The space of sequences which are eventually zero in $l^2$ is not a Hilbert space.

Question

Define $V$ to be the space of sequences which are eventually zero, i.e.

$$V=\bigcup_{N=1}^{\infty}\{(x_n)_{n=1}^{\infty}\in l^2: x_n=0 \; \text{for}\; n\ge N\}.$$

Is $V$ a Hilbert space with respect to the $l^2$ inner product?

I don't think this is a Hilbert space. The example I came up with is $x^n=(1,1/2,\dots,1/n,0,\dots)$. Then $x^n\in V$, and $||x^n-x^m||_2\to 0$ as $m,n \to \infty$, by the tail convergence as $\sum 1/n^2$ converges. Also, this sequence clearly converges to $(1,1/2,1/3,\dots, 1/n, 1/(n+1),\dots)$, by the same reasoning as above, yet this element is not in $V$.

This is the simplest example I could come up with, but what other examples aside from the harmonic series can we think of? I would appreciate any suggestions.

Answer 1

All the answers are very nice, but I just want to add the fact that -: "Every infinite dimensional Banach space can't have countable basis." (Baire Category Theorem)

Answer of your question is not too far from here.