linear algebra – minimum number of elements of a spanning set of vectors of $(mathbb Z / n mathbb Z)^m$

A spanning set of $(mathbb Z / n mathbb Z)^m$ is a set $V = {v_1, ldots, v_{ell}}$ of $m$-coordinates vectors with coefficients in $mathbb Z / n mathbb Z$ such that, for each vectors $u$ of $(mathbb Z / n mathbb Z)^m$, there exists a linear combination of $V$ that gives $u$, i.e. : $$ (mathbb Z / n mathbb Z)^m = left {a_1v_1 + a_2v_2 + ldots + a_{ell}v_{ell} mid a_1, ldots, a_{ell} in mathbb Z / n mathbb Z right }$$
It is obvious that, for every $n geq 2$ and $m geq 2$, there always exists such $V$ with $m$ elements (with the vectors $(0, ldots, 0, 1), (0, ldots, 0, 1, 0)$ and so on) but for a given spanning set $V$ of $(mathbb Z / n mathbb Z)^m$ could we always reduce $V$ to have $n$ elements (that is $ell = m$) ? and if not, on what condition?

Thank you in advance for your help.

reference request – minimum ratio between the shortest and longest distances between $m$ points in $mathbb R^n$

Let there be $m$ points in $mathbb R^n$. Let $D$ be the longest distance between two of these points and let $d$ be the smallest. What is the smallest possible value of $frac{D}{d}$ for each value of $n$ and $m$, and which configurations reach it?

It is clear when $n$ is $1$ the solution is reached when the points are evenly spaced in a line.

For $n=2$ I think the answer may be the regular polygons. For higher dimensions maybe it is reached by setting all the points in a sphere and trying to spread them out as best possible?

real analysis – Example of a condensation point of $mathbb Q times mathbb Q$ in metric space $mathbb Q times mathbb Q$

The following question came to my mind while reading the definition of a condensation point from Exercise (2.)27 on Page 45 of Walter Rudin’s Principles of Mathematical Analysis, 3rd Edition:

Let the metric space $mathbb X$ be the set of all ordered pairs $(alpha_1, alpha_2)$ where $alpha_1, alpha_2 in mathbb Q$ (rational numbers). Is each element in $mathbb X$ a condensation point of $mathbb X$?

I tend to think that the answer is “no”. That’s because, $mathbb Q times mathbb Q$ is countable $implies$ the neighborhood of each point $p$ in $mathbb X$ does not include uncountably many points of $mathbb X$ $implies$ $p$ fails to satisfy the requirement of being a condensation point of $mathbb X$. I’m not sure somehow if this argument is correct, and would appreciate help. Thanks.

real analysis – $f:[a,b]tomathbb R$ is continuous $implies {(x,f(x)): xin [a,b]}$ is compact in $mathbb R^2$

Given $f: (a,b)tomathbb R$, define $G: (a,b)tomathbb R^2$ by $G(x) = (x,f(x))$ for all $xin (a,b)$. Show that the following are equivalent:

  1. $f$ is continuous.
  2. $G$ is continuous.
  3. The graph of $f$ is a compact subset of $mathbb R^2$.

I have shown $1implies 2$ and $2implies 3$. Proving $3implies 1$ would help me complete the argument full circle. Perhaps there are other ways too, such as showing $3implies 2$ and $2implies 1$ instead. Please let me know if the following makes sense (point out errors if any), and help me complete the proof.

$(1implies 2)$: I used the sequential definition of continuity to show that if $x_nto x$ and $f$ is continuous, then $f(x_n)to f(x)$, and hence $G(x_n)to G(x)$. Does this need more argument?

$(2implies 3):$ $(a,b)$ is compact by the Heine-Borel theorem, and the continuous image of a compact set is compact. Hence, $G((a,b))$, i.e. the graph of $f$ is compact in $mathbb R^2$.

Any hints for $3implies 1$, or $3implies 2$ and $2implies 1$? Thanks a lot!

co.combinatorics – Largest set of $k$-wise linearly independent vectors in $mathbb F_q^n$?

The largest size of a $3$-wise independent set is at most $$frac{q^{n-1}-1}{q-1}+1.$$ To see why, if $S$ is $3$-wise independent, and ${bf v}_0in S$, then the set of vectors of the form $a{bf v}_0+b{bf v}$ with $ain mathbb F_q, bin mathbb F_q^{times}, {bf v}in Ssetminus{v_0}$ are pairwise distinct. This is achieved when $q=2$ by the set of vectors with an odd number of ones.
The next simplest case is $q=3$, where I am stuck. It seems to be that finding a large independent set is related to finding a large cap-set in $mathbb F_3^{n-1}$, which is a hard open problem. Specifically, by multiplying all the vectors in $S$ with an appropriate scalar, you can assume their leftmost nonzero coordinate is $1$, and then if $u,v,w$ all have their first coordinate equaling $1$, then ${u,v,w}$ being linearly independent is equivalent ${u’,v’,w’}$ being non-colinear, where $u’$ is obtained by deleting the initial $1$ from $u$, and similarly for $v’,w’$.

Any $4$-wise independent set $S$ must satisfy
$$
(q-1)^2binom{|S|}2le q^n,
$$

which follows by realizing that the vectors of the form $a{bf v}_1+b{bf v}_2$, with $a,bin mathbb F_q^times$ and ${bf v}_1,{bf v}_2in S$, are pairwise distinct (for this to work, each pair ${{bf v}_1,{bf v}_2}$ must appear at most once). This gives $|S|lesssim sqrt{2}q^{n/2}/(q-1)$. I am not sure if this the true growth rate, even for $q=2$.

set theory – Rational bijections $mathbb Rto(0;1)$

Notation:
$$ (0;1) := {xinmathbb R: 0<x<1}$$

=====================================

There are simple rational stretches $ f colon (0; , 1)tomathbb R, $ e.g. let $ sin(0;, 1); $ then
$$ f(x) := frac{1-s}{1-x}-frac sx $$
is an increasing bijection $ f colon (0;,1)tomathbb R $ such that $ f(s)=0.$

In the other direction, there are rational surjections, such as $ g colon mathbb Rto(0; , 1) $ given by
$ g(x)=frac 1{1+x^2}. $

Question. Does there exist a rational bijection $ b colon mathbb Rto(0; , 1) $?

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 ?

algebraic geometry – Global description of holomorphic foliations of $mathbb P^n$

The nice thing about working with a projective variety $V subset mathbb P^n$ is that, to study it, you do not need to dance the annoying dance of taking affine charts and applying transition functions all the time. Instead, you work with the system of equations that defines its affine cone $C(V) subset mathbb A^{n+1}$, and remember that you are actually interested in the orbits of the rescaling $mathbb C^star$-action (after removing the origin).

However, I am working with 1-dimensional foliations of $mathbb P^n$, rather than subvarieties of $mathbb P^n$. From what I have seen, such foliations are described in terms of a polynomial vector field on the affine part $mathbb A^n$:

$$mathscr F : P_1 , frac partial {partial x_1} + dots + P_n , frac partial {partial x_n}$$

Suppose I want to get a description of $mathscr F$ on another affine chart of $mathbb P^n$, say,

$$y_1 = frac 1 {x_1}, qquad y_2 = frac {x_2} {x_1}, qquad dots qquad y_n = frac {x_n} {x_1}$$

Then I need to compute the pushforward of the change of coordinates and then apply it to the vector field. It is not a terribly difficult operation to perform, but it is still annoying to have to do it all the time. What I wish I could do is define a single polynomial vector field $X$ on $mathbb A^{n+1}$ such that the differential of the projection $pi : mathbb A^{n+1} setminus { 0 } to mathbb P^n$ sends $X(p)$ to a generator or $mathscr F_{pi(p)}$. In particular, $X(p)$ should be tangent to the $mathbb C^star$-orbit if and only if $pi(p)$ is a singular point of $mathscr F$.

  1. Is this always possible?

  2. If so, how?

  3. If not, why?