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Tychonoff-ization and Urysohn (functionally Hausdorff) topological spaces

Math Overflow Recent Questions - Fri, 06/07/2019 - 07:31

Let me first make sure I have the correct definitions because my question will be about the difference about the two and there may be some massive confusion on my part.

A topological space $X$ is said to be completely regular or Tychonoff when it is Hausdorff and satisfies the following equivalent conditions:

  • For every $x \in X$ and closed $F \subseteq X$ such that $x\not\in F$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f|_F = 1$.

  • The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ defines a homeomorphism of $X$ to its image.

  • The Stone-Čech compactification map $X \to \beta X$ defines a homeomorphism of $X$ to its image.

  • There exists a compact [Hausdorff] space $K$ such that $X$ is homeomorphic to a subspace of $K$.

On the other hand, a (necessarily Hausdorff) topological space $X$ is said to be functionally Hausdorff (or Urysohn, but some people use this to mean something different, so it's probably best to avoid this terminology) when it satisfies the following equivalent conditions:

  • For every $x,y \in X$ such that $x\neq y$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f(y) = 1$.

  • The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ is injective.

  • The Stone-Čech compactification map $X \to \beta X$ is injective.

  • There exists a continuous injective map $X \to K$ with $K$ a compact [Hausdorff] space.

I note that example 91 (the “deleted Tychonoff corkscrew”) in Steen & Seebach's Counterexamples in Topology gives an example of a functionally Hausdorff space which is not completely regular, showing that the two notions are not equivalent.

Since until recently I thought these two notions were equivalent (I somehow thought that $X \to \beta X$ was automatically an embedding when it is injective), my goal is essentially to dispel the confusion I had; I first have to ask:

Question 0a: Is the above account correct? (Are the properties I claim to be equivalent indeed equivalent, and equivalent to standard definitions for the terms they claim to define?)

Every topological space $X$ has a complete regularization or Tychonoff-ization, namely a continuous map $X \to X'$ with $X'$ a completely regular space, such that every continuous map $X \to Y$ with $Y$ completely regular uniquely factors as $X \to X'\to Y$. (I.e., the functor $X \mapsto X'$ is left adjoint to the inclusion functor from the full subcategory of completely regular spaces to that of topological spaces.) This $X'$ can be defined as the image of the Stone-Čech compactification map $X \to \beta X$ with the subspace topology; in particular, $X \to X'$ is always surjective.

Question 0b: Is this still correct?

Now I thought $X'$ was a quotient space of $X$. This can't be the case because, if what I wrote above is correct, the equivalence relation (“having the same image in $X'$”) is simply “having the same image under every continuous function $X\to\mathbb{R}$ (or equivalently $X\to[0,1]$)”, which is trivial for a functionally Hausdorff space, yet the latter is not necessarily completely regular.

But this is problematic because in section d-2 (“Higher Separation Axioms”, p.158–159) of the Encyclopedia of General Topology (Hart, Nagata & Vaughan eds.) one reads:

“To every space $X$ one can associate a Tychonoff space $Y$ as follows. Two points $x$ and $y$ in $X$ are equivalent if $f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$. The corresponding quotient space $Y$ is Tychonoff and the rings $C(X)$ and $C(Y)$ of real-valued continuous functions are isomorphic; the same holds for the rings $C^*(X)$ and $C^*(Y)$ of bounded real-valued continuous functions.”

It would seem to me that this assertion is contradicted by the existence of the aforementioned counterexample in Steen & Seebach.

Question 0c: Am I correct in believing that the above quote is in error? (Or did I miss some fine print or hidden assumption?)

Now assuming all of the above is correct, there are two natural questions which are left open:

Question 1a: Does every topological space $X$ have a “functional Hausdorffization” (or “Urysohnization”), namely, does the inclusion functor from the full subcategory of functionally Hausdorff spaces to that of topological spaces have a left adjoint? • Question 1b: If so, is it given by quotienting by the equivalence relation “$f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$” or is there some subtlety?

Question 2a: Even if complete regularization is not given by a quotient, is there still a unique coarsest equivalence relation $R$ on any topological space such that $X/R$ is completely regular? • Question 2b: If so, can we describe $R$ concretely, and can we describe the natural continuous map $X' \to X/R$ (where $X'$ is complete regularization as defined above)?

Unstable Integers

Math Overflow Recent Questions - Fri, 06/07/2019 - 07:07

There is a question that has been bothering my mind for quite a while now. I will present it and my current thoughts and progress on it.

Let the prime factorization of an integer $n$ be $$n = p_1^{e_1}\cdot p_2^{e_2}\cdot p_3^{e_3}...$$ And define $n$ to be "unstable" if there exists a prime triplet in its factorization $p_i<p_j<p_k$ $(i<j<k)$ such that $e_j>e_i,e_k$ or $e_j<e_i,e_k$, and the primes are not necessarily consecutive. The exponents of the primes are required to be positive. For example, the numbers $90=2\cdot3^2\cdot5$ and $300=2^2\cdot3\cdot5^2$ are unstable. So is $1361250 = 2\cdot3^2\cdot5^4\cdot 11^2$. It can be seen that $90$ is the only number below $100$ that is unstable. Let $Z(n)$ be the count of unstable integers that do not exceed $n$. So $Z(100)=1$.

I was wondering if there is an efficient way to enumerate such integers, or in other words, to evaluate $Z(n)$. Because I am interested in counting these integers rather than summing them, and because they solely depend on the exponents of the primes in their factorization, I assume the 'right' way to count them would be combinatorial. Due to their "unstable" nature, I thought it would be better to count them in an indirect way:

Define descending integers to be those in which the exponents of the primes in their factorization are weakly decreasing, i.e $e_j\leq e_i$ for all $i<j, p_i<p_j$. For example, $940896=2^5\cdot3^5\cdot11^2$ is one such integer, whereas $2178 = 2\cdot3^2\cdot11^2$ is not. To avoid confusion, let's arbitrarily choose all prime numbers and their powers to be 'descending'.

Let $D(n)$ be the count of descending integers not exceeding $n$

Let $A(n)$ be the count of integers not exceeding $n$ that are neither descending nor unstable

This gives the general formula:

$$n = A(n)+D(n)+Z(n)$$ Because all numbers belong to one of these groups.

The reason for further defining these functions is that I suspect they are easier to evaluate than $Z(n)$, but $Z(n)$ can be directly evaluated using them. Is this the right way to approach this? If so, how can I efficiently calculate $D(n)$ and $A(n)$? Can any of these functions be efficiently calculated?

Prove that $\text{Hom}_R (I(M)/M, I(M)) = 0$?

Math Overflow Recent Questions - Fri, 06/07/2019 - 07:02

Problem: Suppose $I(M)$ is an injective hull of $R-$module $M$. Prove that each isomorphism $\varphi \colon I(M) \rightarrow I(M)$ has this property $\varphi(x)=x, \forall x \in M$ is the identity function $\Longleftrightarrow$ $\text{Hom}_R (I(M)/M, I(M)) = 0$?

Could you give me some hint to solve this problem. Thank all! In this case, we work with the injective hull $I(M)$, should we use the homomorphism theorem to solve this?

Graceful graphs all of whose vertices are labelled with primes or squares

Math Overflow Recent Questions - Fri, 06/07/2019 - 06:07

Do graceful graphs exist with more than any arbitrarily large number of vertices, all of which are labelled with a prime or non-negative square number.

Recall that a graceful graph is a graph with m edges whose vertices can be labelled with some subset of the integers between 0 and m inclusive, no two vertices sharing a label, and each of its edges uniquely identified by the absolute difference between its end points (so that this magnitude lies between 1 and m inclusive).

There is evidence ( that the answer to this is no, but a proof is lacking.

Enumeration and structure of abelian 2-subgroups of a symmetric group

Math Overflow Recent Questions - Fri, 06/07/2019 - 06:02

I am struggling with a group theoretic problem arising in my research. Given a symmetric group $\Sigma_{n}$, let's consider all its abelian 2-subgroups up to conjugation. Is it possible to give a classification of all these subgroups and count how many of them for arbitrary $n$?

Waring's problem: bounds and asymptotics [on hold]

Math Overflow Recent Questions - Fri, 06/07/2019 - 05:23

I'm trying to improve my question but got locked out my account. So here it goes. In Waring's problem (WP) we write $$n=x_1^k+...+x_s^k.\quad (1)$$ We let $g(k)$ be the smallest $s$ where every $n$ can be written like this.

Let $G(k)$ to be the smallest $s$ with every large n written as in (1). So if we can write all $n$ this way, then we can definitely write all large $n,$ so $G(k)\leq g(k).$

Now there is also the asymptotic formula, and $\tilde{G}(k)$ is the smallest $s$ with this formula true. Is there any relation $G(k) \leq \tilde{G}(k)$? or $G(k) \geq \tilde{G}(k)$? or are they independent?

A sufficient and necessary condition for the boundedness of linear operator on Hilbert space [on hold]

Math Overflow Recent Questions - Fri, 06/07/2019 - 03:03

Show that a linear operator $T:H \rightarrow H$ where $H$ is Hilbert Space is continuous if and only if it maps weakly convergent sequences to weakly convergent sequences.

How to prove this Theorem?

Why is it difficult to solve the Monge problem directly?

Math Overflow Recent Questions - Thu, 06/06/2019 - 18:51

I'm trying to understand something about the Monge problem. The Monge problem is:

Let $c(x,y): \mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}^d$ and $$\mathcal{T}(\mu_1,\mu_2) = \{ T: \mathbb{R}^d \rightarrow \mathbb{R}^d | \text{ Borel maps with condition} \, \, T\#\mu_1 = \mu_2 \}$$ where $\mu_1$ and $\mu_2$ are given, compactly supported measures which are absolutely continuous with respect to Lebesgue measure. The Monge problem is to find: $$\inf_{T\in\mathcal{T}(\mu_1,\mu_2)}C_M[T] = \int_{\mathbb{R}^d}c(x,Tx) \mu_1(dx)$$

A book I am reading provides the following discussion on why finding a minimizer directly using "usual" methods (ie taking a minimizing sequence) is tough:

Usually, what one does is the following: take a minimizing sequence $T_n$, find a bound on it giving compactness in some topology (here, if the support of $\mu_2$ is compact, the maps $T_n$ take value in a common bounded set $\text{spt}\mu_2$, and so one can get compactness of $T_n$ in the weak-* $L^{\infty}$ convergence), take a limit $T_n \rightharpoonup T$, and prove that $T$ is a minimizer. This requires semicontinuity of the functional $C_M$ with respect to this convergence (which is true in many cases, for instance, if $c$ is convex in its second variable): we need $T_n \rightharpoonup T \implies \liminf_nC_M[T_n] \geq C_M[T]$, but we also need that the limit $T$ still satisfies the constraint. Yet, the nonlinearity of the pushforward condition prevents us from proving this stability when we only have weak convergence.

Loosely, this all makes sense to me: but I'm getting turned around working the details.

  1. Explicitly: Suppose $\text{spt}\mu_2$ is compact, why does $T_n \rightarrow T$ weak-* in $L^{\infty}$.

  2. Explicitly: what are the sufficient conditions on $C_M[T]$ to ensure the implication in the above statement $$T_n \rightharpoonup T \implies \liminf_nC_M[T_n] \geq C_M[T]$$ Do I just need to say $C_M[T]$ is weakly lower semicontinuous and bounded from below? (i.e. I think this is the same as convexity)

  3. Suppose I show that a minimizer of $C_M[T]$ exists. How do I see that minimizing sequence isn't preserving the pushforward condition in the limit? Apprently "non-linearity" is preventing this -- but I don't see how.


For 1&2: Okay then combining all your comments:

The statement: Let $C_M[T]: L^{\infty} \rightarrow \mathbb{R}$ be weak-* lower semicontinous and bounded from below. Then $C_M[T]$ has a minimizer.

Pf: $C_M[T]$ is bounded from below so $\inf C_M[T]$ exists. Let $T_n \in L^{\infty}$ be a minimizing sequence and let $T$ denote the minimizer. Consider a closed ball in the weak-* topology of positive radius centered at $T$. This ball is compact in weak-* topology by banach-aloglu - so there exists a subsequence of $T_n$, call it $T_{n_k} \rightarrow T$. This gives us $$\liminf C_M[T_{n_k}] \leq C_M[T_i] \quad \forall T_i \in L^{\infty}$$ -- in particular $\liminf C_M[T_{n_k}] \leq C_M[T]$. Weak-* LSC gives the other inequality, so $C_M[T]$ achieves its min.

for 3

It doesn't make sense to write: $$(aT_1+bT_2)_{\#}\mu = a(T{_1}_{\#}\mu) + b(T{_2}_{\#}\mu)$$ Let $a=b=1$, $T_1 = 2x$, $T_2 = x^2$, $\mu$ to be leb msr, and consider a set $A = [0,-1]$. Then $$(x^2 + 2x)_{\#}\mu\big([0,-1]\big) = \mu \bigg((x^2+2x)^{-1}\big([0,-1]\big)\bigg)$$ But $(x^2+2x)^{-1}[0,-1]$ can be $[0,1]$ or $[1,2]$ so LHS of such a statement is not well defined. So in what sense is the push forward non linear?

Is there a set of positive integers of density 1 which contains no infinite arithmetic progression?

Math Overflow Recent Questions - Thu, 06/06/2019 - 15:28

Let $V$ be a set of positive integers whose natural density is 1. Is it necessarily true that $V$ contains an infinite arithmetic progression?—i.e., that there are non-negative integers $a,b,\nu$ with $0\leq b\leq a-1$ so that: $$\left\{ an+b:n\geq\nu\right\} \subseteq V$$

In addition to an answer, any references on the matter would be most appreciated.

Upper bounds for the second largest eigenvalue in terms of degree?

Math Overflow Recent Questions - Wed, 06/05/2019 - 22:25

I am looking for upper bounds on the second largest eigenvalue, $\lambda_2(G)$ of a given graph $G$, with respect to minimum/maximum degrees of the graph. I looked around for some existing bounds most I found were with respect to the order of $G$.

Can anyone point me to something?

Simple description of a Grothendieck topology on the opposite of f.p. complex algebras

Math Overflow Recent Questions - Wed, 06/05/2019 - 15:54

Let ${\cal A}$ be the category of finitely presented $\mathbb{C}$-algebras. Let $J$ be the largest subcanonical Grothendieck topology on ${{\cal A}^{op}}$ such that the local algebras in $\cal A$ are cocovered only by the maximal sieve. Is there a more explicit description of $J$? (Something analogous to the usual description of the Zariski topology in terms of a finitary basis.) Note: it contains the Zariski topology.

Maximum conjugacy class size in $S_n$ with fixed number of cycles

Math Overflow Recent Questions - Wed, 06/05/2019 - 10:08

Context: It is well known that given a permutation in $S_n$ with $a_i$ $i$-cycles (when written as a product of disjoint cycles), the size of the conjugacy class is given by $$ \frac{n!}{\prod_{j=1}^n (j)^{a_j}(a_j !)}$$

It is also known that the conjugacy class containing permutations with exactly one fixed point and exactly one $n-1$ cycle (hence, $a_1 = 1$ and $a_{n-1} = 1$) has maximum conjugacy class size. (see this post)

Question: Now let's fix the number of cycles, $m$ (this count includes the trivial ones). Then how would one go about finding the cycle type with the maximum conjugacy class size, among the ones with $m$ cycles?

Note that the case for $m=2$ is true for $n \geq 3$ by the assertion above in the context.

My "conjecture" is that any such cycle type needs to have $a_1 \geq \frac{m-1}{2}$. (i.e. at least about half of the cycles are trivial). And that about a fourth are 2 cycles and so on.


The conjecture as it stands is not true, as per one of the answers posted below. To have roughly 1/2 of the parts as 1's, the number of parts also needs to be roughly a 1/2 of $n$ or so, it seems.

Upward reflection of rank-into-rank cardinals

Math Overflow Recent Questions - Tue, 06/04/2019 - 16:04

Rank-into-rank cardinals have the rather intriguing property that they reflect upwards. I would be interested to know how far the upward reflection goes:

1) Does "There exists a rank-into-rank cardinal (of type I3, say)" imply that "There exists an unbounded class of rank-into-rank cardinals (in V) ?"

2) If 1) is false, then can we say something interesting about the supremum of this upward reflecting sequence ? For instance, maybe the supremum has some large cardinal properties of interest ?

3) Finally, existence of a rank-into-rank implies the existence of pretty much all of the other large cardinals. But can we strengthen this along the lines of "If there exists a rank-into-rank cardinal, then there exists an unbounded class of cardinals with property P (in V)" for interesting some property P? By interesting property P, I mean something like "inaccessible" or "Mahlo" or hopefully even stronger like "measurable".

Shapiro lemma for Bloch Kato Selmer group

Math Overflow Recent Questions - Tue, 06/04/2019 - 15:07

Let $K$ be a finite of $\mathbb{Q}$. Let $T$ be a finite dimensional $G_{K}$ module over $\mathbb{Z}_p$. Does the Bloch Kato Selmer group $H^{1}_{f}(\mathbb{Q},Ind^{G_{\mathbb{Q}}}_{G_K}T) = H^{1}_{f}(K,T) $?

If not then can you tell me what kind of condition on $K/\mathbb{Q}$ as well as the condition at the prime $p$ do I need for that equality to hold? Thank you.(I was trying to use $Ind(Res V\otimes W)\simeq V\otimes Ind W$ for each prime but I got stuck.)

Can the cardinality of the set of all intervening cardinals between sets and their power sets be always singular?

Math Overflow Recent Questions - Tue, 06/04/2019 - 14:37

This is a question that I've posted to Mathematics Stack Exchange, that was un-answered.

To re-iterate it here:

Is the following known to be consistent relative to some large cardinal assumption?

$\forall \kappa [\kappa >2 \to \kappa < \kappa^* < 2^\kappa \wedge singular(\kappa^*)]$

where $\kappa$ is a cardinal and $``<"$ is strict cardinal smaller than, and $\kappa^* = |\{\lambda|\kappa < \lambda < 2^\kappa\}|$

Automorphisms of a first order infinitesimal deformation which pull back to the identity morphism

Math Overflow Recent Questions - Tue, 06/04/2019 - 14:00

Let $X$ be a scheme over an algebraically closed field $k$ and let $X'$ be a deformation of $X$ over the dual numbers $k[\epsilon]/(\epsilon)^2$. Here we are assuming that $X$ has non-trivial first order infinitesimal deformations.

I am interested in classifying automorphisms of $X'$ which pullback to the identity on $X$. Let $U_i$ be an open affine cover of $X$ so that $U_i' \cong X' \vert_{U_i}$ is an open cover of $X'$ by trivial first order inf. deformations of $U_i$.

Now let $\phi$ be an automorphism of $X'$ which pulls back to the identity morphism on $X$. Then $\phi \vert_{U_i'}$ is an automorphism of $U_i'$. Since $U_i'$ is trivial we know that its automorphism group is isomorphic to $\Gamma(U_i, \mathcal{T}X \vert_{U_i})$. In the sense that if $b \in k[U_i']$ then $\phi(b) = b + \epsilon db$ where $d \in \Gamma(U_i, \mathcal{T} X \vert_{U_i})$.

Is it possible to glue together these local automorphisms (i.e global sections of the associated tangent sheaf) to form a global automorphism? Based on the description given above, I think that the group of automorphisms of $X'$ which pullback to the identity on $X$ is isomorphic to $\Gamma(X, \mathcal{T}_{X'/k[\epsilon]})$.

Is this correct?

Quasi-compact quasi-separated induction?

Math Overflow Recent Questions - Tue, 06/04/2019 - 13:50

I believe I've encountered the statement below, but I've lost my reference and am unable to find another one. So, I'm posting this question to see if someone can give a reference, or at least confirm the statement is true (or make a correction).

Proposition: Let $\mathcal{C}$ be a class of schemes with the following properties:

  • $\mathcal{C}$ contains all affine schemes
  • If $X$ is a scheme, $\{ U, V \}$ is an open cover of $X$, and $\mathcal{C}$ contains the three schemes $U$, $V$, and $U \cap V$, then $\mathcal{C}$ contains $X$.

Then $\mathcal{C}$ contains every quasi-compact quasi-separated scheme. Conversely, the class of quasi-compact quasi-separated schemes has the properties above.

Good upper bounds for $\sum_{t=1}^N\gamma^t e_{N-t}$ given that $\gamma \in [0, 1)$ and $(e_t)_t$ is a decreasing sequence of positive numbers

Math Overflow Recent Questions - Tue, 06/04/2019 - 13:45

Given $\gamma \in [0, 1)$, an integer $N \ge 2$, and a decreasing null sequence of positive numbers $e_1,e_2,\ldots,e_t,\ldots$, I'm interested in estimating the sum $S_N := \sum_{t=1}^N\gamma^t e_{N-t}$.


What is a good upper bound for $S_N$ for large $N$ ?


Empirically, I'm observing (by plotting graphs) that $S_N \sim \dfrac{e_N}{1-\gamma}$, but I'm not able to prove this in general. My experiments have been for $e_t=at^{-b}$ (with $a,b>0$), $e_t = \ln(t)/t$, $e_t=1/\ln(t)$, $e_t=1/\ln(t)^2$, etc.

The case $e_t = at^{-b}$ can be established analytically. Indeed, a tedious computation reveals that $S_N \sim \frac{1}{1-\gamma}N^{-b} \sim \frac{1}{1-\gamma}e_N$.

  • In my (abuse of notations), it's fine for $\sim$ to hide global multiplicative constants (e.g $e_1$, etc.).

A curious prime counting approximation or just data overfitting?

Math Overflow Recent Questions - Tue, 06/04/2019 - 12:05

I am not sure, if this is a research problem. If not I will move this question to ME: Let $\Omega(n) = \sum_{p|n} v_p(n)$, which we might view as a random variable. Let $E_n = \frac{1}{n} \sum_{k=1}^n\Omega(k)$ be the expected value and $V_n=\frac{1}{n} \sum_{k=1}^n(E_n-\Omega(k))^2$ be the variance. Then $$\pi(n) \approx \frac{n\gamma(\frac{V_n}{E_n},1.4854177\cdot \frac{V_n}{E_n^2})}{\Gamma(\frac{V_n}{E_n})}$$ where $\Gamma=$ Gamma function, $\gamma=$ lower incomplete gamma function.

Background: I was trying to fit the gamma distribution to the random variable $\Omega(k)$ ,$1 \le k \le n$. The value $1.4854177$ is fitted to some data. My question is, if there is any heuristic why this approximation should be good, if at all, or if this is just an overfitting problem?

Below you can find some sage code which implements this:

def Omega(n): return sum([valuation(n,p) for p in prime_divisors(n)]) means = [] variances = [] xxs = [] omegas = [Omega(k) for k in range(1,10^4)] for nn in range(10^4,10^4+3*10^3+1): n = nn omegas.append(Omega(n)) print "---" m = mean(omegas[1:-1]) v = variance(omegas[1:-1]) shape,scale = m^2/v,v/m xx = find_root(lambda xx : n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N()-prime_pi(n),1,2) xx = 1.4854177706344873 approxPrimePi2 = n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N() primepi = prime_pi(n) print primepi, approxPrimePi2,shape.N(),scale.N(),xx print "---" print "err2 = %s" % (abs(primepi-approxPrimePi2)/primepi) xxs.append(xx) means.append(m.N()) variances.append(v.N())

The below is the math derivation for Deep Learning. Could someone explain me the differential calculus [on hold]

Math Overflow Recent Questions - Tue, 06/04/2019 - 11:37

enter image description here The image depicts the differential caluclus. It would be helpful if someone could help in that. Also, I think it is based on chain rule

enter image description here


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