User toby bartels - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:45:17Z http://mathoverflow.net/feeds/user/8508 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121633/what-is-an-ideal-supporting-algebra/121648#121648 Answer by Toby Bartels for What is an ideal-supporting algebra? Toby Bartels 2013-02-12T20:56:01Z 2013-02-12T21:09:34Z <p>It looks like the term ‘ideal-supporting algebra’ was written by me and survived slightly more than a decade on Wikipedia without being altered. (Well, somebody added a hyphen, a change that I agree with.) Since I put brackets around it, I'm sure that I must have heard the term somewhere, but I couldn't tell you now. Now that I think of it, a more precise term would be ‘ideal-supporting variety [of algebras]’.</p> <p>And if I search for <em>that</em> phrase, I find it in Eric Schechter's 1996 <em>Handbook of Analysis and its Foundations</em> (which for some reason Google Books has classified under Business &amp; Economics). Since I was reading this book a lot a decade ago, that's probably what I meant all along. Shechter often invented terminology for his book, when terminology in the literature was inconsistent or missing, so I wouldn't be surprised if it's essentially unique to him.</p> <p>At this point, probably the best thing for me to do is to edit Wikipedia for a little bit, finishing what I started in 2002.</p> http://mathoverflow.net/questions/112318/is-the-generalized-baire-space-complete/112321#112321 Answer by Toby Bartels for Is the generalized Baire space complete? Toby Bartels 2012-11-13T22:26:27Z 2012-11-14T15:18:24Z <p>As has already been remarked by Lasse Rempe-Gillen, you need to know what a Cauchy net is, say from a uniform structure. But since you want $\kappa$ to be discrete [edit: I had ‘complete’ here once] as a topological space, surely you just want it to be discrete as a uniform space, so let's do that. And then give $\kappa^\kappa$ the product uniformity. (For $\kappa := \omega$, this is the uniform structure that underlies the usual metric on Baire space.)</p> <p>So the answer is Yes, $\kappa^\kappa$ is complete. This is because every discrete uniform space is complete, and (as Lasse has remarked) any product of complete spaces is complete. Also, note that the underlying topology of the product uniformity is the product topology, so $\kappa^\kappa$ has the topology that you originally wanted.</p> <p>The cited facts are all part of my general knowledge; I will try to find specific references for them.</p> http://mathoverflow.net/questions/60474/is-there-a-convenient-differential-calculus-for-cojets Is there a convenient differential calculus for cojets? Toby Bartels 2011-04-03T18:55:35Z 2012-09-19T09:12:10Z <p>I understand the basics of exterior differential geometry and how to do calculus with exterior differential forms. I know how to use this to justify the notation d<i>y</i>/d<i>x</i> as a literal ratio of the differentials d<i>y</i> and d<i>x</i> (by treating <i>x</i> and <i>y</i> as scalar-valued functions on a 1-dimensional manifold and introducing division formally). I would like to extend this to <em>second</em> derivatives. Ideally, this would justify the notation d<sup>2</sup><i>y</i>/d<i>x</i><sup>2</sup> as a literal ratio.</p> <p>I can't do this with the exterior differential, since both d<sup>2</sup><i>y</i> and d<i>x</i>&nbsp;&and; d<i>x</i> are zero in the exterior calculus. It occurs to me that this would work if, instead of exterior differential forms (sections of the exterior bundle), I used sections of the cojet bundle (cojet differential forms). In particular, while degree-2 exterior forms may be written in local coordinates as linear combinations of d<i>x</i><sup><i>i</i></sup>&nbsp;&and; d<i>x</i><sup><i>j</i></sup> for <i>i</i>&nbsp;&lt; <i>j</i> (so on a 1-dimensional manifold the only exterior 2-form is zero), degree-2 cojet forms may be written in local coordinates as linear combinations of d<sup>2</sup><i>x</i> and d<i>x</i><sup><i>i</i></sup>&nbsp;&middot; d<i>x</i><sup><i>j</i></sup> for <i>i</i>&nbsp;&le; <i>j</i> (so on a 1-dimensional manifold the cojet 2-forms at a given point form a 2-dimensional space).</p> <p>I know some places to read about cojets (and more so about jets) theoretically, but I don't know where to learn about practical calculations in a cojet calculus analogous to the exterior calculus. In particular, I don't know any reference that introduces the concept of the degree-2 differential operator d<sup>2</sup>, much less one that gives and proves its basic properties. I've even had to make up the notation &lsquo;d<sup>2</sup>&rsquo; (although you can see where I got it) and the term &lsquo;cojet differential form&rsquo;. I can work some things out for myself, but I'd rather have the confidence of seeing what others have done and subjected to peer review.</p> <p>(Incidentally, I don't think that it is <em>quite</em> possible to justify d<sup>2</sup><i>y</i>/d<i>x</i><sup>2</sup>; the correct formula is d<sup>2</sup><i>y</i>/d<i>x</i><sup>2</sup>&nbsp;&minus; (d<i>y</i>/d<i>x</i>)(d<sup>2</sup><i>x</i>/d<i>x</i><sup>2</sup>); we cannot let d<sup>2</sup><i>x</i>/d<i>x</i><sup>2</sup> vanish and retain the simplicity of the algebraic rules. It would be better to write &part;<sup>2</sup><i>y</i>/&part;<i>x</i><sup>2</sup>; the point is that this is the coefficient on d<i>x</i><sup>2</sup> in an expansion of d<sup>2</sup><i>y</i>, just as &part;y/&part;x<sup><i>i</i></sup> is the coefficient of d<i>y</i> on <i>x</i><sup><i>i</i></sup> when <i>y</i> is a function on a higher-dimensional space. The coefficient of d<sup>2</sup><i>y</i> on d<sup>2</sup><i>x</i>, which would be &part;<sup>2</sup><i>y</i>/&part;<sup>2</sup><i>x</i>, is simply d<i>y</i>/d<i>x</i> again.)</p> http://mathoverflow.net/questions/105502/which-w-algebras-are-the-duals-of-c-coalgebras Which W*-algebras are the duals of C*-coalgebras? Toby Bartels 2012-08-26T08:08:59Z 2012-08-26T08:43:10Z <p>A Banach algebra (assumed associative and unital) is precisely a monoid object in the monoidal category of Banach spaces, short linear maps, and the projective tensor product. A Banach coalgebra is then a comonoid object in this monoidal category. It's straightforward to write down what a Banach *-coalgebra is too. It's a little less obvious what a C*-coalgebra is, and I don't know if that term appears in the literature, but <a href="http://ncatlab.org/nlab/show/C-star-coalgebra" rel="nofollow">I've written down my definition</a>.</p> <p>Generally, the dual space of a coalgebra is an algebra (but not conversely), and that works here too: the dual of a C*-coalgebra is a C*-algebra. But not every C*-algebra arises in this way; obviously, since all of these C*-algebras have preduals (having been explicitly constructed so), they are W*-algebras. But I don't know what other conditions must be satisfied.</p> <p>So my question is, and I'll be grateful for incomplete answers: Which W*-algebras arise (up to abstract isomorphism) as duals of C*-coalgebras?</p> <p>Partial answers: The sequence space $l^\infty$ is the dual of $l^1$, and $l^1$ is a C*-coalgebra. But this doesn't work for $L^\infty(R)$; this is the dual of $L^1(R)$, but I can't make $L^1(R)$ into even a Banach coalgebra (in an appropriate way), essentially because the diagonal in $R^2$ has measure zero. (Unless I've miscalculated something, and I'm trying to do the wrong thing here.) Of course, these are quite limited examples: they're (co)commutative. I'd be grateful for more.</p> http://mathoverflow.net/questions/93435/chomp-without-the-law-of-the-excluded-middle/93447#93447 Answer by Toby Bartels for Chomp! without the law of the excluded middle Toby Bartels 2012-04-07T19:30:34Z 2012-04-07T19:30:34Z <p>I don't think that there is anything interesting about this constructively, as long as we limit ourselves to <a href="http://ncatlab.org/nlab/show/finite+set" rel="nofollow">finite sets</a> in the strictest sense. The proof that a winning strategy exists gives us no help in finding the strategy, so in this sense it is nonconstructive, but we never needed help in finding a strategy. Only finitely many moves are possible, with only finitely many options for each move, so all that we ever had to do to find a winning strategy (if one exists) is to enumerate all of the possibilities. Practically, this is impossible, so it would be nice if an existence proof for such a strategy would shorten the search, but this is a separate issue from the acceptability of the proof for constructive mathematics. Put another way, we can already prove constructively, since there are only finitely many possible plays of the game, that one player or the other must have a winning strategy, so any proof that relies on this is still constructive.</p> http://mathoverflow.net/questions/93437/math-blog-directory/93441#93441 Answer by Toby Bartels for Math blog directory Toby Bartels 2012-04-07T18:30:58Z 2012-04-07T18:30:58Z <p>There's this nLab page: <a href="http://www.ncatlab.org/nlab/show/math+blogs" rel="nofollow">http://www.ncatlab.org/nlab/show/math+blogs</a> and some other lists that it links to.</p> http://mathoverflow.net/questions/85717/segals-original-definition-of-a-topological-category/85718#85718 Answer by Toby Bartels for Segal's Original Definition of a Topological Category Toby Bartels 2012-01-15T06:55:53Z 2012-01-15T06:55:53Z <p>I would call this an <a href="http://ncatlab.org/nlab/show/internal+category" rel="nofollow">internal category</a> in the category of topological spaces and continuous maps.</p> http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/85638#85638 Answer by Toby Bartels for Why is a topology made up of 'open' sets? Toby Bartels 2012-01-14T07:11:31Z 2012-01-14T07:11:31Z <p>There are several interpretations of the original question, but one is, why focus on open sets rather than closed sets? I have an unusual answer.</p> <p>Suppose you want to do constructive mathematics. (Don't ask me why, you just do.) So you abstract the properties of open and closed subsets from the real line. Then you see that open subsets are closed under arbitrary union but only finitary intersection, OK. Dually, you see that closed sets are closed under arbitrary intersection but … <em>not</em> under finitary union! For example, the union of $[ 0 , 1 ]$ and $[ 1 , 2 ]$ cannot be proved to be closed. (The closure of the union is $[ 0 , 2 ]$, but to prove that the union itself is all of $[ 0 , 2 ]$ requires the lesser limited principle of omniscience. Or less formally, there is no definite method to decide whether a number is near $1$ is in $[ 0 , 1 ]$ or in $[ 1 , 2 ]$.) So open sets are better behaved and naturally you prefer to axiomatise them.</p> <p>But as you continue with constructive topology, more advanced things fail, such as the Tychonoff Theorem (which implies the axiom of choice and thus excluded middle). Then you learn that this stuff works in locale theory, so you abandon traditional topological spaces for locales. And here the duality between open and closed is restored; a locale's frame of opens can just as well be interpreted as a coframe of closeds, and only tradition tells us to do the first.</p> <p>So this answer only works in a very unusual frame of mind: setting off down an unusual path but not going all the way.</p> http://mathoverflow.net/questions/83965/interesting-applications-of-the-classical-stokes-theorem/83966#83966 Answer by Toby Bartels for Interesting Applications of the Classical Stokes Theorem? Toby Bartels 2011-12-20T21:30:11Z 2011-12-20T21:50:57Z <p>In the theory of electromagnetism, the classical Stokes Theorem moves between the differential and integral forms of two of Maxwell's four equations; see <a href="https://en.wikipedia.org/wiki/Stokes%27_theorem#In_electromagnetism" rel="nofollow">https://en.wikipedia.org/wiki/Stokes%27_theorem#In_electromagnetism</a> for discussion. Note that the integral forms may be directly interpreted using classical physical intuition, while the differential forms give us differential equations that we might solve, so it is important that we can switch between them.</p> <p>ETA: I think that Wikipedia's discussion is a little vague, although possibly appropriate in that context. So here is more detail, looking at Faraday's Law. In terms of physically observable quantities, the law states that the rate of change of the magnetic flux through a stationary surface is proportional to the electromotive force around the boundary of the surface. The magnetic flux is the surface integral of the magnetic field $\vec H$, and the EMF is the line integral of the electric field $\vec E$, so we have $$\oint _ { \partial S } \vec E \cdot \mathrm d \vec r = - \frac { \mathrm d } { \mathrm d t } \iint _ S \vec H \cdot \mathrm d ^ 2 \vec A$$ using standard units and sign conventions. Applying the classical Stokes Theorem on the left and using that $S$ is stationary on the right, this becomes $$\iint _ S ( \nabla \times \vec E ) \cdot \mathrm d ^ 2 \vec A = - \iint _ S \frac { \partial \vec H } { \partial t } \cdot \mathrm d ^ 2 \vec A \text ;$$ since this holds for arbitrarily small surfaces, we conclude that $$\nabla \times \vec E = - \frac { \partial \vec H } { \partial t } \text ,$$ a differential equation. (The argument in reverse is even easier, since you don't have to worry about arbitrarily small surfaces.)</p> http://mathoverflow.net/questions/82604/uniform-continuity-and-boundedness/82609#82609 Answer by Toby Bartels for Uniform continuity and boundedness Toby Bartels 2011-12-04T10:39:49Z 2011-12-04T10:39:49Z <p>I'm afraid that I don't like your proposed proof. You derive a bound on $f(x)$, namely $\epsilon + f(y)$, but this is not fixed. Although you may choose any positive $\epsilon$ you wish (which then gives you $\delta$), $y$ must be chosen to be within $\delta$ of $x$. So as you vary $M$, you vary $x$ (to keep $f(x) > M$), but then (to keep it close enough to $x$) you vary $y$, and it seems possible that $f(y)$ would grow fast enough that $\epsilon + f(y) > M$ would be maintained.</p> http://mathoverflow.net/questions/80056/using-slides-in-math-classroom/80282#80282 Answer by Toby Bartels for Using slides in math classroom Toby Bartels 2011-11-07T09:09:20Z 2011-11-07T09:09:20Z <p>Like Terry Tao, I find the transience of slides to be a problem. This is one reason why I stopped using slides as such and began using a single continuous-scroll page for each topic. I lecture from the bottom of the page, so students who are behind can still see the top. (I'm also one of those people who mixes the projector and the board, with bullet points and formulas on the projector and worked-out examples on the board, so I don't scroll down the page very quickly. Fortunately I work in a facility where the lighting allows this.)</p> http://mathoverflow.net/questions/72217/sober-except-not-t-0 Sober except not $T_0$? Toby Bartels 2011-08-06T01:05:05Z 2011-10-22T19:22:12Z <p>tl;dr: Is there an accepted or proposed term for a topological space whose $T_0$ quotient is sober?</p> <p>The condition that a topological space be sober (and therefore equivalent to a locale) may be broken into two parts:</p> <ul> <li>there are enough points,</li> <li>and there aren't too many.</li> </ul> <p>The condition that there are enough points is somewhat complicated: we note that every point gives rise to a completely prime filter in the lattice of open subsets, and require that every such filter arise in this way. I don't know any simpler way to say that, and I don't know any simple term for it.</p> <p>The condition that there aren't too many points may be treated similarly: if two points give rise to the same filter, then they must be equal. But this can be more simply stated: if two points have the same (open) neighbourhoods, then they are equal. And in this guise, the condition is well known: the $T_0$ separation axiom.</p> <p>I'd like to know if there's an accepted or proposed term for the first condition, of having enough points, simpler than "with a sober space as $T_0$ quotient". (The $T_0$ quotient is the quotient space found by identifying points with the same neighbourhoods, and it's easy to prove that this is equivalent.) Another obvious possibility is "with enough points", but this seems rather vague without context. (It could also lead to confusion, as it's not quite symmetric to locales; a locale with enough points is equivalent to a topological space, while a topological space with enough points is only equivalent to a locale if it's also $T_0$.)</p> <p>Does anybody know terminology for this?</p> http://mathoverflow.net/questions/35681/terminology-for-topological-base-closed-under-intersection Terminology for topological base closed under intersection? Toby Bartels 2010-08-15T19:30:42Z 2011-10-08T19:16:59Z <p>Is there an established or well justified terminology for a topological base that is closed under finitary intersections?</p> <p>As motivation, recall these conditions on a collection of subsets of a given set:</p> <ol> <li>closed under finitary intersections and arbitrary unions,</li> <li>closed under finitary intersections,</li> <li>filtered downwards,</li> <li>arbitrary.</li> </ol> <p>Anything that satisfies one condition satisfies any later condition; conversely, anything that satisfies one condition generates something that satisfies any earlier condition. I know names for (1,3,4): ‘topology’, 'topological base' (or ‘base for a topology’), and ‘topological subbase’ (at least when thought of in this context). So I'm asking for a name for (2). And one reason that this is interesting is that the obvious way to generate (3) from (4) already gives you (2), so it really does come up.</p> http://mathoverflow.net/questions/72217/sober-except-not-t-0/77552#77552 Answer by Toby Bartels for Sober except not $T_0$? Toby Bartels 2011-10-08T19:05:14Z 2011-10-08T19:05:14Z <p>Well, I've decided to go ahead and use ‘with enough points’. There are a lot of reasons to restrict to $T_0$ spaces, over and above reasons to restrict to sober spaces, and at least within that context having enough points is perfectly symmetric between topological spaces and locales.</p> http://mathoverflow.net/questions/74938/foundations-existence-of-uncountable-ordinals Foundations: Existence of uncountable ordinals. Toby Bartels 2011-09-08T21:03:09Z 2011-09-08T21:37:18Z <p>This isn't really a research question, but at least it's research-level mathematics. I'm talking with some other people about the first uncountable ordinal, and I want some facts to inform this discussion. Specifically, what useful or interesting foundations of mathematics do or don't allow one to prove the existence of an uncountable ordinal?</p> <p>If you don't have a better interpretation, then for "useful", you can probably take "capable of encoding most if not all rigorous applied mathematics"; for "interesting", you can probably take "popular for study by researchers in foundations". For "existence of an uncountable ordinal", you could take "existence of a well-ordered uncountable set", "existence of a set whose elements are precisely the countable ordinals", etc.</p> <p>Hopefully there is a body of known results or obvious corollaries of such, since it could be a matter of some work to apply this question to foundational system X, and I don't expect anybody to do that.</p> http://mathoverflow.net/questions/71816/w-completion-of-a-c-algebra W*-completion of a C*-algebra? Toby Bartels 2011-08-01T18:10:22Z 2011-08-01T21:35:33Z <p>tl;dr: Is there such a thing as a W*-completion of a C*-algebra, and if so, where can I read about it?</p> <p>I'm wondering about the relationship between (abstract) C*-algebras and W*-algebras. On the one hand, every W*-algebra is a C*-algebra. On the other hand, it seems to me that it should be possible to complete any C*-algebra to a W*-algebra. (Categorially, this would be a reflection.) In the case of commutative algebras, I even think that I know how this works: every commutative C*-algebra is the algebra of continuous functions on some compact Hausdorff space, and we extend this to the W*-algebra of essentially bounded Borel-measurable functions on the space (considered up to equality almost everywhere).</p> <p>So, is this correct? Does it work for noncommutative algebras as well? Is there a good algebro-analytic (without passing through topology) description of this? Is there a good reference, especially online?</p> <p>Also, in the commutative case, it seems that every state (positive normal linear functional) on a C*-algebra extends uniquely to its W*-completion, so they have the same space of states. Is this correct? Does it extend to the noncommutative case?</p> <p>Another question is how this relates to concrete algebras (those given as algebras of operators on some Hilbert space). One way to complete a C*-algebra would be to pick a concrete representation and take its weak closure (or double commutant). But I expect that this will depend on the representation chosen. (And my analysis is bad enough that I can't check this for even the commutative case.)</p> <p>I'd appreciate any help even for the main question, never mind this stuff about states and representations!</p> http://mathoverflow.net/questions/58193/leibnizian-calculus-textbook/60506#60506 Answer by Toby Bartels for Leibnizian calculus textbook Toby Bartels 2011-04-03T23:57:11Z 2011-04-04T21:17:52Z <p>This approach is suggested by Tevian Dray and Corinne Manogue in their program of <a href="http://www.physics.orst.edu/bridge/" rel="nofollow">Bridging the Vector Calculus Gap</a>. They focus on multivariable calculus and differential forms, but they <a href="http://www.physics.orst.edu/bridge/papers/CMJdifferentials.pdf" rel="nofollow">discuss single-variable calculus</a> (pdf) once. Unfortunately, they don't seem to have a textbook for that.</p> http://mathoverflow.net/questions/40082/why-do-we-teach-calculus-students-the-derivative-as-a-limit/60505#60505 Answer by Toby Bartels for Why do we teach calculus students the derivative as a limit? Toby Bartels 2011-04-03T23:23:22Z 2011-04-03T23:23:22Z <p>I'm interested in the differentials-based approach advocated by Dray and Manogue at <a href="http://www.physics.orst.edu/bridge/" rel="nofollow">Bridging the Vector Calculus Gap</a>. This is for multivariable calculus, but they do <a href="http://www.physics.orst.edu/bridge/papers/CMJdifferentials.pdf" rel="nofollow">discuss the one-variable version</a> (pdf). As they mention, people have <em>reviewed</em> calculus (especially for science courses) in these terms, but has anybody lately <em>taught</em> it this way?</p> <p>(Also, the theory behind this approach is a little unclear beyond the first derivative, which is what led me to <a href="http://mathoverflow.net/questions/60474/is-there-a-convenient-differential-calculus-for-cojets" rel="nofollow">this question</a>.)</p> http://mathoverflow.net/questions/50143/cardinal-numbers-vs-collections-of-cardinal-numbers Cardinal numbers vs collections of cardinal numbers Toby Bartels 2010-12-22T09:14:08Z 2011-01-03T10:48:02Z <p>Is there any literature, especially when doing mathematics without the axiom of choice, that discusses using collections of cardinal numbers in place of individual cardinal numbers, when discussing cardinal numbers with certain properties?</p> <p>That's a little vague, and I will presently give a few motivational examples that should help to clarify what I'm thinking of. But first, a quick definition of the translation between cardinal numbers and collections of cardinal numbers assuming the axiom of choice: If <i>K</i> is a cardinal number, then {<i>L</i> | <i>L</i>&nbsp;&lt; <i>K</i>} is a collection of cardinals which is small and downward-closed. Conversely, if <i>C</i> is a collection of cardinals which is small and downward-closed, then there exists a unique cardinal <i>K</i> (to wit, the smallest cardinal that does not belong to <i>C</i>) such that <i>C</i> = {<i>L</i> | <i>L</i>&nbsp;&lt; <i>K</i>}. So the correspondence between these two perspectives is straightforward —<em>if</em> we assume choice.</p> <p>Now I'll consider some types of cardinal numbers. A <em>weak limit cardinal</em> is an infinite cardinal <i>K</i> such that <i>L</i><sup>+</sup>&nbsp;&lt; <i>K</i> whenever <i>L</i>&nbsp;&lt; <i>K</i>. A <em>strong limit cardinal</em> is an infinite cardinal <i>K</i> such that 2<sup><i>L</i></sup>&nbsp;&lt; <i>K</i> whenever <i>L</i>&nbsp;&lt; <i>K</i>. A <em>regular cardinal</em> is an infinite cardinal <i>K</i> such that &Sigma;<sub><i>i</i>&nbsp;&isin;&nbsp;<i>I</i></sub> <i>L</i><sub><i>i</i></sub>&nbsp;&lt; <i>K</i> whenever |<i>I</i>|&nbsp;&lt; <i>K</i> and each <i>L</i><sub><i>i</i></sub>&nbsp;&lt; <i>K</i>. An <em>inaccessible cardinal</em> is an uncountable regular limit cardinal. A <em>weakly compact cardinal</em> is an inaccessible cardinal <i>K</i> such that the height of a tree is less than <i>K</i> whenever every level has width less than <i>K</i> and every branch has length less than <i>K</i> (the tree property). Etc.</p> <p>Although these are all properties of an individual cardinal number <i>K</i>, they all refer to that cardinal only through the collection of smaller cardinal numbers. (The exceptions are the adjectives ‘infinite’ and ‘uncountable’, which are really only there to rule out trivial cases.)</p> <p>If you don't assume the axiom of choice, then it's easy to still consider these conditions on a small, downward-closed collection of cardinals, but now this is more general than a collection of the form {<i>L</i> | <i>L</i>&nbsp;&lt; <i>K</i>}. If you go beyond doubting the axiom of choice and drop the law of excluded middle (so doing constructive mathematics, in a moderate sense), you can get more flexibility by messing with the interpretation of ‘downward-closed’; for example, the collection {0, 1, 2, …} of all finite (in the strictest sense) cardinal numbers is closed under taking decidable sub-cardinals but not arbitrary sub-cardinals and so gives a ‘regular’ collection of cardinals which is different from {<i>L</i> | <i>L</i>&nbsp;&lt; ℵ<sub>0</sub>}.</p> <p>I'm interested in understanding what regular cardinals and inaccessible cardinals should be in constructive mathematics. (Bigger than that, I don't really even understand them in classical mathematics.) It seems to me that they have to be collections of cardinals rather than individual cardinal numbers. But this perspective already makes sense in classical mathematics, especially without the axiom of choice. Has anybody studied this?</p> http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130928#130928 Comment by Toby Bartels Toby Bartels 2013-05-18T22:29:16Z 2013-05-18T22:29:16Z I learnt the epsilon-delta definition of limits in 1990 in the regular calculus class at a regular high school in Nebraska. (I can't recall if we learnt the definition of the Riemann integral or did all of the proofs, but these were in the book.) I didn't learn that this was unusual until years later. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130907#130907 Comment by Toby Bartels Toby Bartels 2013-05-18T22:24:59Z 2013-05-18T22:24:59Z Another way to see the uses of choice in the theorem is to use nonstandard analysis. This works just like ultrafilters, except that you talk about a hyperpoint in the infinitesimal neighbourhood of a standard point <i>p</i> instead of an ultrafilter that converges to <i>p</i>. Like the ultrafilter approach to topology, nonstandard analysis simply doesn't work without the ultrafilter lemma, so that's enough for the Hausdorff case; but the non-Hausdorff case requires an arbitrary choice of <i>p</i> in each space, just as with ultrafilters. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130905#130905 Comment by Toby Bartels Toby Bartels 2013-05-18T22:03:49Z 2013-05-18T22:03:49Z I had learnt the ‘first’ and ‘second’ fundamental theorems under the opposite names, and I'd almost forgotten that I once read a calculus textbook that had them switched, until I read your post now that also has them switched (switched compared to how I learnt them, I mean). I wonder how that came about? http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/131060#131060 Comment by Toby Bartels Toby Bartels 2013-05-18T22:00:06Z 2013-05-18T22:00:06Z Thanks! I never understood the Sylow theorems either (except for purposes of exams, of course), but now that you've explained how to look at them, I think that I'll try again. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130892#130892 Comment by Toby Bartels Toby Bartels 2013-05-18T21:58:01Z 2013-05-18T21:58:01Z To prove the FTC, you first need a definition of the integral, and some textbooks don't even have that! http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130907#130907 Comment by Toby Bartels Toby Bartels 2013-05-18T16:58:33Z 2013-05-18T16:58:33Z You don't need the ultrafilter lemma to set up the basic theory of topology via nets or filters, but you need it to set up the basic theory of topology via universal nets or ultrafilters. And the proof is much cleaner using universal nets or ultrafilters, since you merely need to show that each of either converges. (If instead you try to show that each net has a convergent subnet or that each proper filter has a convergent proper refinement, then you find yourself needing choice again.) And yes, convergence of all nets and proper filters (if convergent at all) is unique in a Hausdorff space http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Toby Bartels Toby Bartels 2013-05-18T16:48:08Z 2013-05-18T16:48:08Z The dot-product proof relies on the equivalence between the geometric meaning of the dot product (which is obviously rotation invariant) and the algebraic formula for it (which gives Dustin's result). But proving this equivalence, which relies on proving that the geometric dot product is distributive, is as complicated as proving the Pythagorean Theorem directly: <a href="https://upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Dot_product_distributive_law.svg/2000px-Dot_product_distributive_law.svg.png" rel="nofollow">upload.wikimedia.org/wikipedia/commons/thumb/a/aa/&hellip;</a> http://mathoverflow.net/questions/9269/category-of-categories-as-a-foundation-of-mathematics/9270#9270 Comment by Toby Bartels Toby Bartels 2013-05-09T17:55:57Z 2013-05-09T17:55:57Z Andrew's link again doesn't work; use this one: <a href="http://ncatlab.org/nlab/show/foundation+of+mathematics#CFCT" rel="nofollow">ncatlab.org/nlab/show/&hellip;</a> In general, if you wish to link to a section within an nLab page, you should first (if it hasn't already been done) edit the page to give the section a permanent title. Otherwise, your link will stop working when the page is edited to add, delete, or rearrange sections. http://mathoverflow.net/questions/129036/counterintuitive-consequences-of-the-axiom-of-determinacy/129038#129038 Comment by Toby Bartels Toby Bartels 2013-05-01T19:05:39Z 2013-05-01T19:05:39Z This really highlights Douglas Zare's point that what's intuitive depends on whether AC is intuitive. I find (1) and (2) quite intuitive, and I gather that many more people did around a hundred years ago. (3) also seems reasonable, although I could imagine its going either way. ((4) and (5) are a little outside where I feel good intuitions.) http://mathoverflow.net/questions/128429/when-is-the-class-of-functions-between-sets-a-set Comment by Toby Bartels Toby Bartels 2013-04-23T16:20:01Z 2013-04-23T16:20:01Z Just for the record, in reply to one of David's comments above, there are set theories in which the class of functions between two given sets is not a set; ZF, ETCS, and SEAR all have strongly predicative versions, in which the power set axiom is removed and (if replaced at all) replaced by something too weak to give function sets. (Of course, if we have classical logic, then function sets are equivalent to power sets, so even removing power set and replacing it with function sets is an option only when using intutitionistic logic.) http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/126846#126846 Comment by Toby Bartels Toby Bartels 2013-04-17T17:56:35Z 2013-04-17T17:56:35Z Yes, the correct chain rule for second derivatives is $d^2y/dx^2 = (d^2y/du^2)(du/dx)^2 + (dy/du)(d^2u/dx^2)$. But unlike the chain rule for first derivatives, you can't derive this by treating $d^2y$ and the rest as if they were elements of an infinitesimal-enriched continuum obeying the ordinary rules of algebra. At the very least, this is annoying (and I have found it so since since high school); but more than that, it suggests that the $dx^2$ that appears in $d^2y/dx^2$ is not really the square of $dx$ in such a continuum. http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/126846#126846 Comment by Toby Bartels Toby Bartels 2013-04-15T03:28:32Z 2013-04-15T03:28:32Z @Vladimir: This does not contradict your point (or indeed any of the text of your latest comment), but one must be careful with $d^2y$, $dx^2$, and the like. Second derivatives in Leibniz's notation don't work as well as first derivatives, because the chain rule $d^2y/dx^2 = (d^2y/du^2) (du/dx)^2$ is false. I would prefer to write $\partial^2y/\partial{x}^2$ myself, on the grounds this is the coefficient on $dx^2$ in an expansion of $d^2y$ (in $d^2x$ and $dx^2$), analogous to the coefficients that are partial derivatives, rather than the ratio of $d^2y$ to $dx^2$. http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/125031#125031 Comment by Toby Bartels Toby Bartels 2013-04-07T19:30:05Z 2013-04-07T19:30:05Z OK, if I write anything up carefully, I'll put it at <a href="http://ncatlab.org/nlab/show/Cauchy+sum+theorem" rel="nofollow">ncatlab.org/nlab/show/Cauchy+sum+theorem</a> (or at least somewhere easily available from there) and put a note here in case anybody else is still following along later, but otherwise I will keep it to private email. http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/125031#125031 Comment by Toby Bartels Toby Bartels 2013-04-04T16:30:20Z 2013-04-04T16:30:20Z And I still think that (6.2) is missing the hypothesis that $n$ is infinite; it must be implicit. The way that it's phrased, it really makes it look like $r_n(x)$ is only required to be infinitesimal when $x$ is infinitesimal, which of course is not the case; but it is the case that $r_n(x)$ is only required to be infinitesimal when $n$ is infinite. So at best, it is confusingly written, although now that you've explained why you wrote it that way, I can understand it. (But earlier I really thought that you had meant to write ‘$n$ infinite’ and had written ‘$x$ infinitesimal’ by mistake.) http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/125031#125031 Comment by Toby Bartels Toby Bartels 2013-04-04T16:23:17Z 2013-04-04T16:23:17Z You offered that formula (apparently on page 57 of the Springer version) as an explanation of the ‘straightforward formalisation’ in Robinson's nonstandard analysis of Cauchy's $x = 1/n$ counterexample to the claim that $\sum_{k=1}^n \frac{sin(k x)}{k}$ converges everywhere, which I had said was difficult to do. And although I put forth a possible formalisation in a later comment, I don't think that it's straightforward, and (6.2) doesn't help.