I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
I solved Exercise 3.E.14 on p.100 as follows.
I want to know if my solution is ok or not.
I also want to know a typical solution to this exercise.
3.E.14
Suppose $U=\{(x_1,x_2,\dots)\in\mathbb{F}^\infty : x_j\neq 0\text{ for only finitely many }j\}$.
(a) Show that $U$ is a subspace of $\mathbb{F}^\infty$.
(b) Prove that $\mathbb{F}^\infty/U$ is infinite-dimensional.
(a)
Obviously, $(0,0,\dots)\in U$.
Suppose that $x,y\in U$ and $\lambda\in\mathbb{F}$.
Let $X:=\{j\in\{1,2,\dots\}\mid x_j\neq 0\}$.
Let $Y:=\{j\in\{1,2,\dots\}\mid y_j\neq 0\}$.
Then, $X$ and $Y$ are finite subsets of $\{1,2,\dots\}$ by the assumption for $U$.
So, $X\cup Y$ is a finite subset of $\{1,2,\dots\}$.
Let $m:=\max X\cup Y$.
If $j>m$, then $x_j=y_j=0$.
So, $x+y\in U$ and $\lambda x\in U$.(b)
Let $p$ be a prime number.
Let $x_p=(x_{p1},x_{p2},\dots)$ be an element of $\mathbb{F}^\infty$ such that $x_{pi}=1$ if $i=p^k$ for some $k\in\{1,2,\dots\}$ and $x_{pi}=0$ if $i\neq p^k$ for any $k\in\{1,2,\dots\}$.
Let $X_p:=\{j\in\{1,2,\dots\}\mid x_{pj}\neq 0\}$.
If $p$ and $q$ are distinct prime numbers, $p^k\neq q^l$ for any $k,l\in\{1,2,\dots\}$ by a famous theorem.
So, $X_p\cap X_q=\emptyset$ if $p$ and $q$ are distinct prime numbers.
Let $\{p_1,\dots,p_n\}$ be a set of $n$ distinct prime numbers.
Then, $x_{p_1}+U,\dots,x_{p_n}+U$ is linearly independent:
Proof:
Suppose that $\lambda_1(x_{p_1}+U)+\dots+\lambda_n(x_{p_n}+U)=0+U$.
Then, $(\lambda_1x_{p_1}+\dots+\lambda_nx_{p_n})+U=0+U$.
So, $\lambda_1x_{p_1}+\dots+\lambda_nx_{p_n}\in U$.
Assume that $\lambda_i\neq 0$ for some $i\in\{1,\dots,n\}$.
Then, $X_{p_i}=\{j\in\{1,2,\dots\}\mid \lambda_ix_{p_ij}\neq 0\}$.
If $j\in X_{p_i}$, then $j\notin X_{p_k}$ for any $k\in\{1,\dots,n\}-\{i\}$.
And $X_{p_i}$ is an infinite set.
So, $\lambda_1x_{p_1}+\dots+\lambda_nx_{p_n}\notin U$.
This is a contradiction.
So, $\lambda_i=0$ for any $i\in\{1,\dots,n\}$.
So, $x_{p_1}+U,\dots,x_{p_n}+U$ is linearly independent.
By a famous theorem, there are infinitely many prime numbers.
So, there exists a list of elements of $\mathbb{F}^\infty/U$ whose length is $n$ for any $n\in\{1,2,\dots\}$ and which is linearly independent.
So, $\mathbb{F}^\infty/U$ is infinite-dimensional.