co.combinatorics – Is there an elementary subexponential upper bound on the size of the stable stems?

This is a question in stable homotopy theory which I will boil down to a pure combinatorics question. If you’re not interested in the homotopy theory, feel free to skip to the end for the combinatorial formulation.

Homotopy theory:

The question is basically whether the stable version of Serre’s method of killing homotopy groups leads directly to a subexponential upper bound on the homotopy groups of a finite spectrum. I learned here that the size of the stable stems (measured as $log |pi_k mathbb S|$) is conjectured to grow roughly linearly, but that no subexponential bound seems to be known.

The motivating observation for the following approach is the simple fact that the dimension $operatorname{dim} mathcal A^k$ of the Steenrod algebra grows subexponentially in $k$ (Proof: by Milnor’s description of the dual $mathcal A_ast$, we have that $dim mathcal A^k$ counts certain partitions of $k$, and the number of partitions grows subexponentially). Since killing homotopy groups just keeps peeling off Eilenberg-MacLane spectra, I have some hope that when one adds up all of these subexponential contributions, the result might still be subexponential.

I know very little about the mod $p^n$-Steenrod algebras for $n geq 2$, but there are recent results by Mathew and by Burklund giving good bounds on the exponents of the stable stems, so for the purposes of this post I’m going to ignore this issue and blithely pretend that all homotopy groups I see have exponent 1.

So let $X = X_{geq 0}$ be a connective $p$-local spectrum, and let $X_{geq k}$ denote the $k$-connective cover of $X$. Assume that $X$ has finite homotopy groups in each degree. Consider the fiber sequence $Sigma^{k-2} H_{k-1} X_{geq k-1} to X_{geq k} to X_{geq k-1}$ (obtained by using Hurewicz and rotating the most obvious fiber sequence). This gives us the bound

$$operatorname{dim} H_n(X_{geq k}) leq operatorname{dim} H_n(X_{geq k-1}) + operatorname{dim} (H_{k-1} X_{geq k-1}) operatorname{dim}(mathcal A^{n-k+2})$$

So let us set $h_{n,k} = operatorname{dim} H_n(X_{geq k})$ and $a_{n} = operatorname{dim}(mathcal A^{n}$). The goal is to get a subexponential bound on $h_{k,k}$, say when $X = M(p)$ is the mod $p$ Moore spectrum so that $h_{n,0} = delta_{n,0}$ is just the Kronecker delta.


Here’s the Question:

Let $h_{n,k}$ be natural numbers defined for $n,k in mathbb N$, where $h_{n,k} = 0$ for $n < k$. Let $a_n$ be natural numbers defined for $n in mathbb N$ satisfying an inequality $a_n leq exp(c log(n)^d)$ for some $c,d>0$ (by convention, $a_n = 0$ for $n < 0$). Suppose that we have the inequality

$$h_{n,k} leq h_{n,k-1} + h_{k-1,k-1}a_{n-k+2}$$

for all $n in mathbb N$ and $k geq 1$. As a boundary condition, suppose that $h_{n,0} = delta_{n,0}$ is just the Kronecker delta. Does there follow an upper bound for $h_{k,k}$ which is subexponential in $k$?


Because of the simplifying assumption made about the exponents of the groups involved, I’m not certain that a positive answer to the combinatorial question would give a subexponential bound on the stable stems, but I suspect the simplifying assumption can only make things worse for us, so it probably would.

elementary set theory – Proving a set of statements is countable or uncountable?

Let the set $ S = { A_{1}, A_{2}, A_{3},… } $ be a countably infinite set of statements. Let the set $V$ be defined as follows:

A) $S subset V$

b) For any $alpha in V,$ the negation of $alpha$ is also in V.

c) For any $alpha, beta in V$, $(alpha vee beta) in V$

d) $V$ does not contain anything else.

Determine whether $V$ is countable or uncountable (with proof).

Unfortunately I’m feeling kind of lost with this question and don’t really know where to start. My first thought was it may be countable since I was thinking of it as a union of countable sets, but I’m not sure if it is a countable union or not. Any guidance or a hint in the right direction would be much appreciated?

elementary number theory – What is the remainder when $1^{2016} + 2^{2016}+ 3^{2016}+…+2016^{2016} $ is divided by $2017$

What is the remainder when $1^{2016} + 2^{2016}+ 3^{2016}+…+2016^{2016} $ is divided by $2017$

I saw a quesiton in stack-exchange : What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$?

When i checking over it , i thought that what if it is divided $2017$ instead of $2016$. The answer was easy to me in the first glance because by using phi function ,the summation must have been equal to $2016$ and $2016 mod(2017)=2016$.

However the answer is equal to $1759$ according to python. What am i missing ?

elementary number theory – Proving there are infinitely many primes in 5 ways

I need to remember for the test 5 different ways of proving that there are infinitely many primes. Now, I have learned to prove the infinity of primes by Euclid’s theorem, Euler’s theorem, Fermat numbers, Dirichlet, and the last I don’t remember. I will be happy to see how do you suggest to prove these proofs, so I can remember this for the test. It is my first course in number theory, so I only got elementary tools.

elementary set theory – How many nonprincipal ultrafilters?

Let $X$ be an infinite set, idemmultiple ($|X|^2 = |X|$) if that helps. I am looking for a proof that $X$ has more than $|X|$-many ultrafilters. It has $|X|$-many principal ultrafilters. Without at least some AC I know of no proof that $X$ must have any non-principal ultrafilters. So let us assume BPI, the Prime ideal theorem. If we assume BPI can we prove $|beta X| > |X|$? ($beta X$ is of course the set of ultrafilters on $X$). If we have AC then it is standard that $|beta X| = 2^{2^{|X|}}$. I don’t much mind which set theory we work in (the question actually arose in NF, with $X = V$) co’s it’s the construction (if any) that I am after.

Summary: i am interested in a proof (no theory specified) that BPI implies that an idemmultiple (infinite) set $X$ has more than $|X|$-many ultrafilters

elementary set theory – How to find all subset $Y in mathbb R$ such that $Y times Y in {(x , y) : x^2 + 4y^2 leq 1}$

How to find all subsets $Y in mathbb R$ such that $Y times Y in {(x , y) : x^2 + 4y^2 leq 1}$

Can anyone please give some idea to proceed?

I think we have to consider all the squares whose diagonals fall on $y = x$ in the ellipse $x^2 + 4y^2 = 1$.

But I can not understand how to see them properly.

Can anyone please help me ?

elementary set theory – Prove that for every $boldsymbol{n} in mathbb{N}$, $boldsymbol{n} subsetneq mathbb{N}$

I have come up with a proof myself. I am not sure if it is a well-written proof from the view of a mathematician.

We shall first show that $boldsymbol{n} subseteq mathbb{N}$ for every $boldsymbol{n} in mathbb{N}$. Note that $boldsymbol{0} = emptyset$. It is immediately clear that $boldsymbol{0} subseteq mathbb{N}$. Next assume that $boldsymbol{n} in mathbb{N}$ and $boldsymbol{n} subseteq mathbb{N}$. According to definition, $boldsymbol{n}^{+} = boldsymbol{n} cup left{boldsymbol{n}right}$. From the induction hypothesis, $boldsymbol{n} subseteq mathbb{N}$. Also, as $boldsymbol{n} in mathbb{N}$, $left{boldsymbol{n}right} subseteq mathbb{N}$. Then $boldsymbol{n}^{+} subseteq mathbb{N}$. As a result, $boldsymbol{n} subseteq mathbb{N}$ implies $boldsymbol{n}^{+} subseteq mathbb{N}$. So we may conclude that $boldsymbol{n} subseteq mathbb{N}$ for every $boldsymbol{n} in mathbb{N}$.

Next, assume that $boldsymbol{m},boldsymbol{n} in mathbb{N}$ and $boldsymbol{m} < boldsymbol{n}$. We shall show that $boldsymbol{n} notin boldsymbol{m}$. Assume $boldsymbol{n} in boldsymbol{m}$. Then $boldsymbol{m} < boldsymbol{n} < boldsymbol{m}$, as $in$ is the same as $<$ for $mathbb{N}$. As $<$ is transistive, we have $boldsymbol{m} < boldsymbol{m}$, and this indicates $boldsymbol{m} in boldsymbol{m}$, which contradicts the fact that $boldsymbol{s} notin boldsymbol{s}$ for all $boldsymbol{s} in mathbb{N}$. As a result, we have $boldsymbol{n} not in boldsymbol{m}$ for any $boldsymbol{n} > boldsymbol{m}$. This indicates that $boldsymbol{n} neq mathbb{N}$ for every $boldsymbol{n} in mathbb{N}$. Thus, for every $boldsymbol{n} in mathbb{N}$, $boldsymbol{n} subsetneq mathbb{N}$.

elementary number theory – Finding $(a,b)$ in some given subset of $mathbb{Z}^2$ with a specific gcd.

Given the subset of $mathbb{Z}timesmathbb{Z}supset S={(9(k+3),4k), kin mathbb{Z}}$ I’m trying to find $(x,y)in S$ such that $mathrm{gcd(x,y)}=18$. The set $S$ arises as the set of solutions to the equation $4x-9y=108,; x,yinmathbb{Z}$. The equation isn’t particularly hard to solve, but I can’t quite figure out how to go about finding the couples $(x,y)$ with gcd 18.

Playing around a little I found that since I want $18$ to be the gcd of $9(k+3)$ and $4k$, 18 has to divide both, which implies $k$ is an odd number and $2k=9q$ with $qinmathbb{Z}$ but this still allows for the possibility of $mathrm{gcd}(x,y)>18$, that is, those are necessary condition but certainly not sufficient. And I’m not entirely sure where to go from there. This feels like it’s very easy and I’m just not seeing something obvious. Anyway, if anyone has a suggestion, a hint or something, I’d appreciate, thanks.

homotopy theory – Elementary proof of the exactness of Čech complex associated to a hypercovering (“Illusie’s Conjecture”)

Let $mathcal{E}$ be a sheaf of abelian groups on a topological space (or a site). For an open covering $mathfrak{U} = (U_i)_i$, it is well known that the augmented Čech complex $0 to mathcal{E} to mathcal{check C}^bullet(mathfrak{U}, mathcal{E})$ is a resolution of $mathcal{E}$. (Depending on the sheaf cohomology of $mathcal{E}$ on the intersections of the members of $mathfrak{U}$, this resolution might compute $mathbb{R}Gamma(mathcal{E})$, but this need not concern us here.)

A simple proof runs as follows: Let $s in mathcal{check C}^{n+1}(mathfrak{U}, mathcal{E})(V)$ such that $ds = 0$. Then $t = (s_{i_text{fix},i_0,ldots,i_n})_{i_0,ldots,i_n} in mathcal{check C}^{n}(mathfrak{U}, mathcal{E})(V cap U_{i_text{fix}})$ is a local preimage of $s$ under $d$, as $(dt)_{i_0, ldots, i_{n+1}} = s_{i_0, ldots, i_{n+1}} – (ds)_{i_text{fix}, i_0, ldots, i_{n+1}} = s_{i_0, ldots, i_{n+1}}$. As $V = bigcup_{i_text{fix}} (V cap U_text{fix})$, that’s all what’s needed.

I’m looking for a similarly elementary proof in the case that $mathfrak{U}$ is a hypercovering. Apparently this fact is sometimes referred to as Illusie’s Conjecture. A proof involving trivial Kan fibrations and an induction reducing to the case of ordinary covers is in Section 01GA of the Stacks Project, but I’d prefer either a more direct proof or some insight as to why a much simpler argument isn’t likely to exist.