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