one of these five. Another related candidate for such a classification is the assertion that the Aronszajn lines are well quasi-ordered by embeddability (if $L_i$ $(i &lt; \infty)$ is a sequence of Aronszajn lines, then there is an $i &lt; j$ such that $L_i$ embeds into $L_j$). These are due to myself and Carlos Martinez, respectively. See a discussion of this (with references) in my <a href="http://www.math.cornell.edu/~justin/Ftp/ICM.pdf" rel="nofollow">ICM paper</a>.</p> http://mathoverflow.net/questions/45774/is-subamenable-the-same-as-amenable Is "subamenable" the same as amenable? Justin Moore 2010-11-12T01:33:02Z 2010-11-15T10:01:38Z <p>Let $G$ be a finitely generated group. Does the following condition imply the amenability of $G$: there is a function $\mu:\mathcal{P}(G) \to [0,1]$ such that:</p> <p><li> (subadditive) $\mu(G) = 1$, $\mu(A \cup B) \leq \mu(A) + \mu(B)$, and $A \subset B$ implies $\mu(A) \leq \mu(B)$,</p> <p><li> (invariant) $\mu(A \cdot g) = \mu(A)$ for all $g$ in $G$,</p> <p><li> (exhaustive) if $\{A_n : n &lt;\infty\}$ is pairwise disjoint, then $\inf_n \mu(A_n) = 0$.</p> <p>A group distinguishing this condition from amenability cannot contain $F_2$. I am aware of the relationship to the (now solved) Maharam problem.</p> <p>I am not expecting an answer so much as asking whether (and where) this question has been studied.</p> http://mathoverflow.net/questions/45781/amenability-versus-the-ideal-of-wandering-sets Amenability versus the ideal of wandering sets Justin Moore 2010-11-12T03:33:39Z 2010-11-12T14:57:14Z <p>Let $G$ be a finitely generated group acting on a set $S$ (on the right). Define the heirarchy of "marginal sets" as follows:</p> <p><li> The emptyset is 0-marginal.</p> <p><li> A set E is $(k+1)$-marginal if $E$ can be covered by finitely many sets $A$ such that for some $k$-marginal set $B$ and some $g$ in $G$, if $a$ is in $A$ and $a \cdot g^i$ is in $A$ for some $i > 0$, then there is a $j &lt; i$ such that $a \cdot g^j$ is in $B$.</p> <p><li> a set is marginal if it is $k$-marginal for some $k$.</p> <p>So if we recall that $A$ is wandering if, for some $g$ in $G$, the sets $A \cdot g^i$ $(i &lt; \infty)$ are pairwise disjoint, then 1-marginal just means that the set is a finite union of wandering sets.</p> <p>The point is that marginal sets are assigned measure 0 by any invariant measure (or even by any invariant exhaustive submeasure --- see my other recent question).</p> <p>Now for the question(s):</p> <p><li> Are there non-amenable actions which are aperiodic but not marginal? (Notice that if $G$ is a non-amenable torsion group acting on itself then the emptyset is the only marginal set.)</p> <p><li> Can $S$ be $(k+1)$-marginal but not $k$-marginal for some $k > 0$?</p> <p><li> Has the notion of marginality been considered (and given a name)? (I used the notion to give estimates in my proof of a lower bound for the Folner function for Thompson's $F$, but it seems this "must" have been considered before.)</p> http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction/128679#128679 Comment by Justin Moore Justin Moore 2013-04-30T13:54:40Z 2013-04-30T13:54:40Z @Douglas: Could you contact me off list? http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction/128679#128679 Comment by Justin Moore Justin Moore 2013-04-30T13:50:42Z 2013-04-30T13:50:42Z @Douglas: Actually you might be right. The formula in my comment is a little cleaner though. http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction/128679#128679 Comment by Justin Moore Justin Moore 2013-04-30T13:47:47Z 2013-04-30T13:47:47Z @Douglas: The cases depending on whether a1 &gt; 1 seems off. I think the (or rather a) formula is 1/2 x = [a0/2,2x'] or [(a0-1)/2,1,1,(x'-1)/2] (depending on parity of a0) and 2x=[2a0,x'/2]. This gives x/2 = [a0/2,2a1,x''/2] or [(a0-1)/2,1,1,(x'-1)/2]. Thanks again; this was exactly what I was looking for. I had dismissed finding a rule this simple for some reason. http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction/128679#128679 Comment by Justin Moore Justin Moore 2013-04-29T17:13:02Z 2013-04-29T17:13:02Z @Douglas: Check your formula for doubling. I think it is not quite correct. http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction/128679#128679 Comment by Justin Moore Justin Moore 2013-04-26T12:07:24Z 2013-04-26T12:07:24Z @Douglas: thanks, but this is not really what I'm asking. The 2[a0;2a1,a2,2a3,a4,...]=[2a0;a1,2a2,a3,2a4,...] part is in the spirit of what I want, but I want something analogous which works for any sequence. I was aware of this observation --- the hard part is of course in dealing with the remainders. http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction/128724#128724 Comment by Justin Moore Justin Moore 2013-04-26T12:05:10Z 2013-04-26T12:05:10Z Thanks, but this is not really what I'm asking. See the comments in the edit portion of the question. http://mathoverflow.net/questions/128676/what-is-the-effect-of-adding-1-2-to-a-continued-fraction/128679#128679 Comment by Justin Moore Justin Moore 2013-04-26T10:46:37Z 2013-04-26T10:46:37Z @David:Actually I think this is not quite true. Unless I am missing something, this is not reversible. You can, however, multiply by 4 by adding 1/2 and reciprocating though. The point is that multiplication by 1/2 corresponds to operations in PSL_2(Z[1/sqrt(2)]), while adding 1/2 comes from PSL_2(Z[1/2]). http://mathoverflow.net/questions/118254/usage-of-set-theory-in-undergraduate-studies/118451#118451 Comment by Justin Moore Justin Moore 2013-01-10T15:31:54Z 2013-01-10T15:31:54Z @ALC: Concerning your historical question, I don't know. I'm not a historian and what I could say would only be speculation. I think the notion that there could be independence of the sort that we now know exists would have been deeply shocking at the time and was, I believe, far from peoples minds. That said, the people at the turn of the century were certainly interested in trying to make some sort of rigorous definition of what mathematics was and what proof was. In light of Godel's works and Cohen's work (not to mention all of modern set theory), that was very much needed. http://mathoverflow.net/questions/118254/usage-of-set-theory-in-undergraduate-studies/118451#118451 Comment by Justin Moore Justin Moore 2013-01-10T15:26:48Z 2013-01-10T15:26:48Z @Toby: Sorry, but I don't know the first thing about ETCS. http://mathoverflow.net/questions/118254/usage-of-set-theory-in-undergraduate-studies/118451#118451 Comment by Justin Moore Justin Moore 2013-01-10T15:25:14Z 2013-01-10T15:25:14Z @Tom and Todd: I have certainly heard people describe category theory as an alternative approach to foundations. In fact this question itself has that character. I think that is incorrect (or, rather, ignorant) to say that set theory and category theory are both part of foundations in the same sense of the word. There are two complementary definitions of foundations at work. Category theory is, as I understand it at least, providing a very useful language. Set theory is providing a rigorous standard. http://mathoverflow.net/questions/51395/a-soft-introduction-to-physics-for-mathematicians-who-dont-know-the-first-thing/51443#51443 Comment by Justin Moore Justin Moore 2012-12-12T15:11:19Z 2012-12-12T15:11:19Z Is this related to: L. D. Faddeev and O. A. Yakubovskii &quot;Lectures on Quantum Mechanics for Mathematics Students&quot;? http://mathoverflow.net/questions/114996/do-distinct-idempotent-measures-on-finite-binary-systems-have-distinct-supports/114999#114999 Comment by Justin Moore Justin Moore 2012-11-30T18:33:21Z 2012-11-30T18:33:21Z Yes, sorry you are right. I deleted my comment. http://mathoverflow.net/questions/114996/do-distinct-idempotent-measures-on-finite-binary-systems-have-distinct-supports Comment by Justin Moore Justin Moore 2012-11-30T16:29:56Z 2012-11-30T16:29:56Z If (S,*) is a group, then the idempotent measures are exactly the uniform measures on the subgroups of S. http://mathoverflow.net/questions/114996/do-distinct-idempotent-measures-on-finite-binary-systems-have-distinct-supports Comment by Justin Moore Justin Moore 2012-11-30T16:28:14Z 2012-11-30T16:28:14Z Any operation. <i>Nothing</i> is assumed. http://mathoverflow.net/questions/107391/applications-of-idempotent-ultrafilters Comment by Justin Moore Justin Moore 2012-11-30T02:36:50Z 2012-11-30T02:36:50Z You are probably aware of this, but my proof that F is amenable has an error.