## linear algebra – Given \$f:mathbb{R}^m supset U rightarrow mathbb{R}^n\$ differential. Relation of rank of \$f’\$ and the property of image of \$f\$?

Let $$f:mathbb{R}^m supset U rightarrow mathbb{R}^n$$ a differential map, $$U$$ open, and $$a in U$$. Are the followings true:

1. If $$m > n$$ and $$f(U)$$ is open set in $$mathbb{R}^n$$ then $$text{rank}f’_a = n$$?

2. If $$m < n$$ and $$f(U)$$ is open set in $$mathbb{R}^m$$ then $$text{rank}f’_a = m$$?

I don’t how to get information of Jacobian of $$f$$ to know about its rank. I try proof by contradiction as follows:

In 1’s problem, If $$text{rank}f’_a = k < n$$ then $$f'(a):mathbb{R}^mrightarrow mathbb{R}^k$$. We have, when $$hrightarrow 0$$, $$mathbb{R}^n ni f(a + h) – f(a)rightarrow f’_a(h)in mathbb{R}^k$$. But, I don’t see any reason to forbid $$f(a + h) – f(a)in mathbb{R}^n$$ close to a point $$mathbb{R}^k$$. So may be the conclusions are wrong.

If they are wrong, could you give an example?

## \$X_n xrightarrow{p} X Rightarrow existshspace{0.2em} X_{n_j} xrightarrow{a.s.}X\$

suppose that $$X_n xrightarrow{p} X$$ and take some decreasing $${epsilon_m}_{m in mathbb{N}}$$ who satisfy $$epsilon_mrightarrow0$$ and define:

• $$A_{j,epsilon}= { omega in Omega : vert X_j(omega)-X(omega)vert
• $$S_{n,epsilon}= bigcap_{j geq n} A_{j,epsilon}$$
• $$S_{epsilon}=bigcup_{n in mathbb{N}}S_{n,epsilon}=liminf A_{j,epsilon}$$
• $$S:= bigcap_{m in mathbb{N}}S_{epsilon_m}$$

is easy to demonstrate that $$S ={ omega in Omega: lim_{jto infty}X_j(omega) = X(omega)}$$

now fixing $$m$$ i can build an increasing $$n : mathbb{N} to mathbb{N}$$ who satisfy $$mathbb{P}(vert X_{n_j}-Xvert>epsilon_m)leq2^{-j}, forall jin mathbb{N}$$

now because $$sum_{j=1}^{infty}mathbb{P}(vert X_{n_j}-Xvert>epsilon_m)leqsum_{j=1}^{infty}2^{-j}, by borel-cantelli lemma

$$mathbb{P}(limsup A_{j,epsilon_m})=0 Rightarrow mathbb{P}(S_{epsilon_m})=1$$
so this demonstate that $$mathbb{P}(S_{epsilon_m})=1$$ do not depend by $$m$$

now $$mathbb{P}(bigcap_{m=1}^{infty}S_{epsilon_m})=lim_{mtoinfty}mathbb{P}(S_{epsilon_m})=1$$, this is because $$epsilon < epsilon’ Rightarrow S_{epsilon} subset S_{epsilon’} Rightarrow S_{epsilon_{m+1}}subset S_{epsilon_{m}}, forall m in mathbb{N}$$

i think that all i used for demonstrate this theorem is legal

## isometry – Isometric isomorphism from \$mathbb{C} rightarrow mathbb{R}^2\$.

$$mathbb{C} cong mathbb{R} times mathbb{R}$$ are identical from set theoretical viewpoint, even as vector spaces or groups. The difference would then be the one considering it as a field, as $$mathbb{R} times mathbb{R}$$ is certainly not a field.
Both underly the same metric, as by the complex modulus we get for two complex numbers $$z_1 = x_1 + iy_1$$ and $$z_2 = x_2 + i y_2$$ the metric:
$$begin{equation} |z_1-z_2| = sqrt{(x_1-x_2)^2 – (y_1-y_2)^2}. end{equation}$$
Clearly this is an Euclidean metric. Thus the complex plane is a $$2$$-dimensional Euclidean space. I wonder if one can construct an isometric isomorphism from $$mathbb{C} rightarrow mathbb{R}^2$$ and how this isomorphism looks like?

## real analysis – Questions about a proof of \$(forall a)[a in mathbb{F} rightarrow -(-a) = a]\$

Proposition:

$$(forall a)(a in mathbb{F} rightarrow -(-a) = a)$$

proof.

Suppose that $$a in mathbb{F}$$. By the additive inverse property, $$-a in mathbb{F}$$ and $$-(-a) in mathbb{F}$$ is the additive inverse of $$-a$$; i.e. $$-(-a) + (-a) = e$$ Since $$-a$$ is the additive inverse of $$a$$, $$(-a) + a = a + (-a) = e$$ which also justifies that $$a$$ is the additive inverse of $$-a$$. From the uniqueness of additive inverses, we conclude that $$-(-a) = a$$.

Are we allowed to state that $$-(-a) + (-a) = e$$ implicitly using commutativity of the reals? I ask that because I thought that we were supposed to state that $$-(-a) + (-a) = (-a) + (-(-a)) = e$$ by definition. Also, is the statement “also justifies that $$a$$ is the additive inverse of $$-a$$ ” simply saying that $$(-a) + a = a + (-a) = e$$ justifies that $$-a$$ is the additive inverse of $$a$$ as well as justifying that $$-a$$ is the additive inverse of $$a$$?

## abstract algebra – Any morphism \$phi:G rightarrow A\$ to an abelian group \$A\$ factors uniquely through the projection \$G rightarrow G /C\$

I’m doing this exercise 7(b) in textbook Algebra by Saunders MacLane and Garrett Birkhoff. Could you please verify if my attempt is fine or contains logical mistakes?

Let $$G$$ be a group and $$C$$ its commutator subgroup. Prove that

My attempt:

For $$a,b in G$$, we have $$aC, bC in G/C$$. It follows from $$b^{-1}a^{-1}ba in C$$ that $$C = (b^{-1}a^{-1}ba)C$$. Then $$(aC)(bC) = (ab)C = (ab)(b^{-1}a^{-1}ba)C=(ba)C = (bC)(aC)$$. Hence $$G/C$$ is abelian.

Next we prove that $$phi(C) = {1}$$. For $$x = b^{-1}a^{-1}ba in C$$, we have $$phi(x) = phi(b^{-1}a^{-1}ba) = phi(b)^{-1} phi(a)^{-1} phi(b) phi(a)$$. On the other hand, $$A$$ is abelian and thus $$phi(a)^{-1} phi(b) = phi(b) phi(a)^{-1}$$. Hence $$phi(x) = 1$$.

To sum up, we have $$C trianglelefteq G$$ and $$phi:G rightarrow A$$ a group morphism and $$phi(C) = {1}$$. Then the result follows from Theorem 26.

## When is \$(x_n,y_n) rightarrow (x,y)\$ equivalent to both \$x_n rightarrow x\$ and \$y_n rightarrow y\$?

Let $$X,Y$$ be metric spaces. What metric is used on the product space to get this statement?

$$(x_n,y_n) rightarrow (x,y)$$ is equivalent to both $$x_n rightarrow x$$ and $$y_n rightarrow y$$

## nt.number theory – Prove that a ring homomorfism \$f: mathbb{Z}_m rightarrow mathbb{Z}_n \$ is injective iff n|m and \$gcd(n/m,m) = 1\$

Prove that a ring homomorfism $$f: mathbb{Z}_m rightarrow mathbb{Z}_n$$ is injective iff n|m and $$gcd(n/m,m) = 1$$

I already did the proof of f being injective under the hypotesis, but I have no idea about the first implication; in particular I dont know how to show that $$mcd(n/m,m)=1$$.

## limits – Show that \$lvert lvert (x,y) rvert rvert rightarrow infty\$ implies \$f(x,y) = x^2-4xy+4y^2+y^4-2y^3+y^2 rightarrow infty\$

I am working on the following exercise:

Consider the function
$$f(x,y) = x^2-4xy+4y^2+y^4-2y^3+y^2.$$

Show that $$lvert lvert (x,y) rvert rvert rightarrow infty quad Longrightarrow quad f(x,y) rightarrow+infty.$$

I do not see how I could prove this in a “nice way”. Is there any way to avoid a lot of case distinctions, so to avoid considering each case like

$$x rightarrow +infty, y rightarrow -infty$$
$$x rightarrow -infty, y rightarrow -infty$$

and so on separately?

## combinatorics – Counting a walk \$i rightarrow j rightarrow k rightarrow j rightarrow l rightarrow j\$ in a graph

This paper gives a procedure for counting redundant paths (which I will refer to as walks) in a graph using its adjacency matrix. As an exercise, I want to count only the walks of the form $$i rightarrow color{blue}j rightarrow k rightarrow color{blue}j rightarrow l rightarrow color{blue}j$$ from node $$i$$ to node $$j$$, with $$i neq j neq k neq l$$. Also see this post.

Let $$A$$ be the adjacency matrix. The notation I use below is: “$$cdot$$” for usual matrix multiplication, “$$odot$$” for element-wise matrix product, “d$$(A)$$” for the matrix with the same principal diagonal as $$A$$ and zeros elsewhere, and $$S = A odot A^T$$.

The matrix for $$i rightarrow color{blue}j rightarrow k rightarrow color{blue}j rightarrow l rightarrow color{blue}j$$ will have its $$i$$, $$j$$ entry: $$a_{ij}cdot a_{jk}cdot a_{kj}cdot a_{jl}cdot a_{lj}$$. I have found this to be:
$$A cdot (d(A^2))^2$$

However, $$i rightarrow color{blue}j rightarrow k rightarrow color{blue}j rightarrow l rightarrow color{blue}j$$ also includes the following walks which repeat undesired nodes and should be subtracted:
$$color{red}i rightarrow j rightarrow color{red}i rightarrow j rightarrow l rightarrow j tag{1}$$
$$color{red}i rightarrow j rightarrow k rightarrow j rightarrow color{red}i rightarrow j tag{2}$$
$$i rightarrow j rightarrow color{red}k rightarrow j rightarrow color{red}k rightarrow j tag{3}$$
$$color{red}i rightarrow j rightarrow color{red}i rightarrow j rightarrow color{red}i rightarrow j tag{4}$$

My calculations for $$(1) – (4)$$ are:
$$S cdot text{d}(A^2) tag{1}$$
$$text{d}(A^2) cdot S tag{2}$$
$$A cdot text{d}(A^2) tag{3}$$
$$S tag{4}$$

Every time one of $$(1) – (3)$$ is subtracted, $$(4)$$ is subtracted as well, since it is included in all three. Since it is not desired in the end, it is added back 2 times. Overall:
$$A cdot (d(A^2))^2 – S cdot text{d}(A^2) – text{d}(A^2) cdot S – A cdot text{d}(A^2) + 2S$$

However, this is wrong and gives incorrect counts, even negatives. What am I missing here?

## regular languages – \$L = {alpha^i beta^j gamma^k vert i,j,k in mathbb{N}_0, (i=1) Rightarrow (j=k)}\$

I am asking this question here, because I am not allowed to comment on the thread that I am actually interested in, but maybe someone can still help me?

I alredy found an anwser to the Problem above (in the post linked to this question), but I still don’t understand, why I can’t just use that one case $$s = alpha beta^p gamma^p$$. I can show for that case, that it doesn’t fit the pumping lemma for regular languages. Isn’t the point of condratiction, that I have to find just one case that doesn’t fit the hypothesis?

Actually I am not even supposed to use the pumping lemma, but the definitions of closure for regular languages. And that is where I started with $$(i=1) Rightarrow (j=k) = i neq 1 vee j = k$$. And then I wanted to use the properties of closure, like in the first anwser in the post I linked.
(I was also thinkg of using a regular expression? It seemed easier)
But if I can’t just find that mentioned one word to proove the language not regular (without the PL)? I am confused.
I hope it makes sense. I am genuinely interested in understanding this problem.

Irregularity of \${a^ib^jc^k mid text{if } i=1 text{ then } j=k }\$