Is the subset of all sequences with finite number of non-zero entries a closed subspace of $l^2$?

Question

Letting $l^2$ be the Hilbert space of all infinite sequences of real numbers $x = (x_1, x_2, ..., x_k, ...)$ with $$\|x\|^2 = \sum_{k=1}^\infty |x_k|^2 < \infty$$ and $$<x,y> = \sum_{k=1}^\infty x_k y_k$$

Then let $V_F$ be the subset of all sequnces with only a finite number of non-zero entries. Is this subset a closed subspace of $l^2$?

I think it's immediately clear that this is indeed a subspace, since any combination of such sequences will return another sequence with a finite number of non-zero entries.

The closed property is harder for me intuitively because of the infinite dimension of the space. I want to say that it would be closed, but I have this nagging idea that would lead to it being open. Since $l^2$ can be thought of as the space of all convergent infinite series because of the norm defined on it, we could consider a sequence of partial sums that converge to one of these infinite series $x^*$. In this sequence, each partial sum would be the result of a sequence in $V_F$ that is just a truncated version of $x^*$

Answer 1

You have the right idea. Given $x\in\ell^2$ and $\varepsilon>0$, there exists $N$ such that $$ \sum_{n>N}x_n^2<\varepsilon^2 $$ hence if we define $y_n=x_n$ for $n\leq N$ and $y_n=0$ otherwise, then $y\in V_F$ and $||x-y||_2<\varepsilon$.

Therefore $V_F$ is dense in $\ell^2$, and since it is a proper subset of $\ell^2$, it cannot be closed.