Suppose U = { $(x_1, x_2, ....) \in \mathbf{F^{\infty}} : x_j \neq 0 $ for only finitely many j }. Prove $\mathbf{F^\infty}/U$ is infinite dim. Suppose  U = { $(x_1, x_2, ....) \in \mathbf{F^{\infty}}  : x_j \neq 0 $  for only finitely many j }.
First you have to prove that U is a subspace of $\mathbf{F^{\infty}}$ which I have done but the second part I am finding quite hard.
The question concerns quotient spaces.
The question is this:
Prove that $\mathbf{F^\infty}/U$ is infinite dimensional.
I think I have proved it but am unsure about my reasoning. I am self-studying so I don't have access to help from lecturers etc.
Consider vectors $v_i \in \mathbf{F^\infty}$. Let $\pi(i)$ denote the ith prime number.
Let $v_j = \sum_{j=1}^{\infty} x_{\pi(i)^j} \space $ Consider the set of vectors, where M is a natural number so finite, $S= \left\{ v_j + U: j \leq M\right\}$. The $v_j$ are linearly independent by construction because they do not share any of the $x_j$.
But if $v_j \in U \space$ then there exists j for any k member of the naturals so that $ k \leq j \space\space\space x_j = 0 $. But we can find $\pi(j)^i >k$ therefore $x_{\pi(j)^i>k} \space\space\space\neq 0$ . This means that $v_j \notin U$. This means that each of the vectors in  S are linearly independent. But you can add another vector to S by considering $v_{(M+1)} = x_{\pi(M+1)} + x_{\pi(M+1)^2} \space+ ....$ But all the vectors in S are still linearly independent. This means that S is unbounded, therefore there are infinitely many linearly independent vectors in $\mathbf{F^\infty}/U$ because each vector in S is a member of it.
thank you for reading
 A: The proof is correct but you assumed or overlooked to prove the following fact.

Proposition. Suppose we have a vector space $V$, subspace $U \subset V$, and infinite dimensional subspace $S \subset V$ such that $S \cap U = \{0\}$. Then $V/U$ is infinite dimensional.

To prove it let $s_1,s_2,\ldots  \in S$ be independent and consider the conjugacy classes $[s_1], [s_2], \ldots $  of the images in $V/U$. We must prove the conjugacy classes are also independent. For consider a finite subset wlog $s_1,\ldots, s_n$ such that $$c_1[s_1] + \ldots + c_n[s_n] \equiv_U 0.$$
This is equivalent to saying
$$c_1 s_1 + \ldots + c_n s_n \in U$$
By the assumption $S \cap U = \{0\}$ the above gives $c_1 s_1 + \ldots + c_n s_n  =0$. Then by linear independence all $c_i =0$ as required.
A: For each $k\in \mathbb{N}$, find a sequence $\{n_{k, m}\}_{m=1}^\infty$ of natural numbers such that the numbers $n_{k, m}$ exhaust $\mathbb{N}$ and whenever $(k,m)\neq (k',m')$ we have $n_{k, m}\neq n_{k', m'}$. Then define a linear map $\Phi: \mathbf{F}^\infty \rightarrow \mathbf{F}^\infty/U$ by
$$\Phi(x)_{n_{k,m}}=x_k$$
Then $\ker(\Phi)=0$ since any non-zero $x\in\mathbf{F}^\infty$ will be mapped to  a sequence which is non-zero infinitely many times. Consequently $\Phi$ is injective so its existence implies that, as vector spaces, we must have $\dim (\mathbf{F}^\infty)\leq\dim (\mathbf{F}^\infty/U).$ In particular $\mathbf{F}^\infty/U$ is infinite-dimensional.
