User adam epstein - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:33:03Z http://mathoverflow.net/feeds/user/15819 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125446/consistency-of-the-concept-of-the-collection-of-all-collection/125450#125450 Answer by Adam Epstein for Consistency of the concept of the collection of all collection Adam Epstein 2013-03-24T10:36:15Z 2013-03-24T12:19:23Z <p>Quine's system New Foundations is now suspected to be consistent with ZFC, and the modified systen NFU has been known to be consistent for decades. These systems certainly allows for a set of all sets, hence do not require any introduction of classes. On the other hand, (sub)set formation is severly restricted. It's perhaps a matter of personal values: do you prefer to have the set</p> <p>$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;${$x: x=x$}</p> <p>or the functions </p> <p>$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;X\ni x\mapsto$ {$x$} $\in\mathcal{P}X$?</p> http://mathoverflow.net/questions/125414/self-containing-structures/125449#125449 Answer by Adam Epstein for Self-containing structures Adam Epstein 2013-03-24T10:25:15Z 2013-03-24T10:25:15Z <p>While you were asking about examples rather than foundations, you did mention the system NF. In the paper</p> <p><a href="http://math.stanford.edu/~feferman/papers/ess.pdf" rel="nofollow">http://math.stanford.edu/~feferman/papers/ess.pdf</a></p> <p>Feferman discusses the merits and demerits of using the modified system NFU as a foundation supporting certain aspects of self-containment.</p> http://mathoverflow.net/questions/125020/partial-linearization-near-a-hyperbolic-fixed-point-classical-scattering/125061#125061 Answer by Adam Epstein for Partial linearization near a hyperbolic fixed point--Classical scattering Adam Epstein 2013-03-20T12:05:52Z 2013-03-20T12:05:52Z <p>From a complex analytic standpoint, an updated discussion is embedded in Lyubich's paper "Feigenbaum-Coullet-Tresser universality and Milnor's Hairiness Conjecture":</p> <p><a href="http://arxiv.org/abs/math/9903201" rel="nofollow">http://arxiv.org/abs/math/9903201</a></p> http://mathoverflow.net/questions/124362/topology-the-board-game/124440#124440 Answer by Adam Epstein for Topology, the board game Adam Epstein 2013-03-13T18:01:41Z 2013-03-13T18:01:41Z <p>This isn't quite topology, but Mike Shulman and John Baez have some interesting things to say about "gamification".</p> <p><a href="http://golem.ph.utexas.edu/category/2012/06/the_gamification_of_higher_cat.html" rel="nofollow">http://golem.ph.utexas.edu/category/2012/06/the_gamification_of_higher_cat.html</a></p> http://mathoverflow.net/questions/121406/where-in-ordinary-math-do-we-need-unbounded-separation-and-replacement/123583#123583 Answer by Adam Epstein for Where in ordinary math do we need unbounded separation and replacement? Adam Epstein 2013-03-05T02:10:32Z 2013-03-06T09:22:01Z <p>This very matter is discussed in depth by Mathias, in Chapter 9 of his <strong>The Stength of Mac Lane Set Theory</strong> <a href="https://www.dpmms.cam.ac.uk/~ardm/maclane.pdf" rel="nofollow">https://www.dpmms.cam.ac.uk/~ardm/maclane.pdf</a>. There he shows that to prove </p> <p>$\;\;$ "for all $n$ there exist $n$ pairwise nonequinumerous infinite sets" </p> <p>requires some use of unbounded Separation, and to prove </p> <p>$\;\;$ "there exists an infinite set of pairwise nonequinumerous infinite sets" </p> <p>requires some use of Replacement. He also discusses various refinements in connection with formula complexity and stratifiability. For example, Coret showed that that the stratified instances of Replacement are already theorems of Zermelo set theory, whence "requires some Replacement" entails "requires some unstratifiable Replacement". </p> <p>Algebraists might prefer these assertions concerning sequences of $\mathbb R$-linear spaces defined though duality: $L_1=\mathbb{R}[t]$ and $L_{k+1}=L_k^*$. In this setting, </p> <p>$\;\;$ "for all $n$ the sequence $L_1,\ldots, L_n$ exists" </p> <p>requires some use of unbounded Separation, and </p> <p>$\;\;$ "the sequence $L_1, L_2, \ldots$ exists"</p> <p>requires some use of Replacement.</p> <p>Whether or not these assertions count as ordinary mathematics, I find them considerably easier to grasp than Borel Determinacy, which seems vastly more intricate. Then again, maybe the second example counts as "there's life beyond $V_{\omega+\omega}$". </p> <p>Meanwhile, some of the motivation of this question resonates with mine in posing these questions:</p> <p><a href="http://mathoverflow.net/questions/120598/when-must-it-be-sets-rather-than-proper-classes-or-vice-versa-outside-of-fou" rel="nofollow">http://mathoverflow.net/questions/120598/when-must-it-be-sets-rather-than-proper-classes-or-vice-versa-outside-of-fou</a></p> <p><a href="http://mathoverflow.net/questions/117910/can-one-exhibit-an-explicit-kuratowski-infinite-set-without-invoking-replacement" rel="nofollow">http://mathoverflow.net/questions/117910/can-one-exhibit-an-explicit-kuratowski-infinite-set-without-invoking-replacement</a></p> <p>Some of the comments on the first (Sets vs Classes) allude to various constructions in homotopy theory involving long-running transfinite recursions - and even large cardinals - so presumably there is some Replacement involved. </p> http://mathoverflow.net/questions/122406/undecidability-and-holomorphic-functions-reference-request/122428#122428 Answer by Adam Epstein for Undecidability and holomorphic functions (Reference request) Adam Epstein 2013-02-20T17:12:30Z 2013-02-20T18:03:17Z <p>There is a famous result of ErdÅ‘s in <em>An interpolation problem associated with the Continuum Hypothesis</em> concerning families $\cal F$ of entire functions $f:\mathbb{C}\rightarrow\mathbb{C}$ such that for every $\zeta\in\mathbb{C}$ the set {$f(\zeta): f\in\mathcal{F}$} is countable.</p> <p><strong>Theorem</strong></p> <p>(1) If $2^{\aleph_0}>\aleph_1$ then every such family is countable.</p> <p>(2) If $2^{\aleph_0}=\aleph_1$ then every such family has cardinality $2^{\aleph_0}$.</p> http://mathoverflow.net/questions/121775/what-is-a-good-example-of-a-hyperspace-where-the-base-space-is-non-hausdorff/121822#121822 Answer by Adam Epstein for What is a good example of a hyperspace where the base space is non-Hausdorff? Adam Epstein 2013-02-14T18:52:09Z 2013-02-14T19:37:12Z <p>I've made use of this in some (unpublished) work in connection with a formalism for discussing so-called geometric limits of holomorphic dynamical systems. The details of the specific application are not so relevant, but I've copied the statement here to give a sense of how what is essentially Fell's Theorem yields a useful compactness statement that was otherwise not possible to even formulate this precisely. Todd and Benjamin's comments are relevant. What saves the day is the fact that the original space, while not Hausdorff, is still sober. </p> <p>A <em>holomorphic dynamical system</em> on a complex manifold $X$ is any collection of open analytic maps, from open subsets to $X$, containing the identity and all implied restrictions and compositions. We say that the systems <code>${\cal F}_\eta$</code> converge <em>geometrically</em> to the system $\cal F$ whenever</p> <p><code>$$\liminf {\cal F}_\eta ={\cal F}= \limsup {\cal F}_\eta$$</code> where $\liminf$ and $\limsup$ are given by the prescription:</p> <p><code>$$\liminf {\cal F}_\eta = \{f: f_\eta\rightarrow f \mbox{ for some } f_\eta \mbox{ chosen from } {\cal F}_\eta\}$$</code></p> <p><code>$$\limsup {\cal F}_\eta = \{f: f_{\eta_\kappa}\rightarrow f \mbox{ for some } f_{\eta_\kappa} \mbox{ chosen from some } {\cal F}_{\eta_\kappa}\}.$$</code></p> <p>By $f_\eta\rightarrow f$ we mean uniform convergence on compact subsets of converging domains: that is, the domain of $f$ contains a given compact set if and only if the domain of $f_\eta$ eventually does. A system $\cal F$ is <em>closed</em> if it contains every $g$ such that $f_\eta\rightarrow g$ for some <code>$f_\eta\in{\cal F}$.</code></p> <p>We denote the set of holomorphic dynamical systems on $X$ by $HDS(X)$, and the subset of closed holomorphic dynamical systems by ${\bf HDS}(X)$. </p> <p><strong>Theorem</strong></p> <p>(1) There is a unique topology on ${\bf HDS}(X)$ such that <code>${\cal F}_\eta\rightarrow{\cal F}$</code> if and only if <code>$\liminf {\cal F}_\eta ={\cal F}= \limsup {\cal F}_\eta$</code>.</p> <p>(2) The space ${\bf HDS}(X)$ is compact and Hausdorff. </p> <p>(3) If $X$ has countably many components then ${\bf HDS}(X)$ is second countable and metrizable.</p> <p>The space ${\bf HDS}(X)$ both generalizes and contains the space of closed subgroups of PSL<code>$_2{\mathbb{C}}$</code> with the Hausdorff-Chabauty topology, but the construction requires closer attention to fine points of general topology. In particular, since the appropriate ambient space is neither Hausdorff nor regular, the proper definition of local compactness is crucial: here it should be in the sense that every open neighborhood of a point contains a compact subneighborhood. For any topological space $X$, Fell's prescription yields a compact topological space $Fell(X)$ whose points are the closed subsets of $X$. and which is $Fell(X)$ is compact. Moreover, if $X$ is locally compact then:</p> <p>(1) The space $Fell(X)$ is Hausdorff.</p> <p>(2) <code>${F}_\eta\rightarrow{F}$</code> if and only if <code>$\liminf {F}_\eta ={F}= \limsup {F}_\eta$</code>.</p> <p>(3) If $X$ is second countable then $Fell(X)$ is second countable and metrizable.</p> http://mathoverflow.net/questions/120598/when-must-it-be-sets-rather-than-proper-classes-or-vice-versa-outside-of-fou When must it be sets rather than proper classes, or vice-versa, outside of foundational mathematics? Adam Epstein 2013-02-02T16:36:55Z 2013-02-03T16:31:16Z <p>Every once in a blue moon it actually matters that some mathematical entity which might <em>a priori</em> only be a class is in fact a set. For clarification, here are some examples of what I do <strong>not</strong> mean: </p> <p>A) Some colleagues of mine once made the following disclaimer: 'The "set" of stable curves does not exist, but we leave this set theoretic difficulty to the reader.' These colleagues (names withheld to protect the innocent) are of course fully aware of the fact that strictly speaking, the class of all stable curves, or topological spaces, or groups, or any the other usual suspects customarily formalized in terms of structured sets, cannot itself be a set. While they recognize that the structure they study is transportable along arbitrary bijections between members of a proper class of equinumerous sets, they also recognize that in their setting this same transportability could justify a technically sufficient <em>a priori</em> restriction to some fixed but otherwise arbitrary underlying set: that is, the relevant large category has a small skeletal subcategory. (Exercise: precisely what makes this work in the example given?) In such cases, the set versus class pecadillo is an essentially victimless one, perhaps barring discussions of the admissibility of Choice, expecially Global Choice. </p> <p>B) There are various contexts in which seemingly unavoidable size issues are managed through the device of Grothendieck Universes. Such a move beyond ZFC might be regarded as cheating, sweeping the issue under the carpet for all the right reasons. Allegations of this nature regarding the use of derived functor cohomology in number theory, as in the proof of Fermat's Last Theorem, can now be laid to rest, as Colin McLarty has nicely shown in "A finite order arithmetic foundation for cohomology" <a href="http://arxiv.org/abs/1102.1773" rel="nofollow">http://arxiv.org/abs/1102.1773</a>.</p> <p>C) Set theory itself is replete with situations where the set versus class distinction is of paramount importance. For just one example, my very limited understanding is that forcing over a proper class of conditions is not for the unwary. I'd be interested to hear some expert elucidation of that, but my question here is in a different spirit. </p> <p>With these nonexamples out of the way, I have a very short list of examples that do meet my criteria. </p> <p>1) Freyd's theorem on the nonconcretizability of the homotopy category in "Homotopy is not concrete" <a href="http://www.tac.mta.ca/tac/reprints/articles/6/tr6abs.html" rel="nofollow">http://www.tac.mta.ca/tac/reprints/articles/6/tr6abs.html</a>. By definition, a <em>concretization</em> of a category is a faithful functor to the category of sets. The homotopy category (of based topological spaces) admits no such functor. The crux of the argument is that while any object of a concretizable category has only a set's worth of generalized normal subobjects, there are objects in the homotopy category - for example $S^2$ - which do not have this property (page 9). The original closing remark (page 6) mentions another nonconcretizability result, for the category of small categories and natural equivalence classes of functors. A purist might try to disqualify the latter as too `metamathematical', but the homotopy example seems unassailable. </p> <p>2) A category in which all (co)limits exist is said to be <em>(co)complete</em>; a <em>bicomplete</em> category is one which is both complete and cocomplete. Freyd's General Adjoint Functor Theorem gives necessary and sufficient conditions for the existence of adjoints to a functor $\Phi:{\mathfrak A}\rightarrow{\mathfrak B}$ with $\mathfrak A$ (co)complete. Let us say that a functor which preserves all limits is <em>continuous</em>, and that one which preserves all colimits is <em>cocontinuous</em>. A <em>bicontinuous</em> functor is one which is both continuous and cocontinuous.</p> <p>Let us say that $\Phi$ is <em>locally bounded</em> if for every $B\in {\rm Ob}\,{\mathfrak B}$ there exists a set $\Sigma$ such that for every $A\in{\rm Ob} \,{\mathfrak A}$ and <code>$b\in{\rm Hom}_{\mathfrak B}(B,\Phi A)$</code> there exist <code>$\hat{A}\in{\rm Ob}\,{\mathfrak A}$</code> and <code>$\hat{b}\in{\rm Hom}_{\mathfrak B}(B,\Phi\hat{A})\cap\Sigma$</code> such that<br> $b=(\Phi \alpha)\hat{b}$ for some $\alpha\in{\rm Hom}$ $_{\mathfrak A}$ $(\hat{A},A)$, and that $\Phi$ is <em>locally cobounded</em> if for every $B\in {\rm Ob}\,{\mathfrak B}$ there exists a set $\Sigma$ such that for every $A\in{\rm Ob}\,{\mathfrak A}$ and <code>$b\in{\rm Hom}_{\mathfrak B}(\Phi A,B)$</code> there exist $\hat{A}\in{\rm Ob}\,{\mathfrak A}$ and <code>$\hat{b}\in{\rm Hom}_{\mathfrak B}(\Phi\hat{A},B)\cap\Sigma$</code> such that $b=\hat{b}(\Phi \alpha)$ for some <code>$\alpha\in{\rm Hom}_{\mathfrak A}(A,\hat{A})$</code>. In the literature these are known as the Solution Set Conditions. </p> <p><strong>Theorem.</strong> Let $\Phi:{\mathfrak A}\rightarrow{\mathfrak B}$ be a functor, where $\mathfrak B$ is locally small. </p> <p>$\star$ If $\mathfrak A$ is complete then $\Phi$ admits a left adjoint if and only if $\Phi$ is continuous and locally bounded.<br> $\star$ If $\mathfrak A$ is cocomplete then $\Phi$ admits a right adjoint if and only if $\Phi$ is cocontinuous and locally cobounded. </p> <p>See pages 120-123 of MacLane's "Categories for the working mathematician". </p> <p>The local (co)boundedness condition has actual content. For example: </p> <p>a) The forgetful functor ${\bf CompleteBooleanAlgebra}\rightarrow{\bf Set}$ is continuous but admits no left adjoint. </p> <p>b) Functors ${\bf Group}\rightarrow {\bf Set}$, continuous but admitting no left adjoint, may be obtained as follows: let $\Gamma_\alpha$ be a simple group of cardinality $\aleph_\alpha$ (<em>e.g.</em> the alternating group on a set of that cardinality, or the projective special linear group on a 2-dimensional vector space over a field of that cardinality) and take the product (suitably construed), over the proper class of all ordinals, of the functors <code>${\rm Hom}_{\bf Group}(\Gamma_\alpha,-)$</code>. </p> <p>c) Freyd proposed another interesting example (see page -15 of the Foreword to "Abelian categories" <a href="http://www.tac.mta.ca/tac/reprints/articles/3/tr3abs.html" rel="nofollow">http://www.tac.mta.ca/tac/reprints/articles/3/tr3abs.html</a>) of a locally small bicomplete category $\mathfrak S$ and a bicontinuous functor $\Phi:{\mathfrak S}\rightarrow {\bf Set}$ which admits neither adjoint: loosely speaking, the category of sets equipped with free group actions, and the evident underlying set functor. </p> <p>Does anyone know of any other examples, especially fundamentally different examples? </p> <p>Finally, one could focus critical attention on the very question posed. To what extent does the strength and flavor of the background set theory matter? Force of habit and comfort have me implicitly working in some material set theory such as ZF, perhaps a bit more if I want to take advantage of Choice, perhaps a bit less if I prefer to eschew Replacement. Indeed, I have actually checked that example b) may be formulated in the absence of Replacement: while the von Neumann ordinals are no longer available, the same trick already used to give a kosher workaround to the illegitimate product over all ordinals further shows that an appropriate system of local ordinals suffices for the task. I am also quite interested in hearing what proponents of structural set theory have to say.</p> http://mathoverflow.net/questions/120534/basic-questions-on-the-homotopy-category/120558#120558 Answer by Adam Epstein for Basic questions on the homotopy category Adam Epstein 2013-02-01T23:28:18Z 2013-02-01T23:28:18Z <p>As Tom points out, many diagrams in the homotopy categories do not even have limits or colimits, so the forgetful functor cannot preserve them!</p> <p>Meanwhile, I too had difficulty chasing this down, until I found the following remark (M. Mather, Pullbacks in homotopy theory, Can. J. Math. 1976):</p> <p>"For example, no essential map between Eilenberg-MacLane spaces of different dimensions has a kernel". </p> <p>My attempt at reconstruction goes as follows. Let $m,n$ be positive integers and let $G,H$ be abelian groups. If $m\neq n$ then any morphism $K(G,m)\rightarrow K(H,n)$ induces the zero morphism $\pi_\ell(K(G,m))\rightarrow\pi_\ell(K(H,n))$ for each $\ell\geq 0$. However, since the set of morphisms $K(G,m)\rightarrow K(H,n)$ is in canonical bijection with $H^n(K(G,m),H)$ which is typically nontrivial, such essential morphisms do exist. To see that such a morphism $K(G,m)\rightarrow K(H,n)$ admits no kernel, note that if $X\rightarrow K(G,m)$ were such a kernel then all the induced maps $\pi_\ell(X)\rightarrow\pi_\ell(K(G,m))$ would be bijections (for $\ell=m$ because $\pi_m(K(H,n))=0$, for $\ell\neq m$ because $\pi_\ell(K(G,m))=0$) so that $X\rightarrow K(G,m)$ would be an isomorphism in the homotopy category, whence $K(G,m)\rightarrow K(H,n)$ must be inessential.</p> <p>I have been meaning to post my own question regarding whether anyone happens to know a simpler example (or a recipe, or a reference) of a diagram which fails to have co/limits.</p> http://mathoverflow.net/questions/119494/examples-of-exotic-induction/119579#119579 Answer by Adam Epstein for Examples of "exotic" induction Adam Epstein 2013-01-22T15:18:11Z 2013-01-22T15:18:11Z <p>Goodstein's Theorem </p> <p><a href="http://en.wikipedia.org/wiki/Goodstein%27s_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Goodstein%27s_theorem</a> </p> <p>is proved using an induction of length $\epsilon_0$.</p> http://mathoverflow.net/questions/119113/essential-uniqueness-of-the-real-analytic-structure-on-mathbb-r Essential uniqueness of the real-analytic structure on $\mathbb R$ Adam Epstein 2013-01-16T20:55:08Z 2013-01-22T11:40:48Z <p>It is well-known that any $C^k$-smooth $1$-manifold homeomorphic to $\mathbb R$ is $C^k$-diffeomorphic to $\mathbb R$. The cases of $k\in{\mathbb N}\cup$ {$\infty$} may all be handled similarly by an elementary argument as follows: using partitions of unity to construct a nowhere vanishing 1-form, integrate to obtain a diffeomorphism to a connected open submanifold of $\mathbb R$, and compose with some elementary diffeomorphism from the submanifold to $\mathbb R$.</p> <p>Presented this way, the argument breaks down in the real-analytic case $k=\omega$ because partitions of unity are no longer available. Even in that case, the assertion is still true, since Grauert-Remmert have shown that $C^1$-diffeomorphic real-analytic manifolds are real-analytically diffeomorphic (assuming paracompactness, there being uncountably many inequivalent real-analytic structures on the long ray). However, this is a very difficult general result. </p> <p>In the case at hand, it is not hard to see that the partitions of unity are merely a device for proving a cohomological vanishing theorem $$H^1({\mathbb R},{\mathcal E})=0$$ where ${\mathcal E}$ is the sheaf of germs of appropriately smooth real-valued functions. Indeed, consider a covering of $\mathbb R$ by open intervals $I_n$, each intersecting only its immediate predecessor and immediate successor, chosen small enough that each is real-analytically diffeomorphic to a standard interval (hence also to $\mathbb R$). Note that any collection of functions defined on the intersections $I_n\cap I_{n+1}$ yields a 1-cocycle. Since the $I_n$ are standard intervals, there exist everywhere positive 1-forms $\eta_n$ defined on $I_n$. Thus, there exist smooth functions $f_n$ defined on $I_n\cap I_{n+1}$ such that $\eta_{n+1}=(\exp f_n)\eta_n$ on that intersection. The vanishing theorem implies that the 1-cocycle {$f_n: n\in{\mathbb Z}$} is a 1-coboundary, that is, for some collection of functions $g_n$ defined on $I_n$, we have that $f_n$ is the restriction of $g_{n}-g_{n+1}$ to $I_{n}\cap I_{n+1}$. By construction, the 1-forms $(\exp g_n)\eta_n$ on $I_n$ are the restrictions of a globally defined positive 1-form $\eta$. </p> <p><strong>Question 1:</strong> How is the vanishing theorem established in the real-analytic case?</p> <p>I imagine this must be well-known, but I've not been able to find such a discussion in the literature. Perhaps I am just looking in the wrong places. In any event, several years ago I put together such an argument. The idea is to consider each consecutive pair of intervals $I_n, I_{n+1}$ each slightly thickened to a complex neighborhood given by a smoothly bounded Jordan domain $D_n$, the intersection of consecutive neighborhoods being another Jordan domain. If the neighborhoods are small enough, the given functions $f_n$ extend complex-analytically to the intersections, the real part yielding values on $\partial(D_n\cap D_{n+1})$, which in turn (suitably extended by 0) yield a function on (say) $\partial D_n$ which we then extend harmonically, hence real-analytically, via the Poisson Integral formula.</p> <p><strong>Question 2:</strong> Is such an argument been written down in the literature, or otherwise well-known?</p> <p>Of course, once we resort to patching suitable complex neighborhoods of the chart images, there is a quick and dirty proof via the Uniformization Theorem: it suffices to glue suitable real-symmetric neighborhoods to obtain a simply connected Riemann surface with an anticonformal involution whose fixed locus is the given 1-manfold.</p> <p><strong>Question 3:</strong> Is this surely well-known argument written down in the literature?</p> <p>To be honest, I started out knowing the argument via Uniformization, and decided to see whether this could be reduced to more elementary considerations. The proposed argument to prove the vanishing theorem succeeds partially, but I was struck by the fact that I am still doing complex analysis, or at least potential theory. Maybe it's unreasonable to expect to be able to produce real-analytic functions without sneaking a peek into the complex plane. </p> http://mathoverflow.net/questions/119317/holomorphic-covering-between-points-in-teichmuller-space/119321#119321 Answer by Adam Epstein for holomorphic covering between points in Teichmuller space Adam Epstein 2013-01-19T11:34:39Z 2013-01-19T11:34:39Z <p>As $g\geq 2$, it follows by Riemann-Hurwitz that any topological covering $X\rightarrow Y$ is a homeomorphism, and any holomorphic homeomorphism is biholomorphic.</p> http://mathoverflow.net/questions/60375/is-r3-the-square-of-some-topological-space/119296#119296 Answer by Adam Epstein for Is R^3 the square of some topological space? Adam Epstein 2013-01-19T00:12:41Z 2013-01-19T00:12:41Z <p>I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.</p> http://mathoverflow.net/questions/118407/forcing-over-the-poset-of-nonempty-open-subsets-of-a-nice-topological-space Forcing over the poset of nonempty open subsets of a nice topological space Adam Epstein 2013-01-08T23:58:13Z 2013-01-09T00:43:43Z <p>Is there anything sensible to be said concerning a notion of forcing given by the poset of nonempty open subsets of the sort of topological space that comes up in ($e.g.$ algebraic) topology? If so, are interesting topological properties somehow coded in the resulting forcing extennsion. For example, would ${\mathbb S}^1$ versus ${\mathbb S}^2$ (or the open interval versus the closed interval versus the Hawaiian earring) yield a detectable difference? I suppose what's really at issue is how much topological information is lost on passage to the complete Boolean algebra of regular open subsets: to what extent can a space be reconstructed from that structure?</p> http://mathoverflow.net/questions/117174/category-of-topological-spaces-with-open-or-closed-maps Category of topological spaces with open or closed maps Adam Epstein 2012-12-25T00:01:21Z 2013-01-08T20:07:47Z <p>Consider the category whose objects are topological spaces and whose morphisms are the open maps (or closed maps, open continuous maps, closed continuous maps ... that is, one whose isomorphisms are precisely the homeomorphisms). How does such a category compare with the usual one whose objects are topological spaces and whose morphisms are continuous maps? For example, what limits and colimits exist?</p> <p>I'm probably missing something obvious, but why don't products typically exist in the category with open maps? The projections from the usual product (in the category with continuous maps) are open, yielding a canonical open map from the usual product to the putative unusual product.</p> <p>Todd's observation is true enough: the product in the usual topology (contiuous maps) typically fails to realize the corresponding universal property in the unusual topology (open maps). Nevertheless, some other object might realize that universal property. Is it even clear that the if such a space exists its underlying set should be naturally identifiable with the underlying set of the factors? After all, while one point spaces are still terminal, maps out of such objects tend not to be open: it seems one would thereby only extract the subset of isolated points. In any event, <a href="http://christianmarks.wordpress.com/category/bagatelle" rel="nofollow">http://christianmarks.wordpress.com/category/bagatelle</a><br> treats the special case of squares. The appropriate space is $X\times X$ with the weakest topology (stronger than the usual) which makes the diagonal embedding open. This construction is clearly not available for products of distinct spaces. My question concerns whether there isn't (as that post suggests there isn't) some devious workaround.</p> http://mathoverflow.net/questions/117910/can-one-exhibit-an-explicit-kuratowski-infinite-set-without-invoking-replacement Can one exhibit an explicit Kuratowski infinite set without invoking Replacement? Adam Epstein 2013-01-02T22:15:45Z 2013-01-03T23:33:11Z <p>The customary formulation of the Axiom of Infinity within Zermelo-Fraenkel set theory asserts the existence of an inductive set: a set $I$ with $\varnothing\in I$ such that $x\in I$ implies $x\cup\{x\}\in I$. Since the intersection of any nonempty set of inductive sets is itself inductive, an instance of the Axiom Schema of Separation implies the existence of a smallest inductive set, namely the set of <em>von Neumann naturals</em> $$\mathbb{N}_{\bf vN} = \{\varnothing, \{\varnothing\},\{\varnothing,\{\varnothing\}\},\ldots\}.$$</p> <p>Any inductive set is infinite (in fact, Dedekind infinite) but this formulation of the axiom asserts more, namely the existence of a specific countably infinite set. Given one such set, the existence of others, for example the set of <em>Zermelo naturals</em> $$\mathbb{N}_{\bf Zer}=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\ldots\}$$ follows from appropriate instances of the Axiom Schema of Replacement. </p> <p>Consider the subsystem of Zermelo-Fraenkel set theory with axioms Extensionality, Separation Schema, Union, Power Set, Pair. Augment this Basic System with an Axiom of Infinity which asserts the existence of an infinite set, but not any particular one. Such a formulation requires that the notion of 'finite' be defined prior to that of 'natural number', following Kuratowski for example. Any infinite set $I$ determines a Dedekind-infinite set of <em>local naturals</em> <code>$$\mathbb{N}_I=\{\mbox{equinumerosity classes of finite subsets of } I\}$$</code> which (duly equipped with initial element and successorship) yields a Lawvere natural number object, as in the Recursion Theorem. The existence of <code>$\mathbb{N}_{\bf vN}$</code> and <code>$\mathbb{N}_{\bf Zer}$</code> then follow from appropriate instances of Replacement.</p> <p>One might wonder if there is some clever way to specify an infinite set without recourse to Replacement. That is, does there exist (in the language of set theory) a formula $\boldsymbol \phi$ with one free variable $x$ such that<br> $$\mbox{Basic+Infinity+Foundation } \vdash\; \exists y ( \forall x (x\in y \leftrightarrow \boldsymbol \phi)\,\wedge \, y \mbox{ is infinite})\;\;?$$</p> <p>I'm inclined to guess no, on the following circumstantial grounds:</p> <ul> <li><p>For <code>$\mathbb{N}_{\bf vN}$</code> and <code>$\mathbb{N}_{\bf Zer}$</code> the use of Replacement is essential: Mathias has shown (Theorem 5.6 of <em>Slim Models of Zermelo Set Theory</em> that there exist transitive models <code>${\mathfrak M}_{\bf vN}$</code> and <code>${\mathfrak M}_{\bf Zer}$</code> of Basic+Infinity+Foundation with <code>${\mathbb N}_{\bf vN}\in {\mathfrak M}_{\bf vN}$</code> and <code>${\mathbb N}_{\bf Zer}\in{ \mathfrak M}_{\bf Zer}$</code>, but such that every element of <code>${\mathfrak M}_{\bf vN}\cap {\mathfrak M}_{\bf Zer}$</code> is hereditarily finite.</p></li> <li><p>The usual definitions of <code>$\mathbb{N}_{\bf vN}$</code> and <code>$\mathbb{N}_{\bf Zer}$</code> involve unstratified formulas. Coret has shown (Corollary 9 of <em>Sur les cas stratifiés du schéma du replacement</em>) that this is unavoidable: $$\mbox{Basic+Infinity } \vdash\; \forall y ( \forall x (x\in y \leftrightarrow \boldsymbol \phi)\,\rightarrow \, y \mbox{ is hereditarily finite})$$ for any stratified $\boldsymbol \phi$. Using the same technique he has shown (Corollary 10) that Basic+Infinity proves every stratified instance of Replacement.</p></li> </ul> http://mathoverflow.net/questions/106943/who-named-it-the-snake-lemma Who named it the Snake Lemma? Adam Epstein 2012-09-11T19:04:37Z 2012-09-12T10:10:32Z <p>What is the history behind the colorful name of this result? Cartan-Eilenberg states it without any particular fanfare.</p> http://mathoverflow.net/questions/105478/finite-dimensional-real-division-algebras finite dimensional real division algebras Adam Epstein 2012-08-25T20:04:42Z 2012-08-25T20:04:42Z <p>A celebrated theorem of Milnor and Kervaire asserts that any finite dimensional division algebra over the real numbers has dimension 1,2,4 or 8. This result is established using methods from algebraic topology, such as K-Theory.</p> <p>Now for any given natural number $n$ the existence of such an algebra of dimension $n$ is expressible as an assertion $\phi_n$ in the first-order language of field theory. Since the theory $RCF$ of real closed fields is complete, it follows from the theorem above that $RCF \vdash \neg \phi_n$ for all $n\not\in$ {1,2,4,8}. Here the universal quantifier on $n$ is in the meta-theory: we might say that for each $n$ there is an elementary proof of $\phi_n$. </p> <p>Given such a theorem scheme, one might wonder whether there might be a uniform elementary proof. Informally this could mean a proof by induction on the relevant complexity parameter: for example, $$RCF \vdash \mbox{ any degree } d \mbox{ polynomial has at most } d \mbox{ roots}.$$ I would like to imagine that there is some first order-theory which suitably contains both RCF and Peano Arithmetic (in particular, so as to enable discussion of finite sequences of field elements) in which the assertion $$\forall n \;\phi_n\leftrightarrow(n=1 \vee n=2 \vee n=4 \vee n=8)$$ can be legfitimately formalized. Are there standard constructions for supporting finite sequences? If so, it should follow from completeness of RCF that this assertion is equivalent (within such a larger theory) to a sentence $\Phi$ in the language of arithmetic. As noted above, via difficult results from topology, $\Phi$ is true in the standard model of Peano Arithmetic. Consequently, it makes sense to ask whether $\Phi$ is provable within Peano Arithmetic. </p> <p>Some questions:</p> <p>(1) Can such a recipe be formalized, and does it reasonably capture the notion of "uniform elementary proof" or "purely algebraic" proof for such theorem schemes? Here I am not necessarily claiming that these conjectural notions are the same.</p> <p>(2) In the given example of the 1,2,4,8 theorem, do we expect $\Phi$ to be provable in Peano Arithmetic? </p> <p>Perhaps I have been looking in the wrong places, but all I have managed to find are a few comments by Kreisel about "unwinding", on pages 67-68 of this note: <a href="http://elib.mi.sanu.ac.rs/files/journals/zr/10/n010p063.pdf" rel="nofollow">http://elib.mi.sanu.ac.rs/files/journals/zr/10/n010p063.pdf</a></p> <p>The situation could be compared with what is known in the special cases of commutative division algebras (dimensions 1,2) and associative division algebras (dimensions 1,2,4). Hopf's proof of the (1,2) theorem also uses some topology, namely that the $n$-dimensional sphere and $n$-dimensional projective space are not homeomorphic when $n>1$; in fact it suffices to show that a specific map between these spaces is not a homeomorphism. Perhaps there is an elementary way to formulate this consideration? On the other hand, there is a different and "purely algebraic" proof, via Bezout's Theorem. I don't have the reference at hand, but it there is a citation (froom the 1950s, as I recall) in the Springer-Verlag book Numbers (Ebbinghaus et. al.). I've seen this proof dismissed as unreadable or unenlightening, but when I examined it years ago it seemed like it might qualify. The Frobenius proof of the (1,2,4) theorem is quite evidently purely algebraic, as is the later extension (1,2,4,8) to alternative division algebras.</p> http://mathoverflow.net/questions/67962/cardinality-of-connected-manifolds Cardinality of connected manifolds Adam Epstein 2011-06-16T15:47:24Z 2011-06-16T23:05:47Z <p>Consider the assertion:</p> <p>Every connected, but not necessarily paracompact, n-manifold is of cardinality $2^{\aleph_0}$ (at least assuming the axiom of choice). </p> <p>For n=1 this may be proved via enumeration of the short list of examples. The essential point is that while there is a Long Line, there is no Extra Long Line.</p> <p>What is the situation for n>1?</p> http://mathoverflow.net/questions/125446/consistency-of-the-concept-of-the-collection-of-all-collection/125539#125539 Comment by Adam Epstein Adam Epstein 2013-03-25T16:11:56Z 2013-03-25T16:11:56Z That sounds nice, I'm looking forward to it. http://mathoverflow.net/questions/125446/consistency-of-the-concept-of-the-collection-of-all-collection Comment by Adam Epstein Adam Epstein 2013-03-24T16:47:18Z 2013-03-24T16:47:18Z When you specialize Cantor's proof (that no function from a set to its power set is surjective) to the identity - from the set of all sets to its power set, that is itself - the witness to nonsurjectivity is the Russell entity {$x: x\notin x$} which, not being a set, contradicts nothing. http://mathoverflow.net/questions/21880/can-one-escape-from-the-mirror-image-of-russells-paradox/22004#22004 Comment by Adam Epstein Adam Epstein 2013-03-24T10:41:24Z 2013-03-24T10:41:24Z In ZF-Foundation you can have as many Quine atoms (sets which are their ownn unique element) as you like. For example, see Chapter IIIA of Felgner's &quot;Models of ZF Set Theory&quot;. http://mathoverflow.net/questions/125414/self-containing-structures Comment by Adam Epstein Adam Epstein 2013-03-24T10:20:53Z 2013-03-24T10:20:53Z +1 for the Gromov-Hausdorff example. http://mathoverflow.net/questions/105478/finite-dimensional-real-division-algebras Comment by Adam Epstein Adam Epstein 2013-03-23T21:50:43Z 2013-03-23T21:50:43Z Thanks, I'll have a look! http://mathoverflow.net/questions/124824/the-mandelbrot-set-is Comment by Adam Epstein Adam Epstein 2013-03-23T17:50:33Z 2013-03-23T17:50:33Z It is also [bounded] http://mathoverflow.net/questions/125020/partial-linearization-near-a-hyperbolic-fixed-point-classical-scattering/125061#125061 Comment by Adam Epstein Adam Epstein 2013-03-20T18:57:43Z 2013-03-20T18:57:43Z Even in a finite diimensional setting there would be the issue of possible resonances among the eigenvalues, and little is known about those numbers http://mathoverflow.net/questions/125020/partial-linearization-near-a-hyperbolic-fixed-point-classical-scattering/125061#125061 Comment by Adam Epstein Adam Epstein 2013-03-20T14:30:39Z 2013-03-20T14:30:39Z I'd look at Section 6. For example, Theorem 6.3 gives a hyperbolic splitting of the tangent space. Subsequent results discuss stable and unstable manifolds. If you want an actual reduction to normal form, this does not seem to be considered, and on reflection I am not aware that such a result has been formally claimed in a complex analytic setting. http://mathoverflow.net/questions/124892/on-the-category-of-virtual-species Comment by Adam Epstein Adam Epstein 2013-03-19T13:00:00Z 2013-03-19T13:00:00Z For clarification, perhaps someone could say something about this matter is (or is not) related to considerations in Blass's &quot;Seven Trees in One&quot; and Schanuel's &quot;Negative sets have Euler Charactieristic and dimension&quot;. http://mathoverflow.net/questions/124875/finding-sun-acceleration-towards-center-of-galaxy Comment by Adam Epstein Adam Epstein 2013-03-19T00:23:55Z 2013-03-19T00:23:55Z I'd upvote the editor for the tag. http://mathoverflow.net/questions/124347/range-is-to-co-domain-as-domain-is-to-what Comment by Adam Epstein Adam Epstein 2013-03-18T20:46:24Z 2013-03-18T20:46:24Z You are of course absolutely right. http://mathoverflow.net/questions/124494/bijective-equivalent-collections-of-proper-classes-in-set-theory Comment by Adam Epstein Adam Epstein 2013-03-14T11:53:28Z 2013-03-14T11:53:28Z Dear Gerard Lang, No worries - it was clear that you didn't mean to separately count the result of precomposing by some transposition :) http://mathoverflow.net/questions/124494/bijective-equivalent-collections-of-proper-classes-in-set-theory Comment by Adam Epstein Adam Epstein 2013-03-14T10:40:28Z 2013-03-14T10:40:28Z There surely won't be a uique bijection between POrd and a proper subclass, but maybe you meant that there is a preferred one in which they are listed in order. http://mathoverflow.net/questions/124347/range-is-to-co-domain-as-domain-is-to-what Comment by Adam Epstein Adam Epstein 2013-03-12T21:33:55Z 2013-03-12T21:33:55Z Co-image, anyone? http://mathoverflow.net/questions/123482/is-there-a-constructive-proof-of-cantorbernsteinschroeder-theorem/123485#123485 Comment by Adam Epstein Adam Epstein 2013-03-04T22:20:59Z 2013-03-04T22:20:59Z The inclusion $g$ seems the more obvious to me. For $f$, I imagine you would start with a Turing machine which loops on input 0, and then for each $n$ attach a subroutine which effectively lets it sit idle that many steps before looping?