User david milovich - MathOverflow

Leibnizian calculus textbook

2011-03-11T19:32:00Z

Where can I find a calculus textbook that emphasizes differentials? Is there such a book that I could realistically require my calculus students to use?

I want a textbook that supports me when I tell my students something like:

$\Delta((x^2+1)^5)\approx5(x^2+1)^4\Delta(x^2+1)\approx5(x^2+1)^4(2x\Delta x)$

$d((x^2+1)^5)=5(x^2+1)^4d(x^2+1)=5(x^2+1)^4(2x\ dx)$

Or:

$\Sigma_{k=1}^n 3x_k^2\Delta x_k\approx\Sigma_{k=1}^n\Delta(x_k^3)=x_n^3-x_1^3$

$\int_{x=0}^{x=4}3x^2\ dx=\int_{x=0}^{x=4}d(x^3)=4^3-0^3=64$

Perhaps I could write this book someday, but it'd be a lot easier for me if my students and I could just buy and/or download a book that takes this approach without neglecting to provide a cornucopia of exercises, examples, and applications similar to what's available in today's most popular calculus textbooks.

Answer by David Milovich for Hyperreal finitely-additive measure on [0,1) assigning $b-a$ to $[a,b)$ or $(a,b]$ and infinitesimals to singletons

2012-09-27T14:57:01Z

Yes, by compactness.

Let $R$ denote your favorite hyperreal ordered field and let $\delta\in R$ be a positive infinitesimal. Let $\mathcal{E}$ denote the set of all (standard) finite Boolean subalgebras of $\mathcal{P}([0,1))$. For every $A\in\mathcal{E}$, let $\lambda_A(I)$ be the (exact) length of $I$ for all half-open intervals $I\in A$; for all open or closed intervals $I\in A$, respectively subtract or add $\delta$ to the length of $I$ to define $\lambda_A(I)$; let $\lambda_A(S)=\delta$ for all singletons $S\in A$. Extend $\lambda_A$ to a probability measure $\mu_A$ on $A$.

(Specifically, for each minimal finite union of intervals $F\in A$, let the connected components of $F$ be $[a_0,b_0),\ldots,[a_k,b_k)$ with $p$ elements of $\{a_i,b_i:i\leq k\}$ added and $q$ removed. Partition $F$ into its atomic subsets $H_0,\ldots,H_n$. Choose a positive $\mu_A(H_i)\in R$ for each $i$, such that $\sum_{i\leq n}\mu_A(H_i)=(p-q)\delta+\sum_{i\leq k}(b_i-a_i)$. Now extend $\mu_A$ from the atoms to all of $A$.)

Let $U$ be a fine ultrafilter on $\mathcal{E}$. ("Fine" means that $\{B\in\mathcal{E}:A\subseteq B\}\in U$ for all $A\in\mathcal{E}$.) The ultraproduct measure $\mu_U$ is $R^U$-valued and has the two properties you seek.

Answer by David Milovich for Is there a statement equivalent to a sentence admitting $(\alpha^{+n},\alpha)$?

2012-05-23T22:32:45Z

Letting $\alpha$ be an infinite cardinal and $3\leq n\lt\omega$, I think Peter Komjath's proposed statement $2^\alpha\geq\alpha^{+n}$ is the simplest and most natural equivalent of a first-order sentence admitting $(\alpha^{+n},\alpha)$: just let $\sigma$ say a binary relation is extensional with domain given by a predicate and range equal to the universe. However, you asked for a GCH example in your comment, so I suggest the following statement, that a generalized kind of Kurepa family exists.

$(*_{\alpha,n})$ says there exists $\mathcal{X}\subset\mathcal{P}(\alpha^{+n-1})$ of size $\alpha^{+n}$ such that $|\{X\cap A: X\in\mathcal{X}\}|\leq\alpha$ for all $A\subset\alpha^{+n-1}$ of size $\leq\alpha$.

To get our first-order sentence $\sigma$, we use the fact that $[\alpha^{+n-1}]^{\leq\alpha}$ has nice cofinal subsets, and the equivalence of $(*_{\alpha,n})$ to its formal weakenings where $A$ is quantified over a cofinal set. Our first-order sentence $\sigma$ says that

the universe is $L_n$,
$(L_0,\lt_0),\ldots,(L_n,\lt_n)$ are linear orders,
$f_i(x,\bullet)\colon L_i \rightarrow \{ y : y\leq_{i+1} x\}$ is always onto,
$g(x,\bullet)\colon L_{n-1}\rightarrow \{0,1\}$ for all $x\in L_n$,
$g(x,\bullet)\not=g(y,\bullet)$ for all $x\not= y$,
$h(x_{n-1},\ldots,x_0,\bullet)\colon L_0\rightarrow 2$ for all $x\in\prod_{i=0}^{n-1} L_i$, and
every $g(x_n,f_{n-2}(x_{n-1},f_{n-3}(x_{n-2},\cdots,f_0(x_1,\bullet)\cdots)))$ equals some $h(x_{n-1},\ldots,x_0,\bullet)$.

$(*_{\alpha,n})$ holds iff $\sigma$ has a model with size $\alpha^{+n}$ with $L_0$ of size $\alpha$.

Jensen proved that something stronger than $(*_{\alpha,n})$ holds if $V=L$.

KH($\kappa,\lambda$) says that there exists $\mathcal{F}\subset\mathcal{P}(\kappa)$ of size $\kappa^+$ such that $|\{X\cap A:X\in\mathcal{F}\}|\leq|A|+\aleph_0$ for all $A\subset\kappa$ of size $\lt\lambda$.

Clearly, KH($\alpha^{+n-1},\alpha^+$) implies $(*_{\alpha,n})$.

Jensen proved that if $V=L$, then KH($\kappa,\lambda$) holds for all regular uncountable $\kappa$ and all uncountable $\lambda<\kappa$. In particular, $(*_{\alpha,n})$ is always true in $L$. Jensen proved that $V=L$ also implies KH($\kappa,\kappa$) for all regular uncountable cardinals that are not ineffable. Jensen and Kunen proved that if $\kappa$ is ineffable, then KH($\kappa,\kappa$) fails. Devlin's exposition of the proofs is available here; see the chapter "Ineffable cardinals and the generalised Kurepa hypothesis."

On the other hand, $(*_{\alpha,n})$ is directly refuted by the Chang conjecture variant $(\alpha^{+n},\alpha^{+n-1})\twoheadrightarrow(\alpha^+,\alpha)$. For regular $\alpha$, we can force this Chang conjecture easily using the method of Levinski-Magidor-Shelah (MR1045371). Assume GCH and epsilon more than a huge embedding: $j\colon V\prec N\supset {}^{\lambda^+}N$ where $\kappa=cp(j)$ and $\lambda=j(\kappa)$. Since $N$ knows that $j''\mathfrak{A}\prec j(\mathfrak{A})$ for all structures $\mathfrak{A}$ of the form $(H(\lambda^+),\in,P)$, we have $(\lambda^+,\lambda)\twoheadrightarrow (\kappa^+,\kappa)$ in $V$ by elementarity of $j$. The two-step iteration $\mathbb{P}=\mathrm{Coll}(\alpha,\kappa)*\mathrm{Coll}(\kappa^{+n-2},\lt\lambda)$ preserves GCH and forces $(\alpha^{+n},\alpha^{+n-1})\twoheadrightarrow(\alpha^+,\alpha)$. The key point is that because there are only $\lambda$-many nice $\mathbb{P}$-names for elements of $\lambda$, this set of names Chang-transfers to a $\kappa$-sized set of names for elements of $\lambda$, implying our desired transfer from $\alpha^{+n-1}$ to $\alpha$ in the generic extension $V[G]$: if in $V$ we have

$\mathbb{P}\in M\prec(H(\lambda^+),\in,\dot{P})$,
$|M|=\kappa^+$, and
$|M\cap\lambda|=\kappa$,

then in $V[G]$ we have

$M[G]\prec(H(\lambda^+),\in,\dot{P}_G)=(H(\alpha^{+n}),\in,\dot{P}_G)$,
$|M[G]|=\kappa^+=\alpha^+$, and
$|M[G]\cap\alpha^{+n-1}|=|M[G]\cap\lambda|=|\kappa|=\alpha$.

Answer by David Milovich for Sizes of bases of vector spaces without the axiom of choice

2012-04-06T21:15:50Z

Yes, ZF+BPIT implies that vector space dimension is well-defined. [Edit: some Googling shows that James Halpern gave the same answer back in the 1960s.]

Working in ZF+BPIT, fix a field $F$ and an $F$-vector space $V$ and bases $A$ and $B$ of V. That is, each element of $V$ is a unique $F$-linear combination of elements of $A$; likewise for $B$. For each $a\in A$, let $S_a$ be the minimal subset of $B$ such that $a$ is spanned by $S_a$. Each $S_a$ is finite; give it the discrete topology. Let $X=\prod_{a\in A}S_a$, which is nonempty by BPIT (and is compact Hausdorff). By Schroeder-Bernstein, it suffices to show that some $f\in X$ is injective. By compactness, it suffices to show that for every finite subset $K$ of $A$, there is an $f\in X$ that is injective on $K$. Since each $\prod_{a\in A\setminus K}S_a$ is nonempty by BPIT, it suffices to show that there is an injection in every $\prod_{a\in K}S_a$. That is a nice little linear algebra exercise you can solve in ZF using the finite case of Hall's marriage theorem.

Cech cohomology and set theory

2012-03-26T22:57:39Z

Can anyone point me to places in the literature where modern set theory has been applied to say something about the Cech cohomology of connected non-metrizable compacta? I'm looking for something deeper than, e.g., the observation that the Cech cohomology detects that the long loop is a loop.

Most of my work is on non-metrizable compacta, and, most of time, whenever I prove something in the zero-dimensional case, it almost effortlessly generalizes to all compacta. By Stone duality, I'm "really" working on uncountable boolean algebras most of the time. So, on days when I'm particularly interested in connectedness properties, I feel that I'm not asking enough of the right questions.

It seems that just as all questions about zero-dimensional compacta are equivalent to questions about boolean algebras, sometimes interesting and difficult infinitary combinatorial questions, many questions about connected non-metrizable compacta should reduce to questions their Cech cohomologies, presumably including some interesting and difficult infinitary combinatorial questions about uncountable abelian groups. I'm curious as to whether it would be fruitful to pursue this analogy further, and therefore curious as to how the analogy has been pursued in the past.

Answer by David Milovich for Ordered sum of posets

2012-02-15T23:28:07Z

I suggest first restricting to, say, suborders $P$ of the plane $\mathbb{R}^2$(with the product ordering where $(x_1,y_1)\leq(x_2,y_2)$ iff $x_1\leq x_2$ and $y_1\leq y_2$). See if you can find a satisfactory decomposition there, focusing on simple counterexamples to a straight generalization of Schoenflies' Theorem.

For example, let $$P=(\{0\}\times[0,1])\cup\bigcup_{n=0}^\infty(\{2^{-n} \}\times\{m\cdot 2^{-n}:m=1,\ldots,2^n\}).$$ Suppose you had a decomposition of $P$ into an ordered sum where the index has some property D reminiscent of denseness and each summand has property S reminiscent of scatteredness. If the linear order $[0,1]$ lacks property S, then you must decompose $P$ into singletons. Hence, either $[0,1]$ has property S or $P$ has property D. (Neither of these options looks appealing to me.)

Answer by David Milovich for The consistency of Martin's Axiom

2012-01-26T22:50:48Z

I like to think about the matter topologically. Let $X=MF(P)$ be the set of all maximal filters of $P$. Give $X$ the topology generated by the sets $N_p=\{U\in X: p\in U\}$. Declare $B(P)$ to be the algebra of regular open subsets of $X$ (that is, sets equal to the interior of their closures).

Given a regular open set $R$, pick a maximal family $A$ of pairwise disjoint subsets of $R$ of the form $N_p$ for some $p\in P$. The union of $A$ is dense in $R$, so $R$ is the smallest regular open set containing this union. In other words, every element of $B(P)$ is the sum of an antichain from $P$.

Answer by David Milovich for Non standard algebraic geometry: shadows of varieties

2011-11-19T05:36:27Z

The answer is "yes" if $V$ is the projective zero-set $Z(f)$ of a single homogeneous polynomial $f$. Perhaps an expert in several complex variables will provide a more general answer.

Without loss of generality, every coefficient of $f$ is finite (i.e., they all have modulus less than some standard $R$) and at least one coefficient is not infinitesimal. Let $st(f)$ be the "standard part" of $f$ obtained by replacing every coefficient with its standard part. Let $W$ be the shadow of $V$. I claim that $Z(st(f))=W$ where $Z(st(f))$ is the set of standard projective zeroes of $st(f)$.

The easy half, which generalizes to all algebraic varieties, is that $W\subseteq Z(st(f))$. Let $x\approx y$ iff $\lvert x-y\rvert$ is infinitesimal. If $[\vec w]\in W$ and (without loss of generality) all coordinates of $\vec w$ are finite, then $f(w)\approx 0$ by continuity of $f$, so $st(f)(\vec w)\approx 0$, so $st(f)(\vec w)=0$ because $st(f)(\vec w)$ is standard.

To prove the other half, suppose $[\vec u]\in Z(st(f))$. Choose a standard unit vector $\vec e$ such that $h(z)=st(f)(\vec u+z\vec e)$ is not constant on any open disc $D_r=\{z\in\mathbb{K}:\lvert z\rvert< r\}$ where $r>0$ is standard. Since $h$ has only finitely many roots, we may choose an arbitrarily small standard $r$ such that $h$ has no root on $\partial D_r$. Let $M$ be the (necessarily standard) minimum modulus of the $h$-image of $\partial D_r$. Setting $p(z)=f(\vec u+z\vec e)$, we have $p(z)\approx h(z)$ for all $z\in\mathbb{K}$. Hence, $\lvert p(z)-h(z)\rvert< M\leq\lvert h(z)\rvert$ for all $z\in\partial D_r$. By Rouche's Theorem, $h$ and $p$ have the same number of roots (counting multiplicities) in $D_r$. By overspill, $h$ and $p$ have the same number of roots in $D_\delta$ for some positive infinitesimal $\delta$. In particular, $p$ has an infinitesimal root, so $f$ has a root infinitely close to $\vec u$.

Answer by David Milovich for Direct construction of the integers

2011-11-07T17:16:19Z

Informally speaking, taking the limit of two's complement as the number of bits goes to $\infty$, the integers are just the eventually constant binary sequences (which are naturally represented by finite binary sequences). For this to work, said sequences must start with the least significant bit, i.e., $1001011\overline{0}$ is interpreted as $2^0+2^3+2^5+2^6$ and $1001010\overline{1}$ is interpreted as $2^0+2^3+2^5-2^7$. The arithmetic and ordering of these strings is natural (and efficient for microprocessors when we restrict from $\mathbb{Z}$ to, say, $\{-2^{63},\ldots,2^{63}-1\}$).

The above can be reinterpreted as the following less direct construction. If $R$ is the inverse limit of rings $\lim_{\infty\leftarrow n}\mathbb{Z}/2^n\mathbb{Z}$, then the diagonal map $\Delta\colon\mathbb{Z}\rightarrow R$ given by $m\mapsto \lim_{\infty\leftarrow n}(m\mod 2^n)$ is an injective ring homomorphism. [Edit: The image is characterized as the set of $\vec x\in R$ for which the truth value of $x(n+1)=x(n)$ is eventually constant.] Moreover, the ordering of $\mathbb{Z}$ is coded via $m\geq 0\Leftrightarrow(m\mod 2^n: n\in\mathbb{N})$ is eventually constant.

Update: I couldn't resist the temptation to write a functional programming implementation.

Answer by David Milovich for Topology generated by the collection of open sets

2011-10-12T17:54:45Z

There is a Hausdorff space with all points $G_\delta$ but without a condensation to a first countable Hausdorff space.

Let $X$ be the space of ultrafilters on $\mathbb{N}$ with the topology generated by sets of the form $\{\mathcal{p}\}\cup A$ where $A\in p\in X$ and I have committed the common abuse of notation that identifies $\mathbb{N}$ with the principal ultrafilters on $\mathbb{N}$. This makes $X$ Hausdorff (but not regular). Moreover, every point is $G_\delta$ because $X$ is $T_1$ and every point has a countable neighborhood. Crucially, $X$ has cardinality $2^{2^{\aleph_0}}$.

Now suppose $Y$ is the set $X$ with a coarser first countable Hausdorff topology. For each $p\in X$, let $(U_n(p):n\in\mathbb{N})$ be a base of $Y$-neighborhoods of $p$; let $V_n(p)=U_n(p)\cap\mathbb{N}$. Because $Y$ is Hausdorff and $\mathbb{N}$ is $X$-dense, we have $\vec{V}(p)\not=\vec{V}(q)$ for all distinct $p,q$. Therefore, $X$ injects into $\mathcal{P}(\mathbb{N})^{\mathbb{N}}$. But $X$ is too big, so actually there is no such $Y$.

Answer by David Milovich for Taking "Zooming in on a point of a graph" seriously.

2011-10-05T17:19:54Z

As AmbroseH commented, Keisler's book takes this pedagogical approach. The catch is that his book is old, out-of-print, and uses the infintesimal approach. He talks of "infinite microscopes" and "infinite telescopes" (the latter for horizontal and vertical asymptotes of curves) and has illustrations (sadly without color or animation). You can download the book from his website (Creative Commons License).

You can find more recent notes (not full textbooks yet) that sketch how to teach calculus using various flavors of infinitesimals (without sacrificing a rigorous mathematical foundation):

Keisler uses Robinson-style nonstandard analysis. A more recent variation on this is relativized nonstandard analysis. Instead of two "levels," standard and nonstandard, it's turtles all the way down. See these slides for an introduction and citations.
A very different flavor is smooth infinitesimal analysis, which uses nilsquare infinitesimals (and may use intuitionistic logic). Here is one exposition.
The idea of nilsquare infinitesimals generalizes to nilpotent infinitesimals. This is discussed here, for example.

Since you tagged your question algebraic geometry, my guess is that you'd prefer option 2 or 3.

Finally, you might find this relevant; there's an even older (but not out of print) calculus book that has wonderful pedagogy but implicitly throws around nilsquares without any pretense of rigor.

Answer by David Milovich for How misleading is it to regard $\frac{dy}{dx}$ as a fraction?

2011-08-23T17:37:21Z

What's most misleading about Leibnizian notation is its implicit context dependence. After you get over that hurdle, it will be easy to safely think of $dy/dx$ as a fraction.

In the context of $y=f(x)$, you think of $dx$ either as an arbitrary nonzero infinitesimal also called $\Delta x$---I did this, using Keisler's book last fall---or as a nonzero real $\Delta x$ small enough for whatever your accuracy you currently need. Either way, $dy$ is defined as $f'(x)dx$, where $f'(x)$ is defined as the usual limit of difference quotients $\Delta y/\Delta x$. Of course, in the $x=g(y)$ context, the meanings of $dx$ and $dy$ switch, as do the meanings of $\Delta x$ and $\Delta y$. In the $z=h(x,y)$ context, the meanings of $dx$, $dy$, $\Delta x$, and $\Delta y$ change yet again.

The "small enough, but not infinitely small" approach is what you'll find in standard calculus textbooks, with a section devoted to the distinction between $\Delta y$ and $dy$ (in the $y=f(x)$ context).

That said, this fall I'm planning to de-emphasize $dy/dx$ as much as I can get away with. Whether I use the little-o notation or not, I will push hard (with lots of numerical examples) on the $\Delta y=f'(x)\Delta x+o(\Delta x)$ definition of $f'(x)$, and how this makes the chain rule true but not trivial.

If $y=f(x)=x^2$ and $dx=\Delta x$ is small (but not infinitely small this time around), then $\Delta(x^2)$ equals $(x+\Delta x)^2-x^2$ equals $2x\Delta x+\Delta x^2$ equals $2x\Delta x+(\mathrm{small})\Delta x$, so $dy=2x\ dx$ and $f'(x)=2x$. In the context of $y=f(u)$ and $u=g(x)$, my presentation of the chain rule will just be that a first-order approximation of a first-order approximation is a first-order approximation:

\begin{align*} \Delta y&=f'(u)\Delta u+(\mathrm{small}_1)\Delta u\\ &=f'(u)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)+(\mathrm{small}_1)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)\\ &=f'(u)g'(x)\Delta x+(\mathrm{small})\Delta x \end{align*} No fractions here!

Consistency strengths related to the perfect set property

2011-07-21T21:49:05Z

I want a model of $\mathrm{MA}_{\sigma\mathrm{-centered}}+\neg\mathrm{CH}$ in which every set of reals in $L(\mathbb{R})$ has the perfect set property. In terms of consistency strength, it is known that I need at least an inaccessible: if $\mathrm{PSP}(L(\mathbb{R}))$, then $\omega_1$ is inaccessible in $L$. I haven't been able to find any other lower bounds on the consistency strength of $\mathrm{MA}+\neg\mathrm{CH}+\mathrm{PSP}(L(\mathbb{R}))$.

The best upper bound I can find is the fact, due to Woodin, that a measurable cardinal above infinitely many Woodin cardinals outright implies $\mathrm{Det}(L(\mathbb{R}))$. So, starting with these large cardinals in the ground, I can get what I want by forcing MA (using a "small" forcing).

My question is, do I really need such strong hypotheses? The ideal answer would be "this is known; the answer can be found in...". One the other hand, if you can tell me with confidence that it's an open problem, then at least I'll know that trying to solve it isn't a waste of time.

If it's easier to answer my question for projective sets, please do!

Back in 1964, Solovay proved that Levy-collapsing an inaccessible $\kappa$ to $\omega_1$ forces every set of reals definable by an omega-sequence of ordinals---this includes every set of reals in $L(\mathbb{R})$---to have the perfect set property. The catch is that the Solovay model also satisfies CH.

There's a 1989 JSL paper by Judah and Shelah (http://www.jstor.org/stable/2275017) that looks at the consistency strength of $\mathrm{MA}_{\sigma\mathrm{-centered}}+\neg\mathrm{CH}$ (and similar forcing axioms) in conjunction with various regularity properties for projective sets: Lebesgue measurability, the Baire property, and the Ramsey property. The perfect set property is (from my point of view) conspicuously absent.

Answer by David Milovich for Consistency strengths related to the perfect set property

2011-07-26T17:30:57Z

$\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}+\mathrm{PSP}(L(\mathbb{R}))$ is equiconsistent with a Mahlo cardinal.

Before Goldstern's comment, I had assumed the perfect set property was important enough to authors to mention if their theorems covered it. With that assumption falsified, I read more carefully, and determined that the Judah-Shelah paper I mentioned in the question had the answer all along. The proof of Lemma 1.1 shows that any forcing extension satisfying the hypotheses of the lemma actually has the same $L(\mathbb{R})$ as some Solovay model. Moreover, the proof of Theorem 3.1 actually builds from a Mahlo cardinal a forcing extension that satisfies $\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}$ and the hypotheses of Lemma 1.1. Therefore, a Mahlo cardinal suffices. A Mahlo cardinal is necessary because $\mathrm{PSP}(L(\mathbb{R}))$ is well-known to imply that $\omega_1^{L[x]}<\omega_1$ for all reals $x$, which in turn implies, by (ii)$\Rightarrow$(i) of 3.1, that if $\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}$ also holds, then a Mahlo cardinal is consistent with ZFC (specifically, $\omega_1$ is Mahlo in $L$ by the proof of (ii)$\Rightarrow$(i)).

(The hypotheses of Lemma 1.1 are that $\kappa$ is inaccessible, $\mathbb{P}$ satisfies the $\kappa$-chain condition, $\mathbb{P}$ forces $\kappa=\omega_1$, and, for every subset $Q$ of $\mathbb{P}$ of size less than $\kappa$, $Q$ extends to $P'\subseteq \mathbb{P}$ such that $|P'|<\kappa$ and the inclusion map completely embeds $P'$ into $\mathbb{P}$.)

Answer by David Milovich for Entropy of a measure

2011-07-22T05:13:59Z

I just wanted to say that topologically, your hand is forced: Tapio's supremum definition is the way to go.

Give the space $X$ of all maps from $\mathcal{P}(\mathbb{N})$ to $[0,1]$ the topology of pointwise convergence. The space $FAM$ of finitely additive probability measures on $\mathbb{N}$ is then a closed subspace of $X$. Let $FSM$ be the set of finitely supported measures in $FAM$. Let $FSA$ be the set of finite subalgebras of of $\mathcal{P}(\mathbb{N})$. Given $G\in FSA$ and $\mu\in FAM$, it is easy to (definably) choose $\mu_G\in FSM$ that agrees with $\mu$ on $G$: $\mu_G({\min(A)})=\mu(A)$ for every atom $A\in G$. In $FAM$, each point $\mu$ has a neighborhood base consisting of sets of the form $U(\mu,G,\varepsilon)$, which I use to denote the set of $\nu\in FAM$ that agree with $\mu$ on $G$ up to error $\varepsilon$. Therefore, $FSM$ is dense in $FAM$.

The Shannon entropy Ent is continuous on $FSM$, and Ent extends uniquely to a continuous map from $FAM$ to $[0,\infty]$ given by Tapio's $Ent(\mu)=\sup{Ent(\mu_G): G\in FSA}$ (using my notation). To see this, the important step is to check that $Ent(\mu_G)\leq Ent(\mu_H)$ if $G\subseteq H$.

Answer by David Milovich for Is there a least-fixed-point formulation of inaccessible cardinals?

2011-07-14T22:35:56Z

There is a monotone (but discontinuous) operation $F$ that does what you want in a way that is quite natural (in my opinion):

$F(\alpha)=(\omega+1)\cup\sup_{f\in{}^{<\alpha}\alpha}\left(\left(\sup_{\gamma\in\mathrm{ran}(f)}|P(f(\gamma))|\right)+1\right)$.

Answer by David Milovich for Nonstandard models of PA of large cardinal size

2011-01-12T17:26:28Z

See Section 4.2 of Harvey Friedman's book draft for the details of a combinatorial, non-metamathematical application of n-Mahlo-sized extensions of structures of the form $(\omega,<,0,1,+,f,g)$.

In this example, $f$ and $g$ are not allowed to grow fast enough for either to be chosen to be the multiplication function. More generally, I am not aware of any "interesting" number-theoretic applications of large cardinals. That said, perhaps you should ask Friedman about large-cardinal-sized models of PA.

Answer by David Milovich for Can Sierpinski's anisotropic bicolouring of the plane, assuming the continuum hypothesis (CH), be extended to three dimensions?

2011-01-09T00:36:55Z

Hamkins' theorem can be pushed down to 3 colors. Give $(x_1,x_2,x_3)$ color $i$ where $x_i$ is the least coordinate according to the $\omega_1$-type ordering. (Break ties by choosing the least possible $i$.) Every plane perpendicular to the $x_i$-axis will be colored $i$, except possibly on countably many lines. Every line parallel to the $x_i$-axis will colored $i$ on at most countably many points; the rest of the line will be monochromatic in another color. If we switch to spherical coordinates, then we get an analogous theorem involving spheres, cones, planes, rays, and circles.

Comment by David Milovich

2013-04-19T21:50:30Z

I could see myself someday using a book like this in the context of teaching graduate students or very strong undergrads. However, I was looking for a friendlier book for the purpose of teaching calculus with infinitesimals.

Comment by David Milovich

2012-09-27T20:27:59Z

@Gerald Edgar: Indeed, we can do much much more with the same technique, which boils down to replacing a single ultraproduct with an iterated ultraproduct to get more control. (In this case, a two-step iteration was sufficient.) To my mind, the most salient barriers are finite obstructions like the Banach-Tarski obstruction to congruency invariance in three dimensions. I don't see a finite obstruction to what you propose regarding Hausdorff measure.

Comment by David Milovich

2012-09-27T16:58:28Z

I just minorly corrected the answer: the parenthetical paragraph now correctly handles cases such as $[0,1/3)\cup[2/3,1)\in A$ but $[0,1/3)\not\in A$.

Comment by David Milovich

2012-08-28T20:22:02Z

@AliReza Start with the ring of integers. Well order the reals. At each stage, add the next real to your growing subring iff the resulting subring does not have 1/2 as an element.

Comment by David Milovich

2012-07-31T18:37:08Z

This answer is over a year old, but for the sake of future readers: This is false: "Let $X$ be a totally ordered set. Let $\omega^X$ be the set of functions $X \to \omega$ which are $0$ for almost all $x \in X$, ordered lexicographically. Then $\omega^X$ is well ordered." Just look in $\omega^\omega$: the characteristic functions of the singletons. The correct version requires $X$ to be well ordered and gives $\omega^X$ the reverse lexicographic order, i.e., compares functions at their greatest coordinate of disagreement.

Comment by David Milovich

2012-07-10T20:10:56Z

Yes, we agree. A more formal version of (7) would start with "for all $x_n,\ldots, x_1$, there exists $x_0$ such that..."

Comment by David Milovich

2012-06-10T19:31:04Z

Details: Let $\mathcal{U}$ be a free ultrafilter on $\omega$; let $X$ be the tree $\omega^{<\omega}$ with the topology generated by the set $\mathcal{B}$ consisting of all subsets $Y$ such that $Y$ has a unique root and $\forall y\in Y\ \forall^{\mathcal{U}}n\ \ y^\frown n\in Y$. Check that $X$ is $T_1$ and $\mathcal{B}$ is a clopen base, implying $X$ is $T_{3.5}$. By Konig's Lemma, if $A\subset X$ is infinite, then $A$ contains an infinite chain or an infinite set of the form $\\{s^\frown n:n\in I\\}$. In either case, check that $A$ is not compact.

Comment by David Milovich

2012-04-12T18:24:29Z

Sure---you can find one my email addresses here: tamiu.edu/~dmilovich

Comment by David Milovich

2012-04-09T17:49:51Z

I'm suggesting $W$ as a test question. Can you cook up a model of ZF with an amorphous set $S$ such that its power set $W$ has an $F_2$-basis not of cardinality $|S|+1$, or does amorphousness of $S$ ZF-imply that all bases of $W$ are too "trivial" for that to happen?

Comment by David Milovich

2012-04-09T17:28:12Z

I can think of one pathology where you can't "avoid" a basis: the power set $W$ of an amorphous set $S$ has an $F_2$-basis: the singletons and $S$ itself. Try this: can you prove $dim(W)=|S|+1$ from the amorphousness of $S$?

Comment by David Milovich

2012-04-09T17:03:32Z

If I was looking for some kind of reversal, I would play with the $F_2$-vector space of all functions from a given set $S$ to $F_2$. There a basis is exactly a minimal family of subsets of $S$ such that every subset of $S$ is a symmetric sum of finitely many sets from the family. In any case, I wouldn't be surprised if the literature has already answered the questions in your comment. I'm just not very familiar with this literature.

Comment by David Milovich

2012-04-09T16:49:48Z

My guess is that it's a very weak assertion. What does it buy you when the dimension is $\infty$ (that is, when there is no basis)? I conjecture that you could have all your favorite "pathological" sets in a model where vector space dimension is well defined; you'd just need to "protect" the pathologies by ensuring that all vector spaces into which they inject have dimension $\infty$.

Comment by David Milovich

2012-03-27T14:36:32Z

Here's an example of what I mean by applying set theory. Kunen has shown that it is consistent with MA+not(CH) that there is a compactum that is hereditarily Lindelof, hereditarily separable, non-metrizable, and locally connected. I believe it is still open whether PFA is consistent with the existence of such compacta. And yes, I am interested in Cech homotopy too.

Comment by David Milovich

2012-02-29T22:45:04Z

You can't find the f without an additional hypothesis. Let A be an indiscrete space and B a metric space. Then C(A,B) is homeomorphic to B and thus metrizable, but distinct points in A are not separated by continuous real-valued functions.

Comment by David Milovich

2012-02-08T16:03:02Z

The paper is on my website tamiu.edu/~dmilovich/conn_amalgams.pdf and has MR #2298632. The part of the paper that your question reminded me of is a ZFC construction of a homogeneous compactum that is not homeomorphic to any product where each factor is dyadic or first-countable. However, the proof is a connectedness argument, and your question is about zero-dimensional spaces.