H is finite dimensional

Question

Let $H$ be a Hilbert space and $\lbrace x_ i \rbrace_ i \subset H$ a countable set of vectors that algebraically spans $H$, i.e. every vector in $H$ is a linear combination of some finite subset of $\lbrace x _i \rbrace_ i$. Show that $H$ is finite dimensional.

My attempt:

In this set of vectors I can find a minimal spanning set, i.e basis of $H$, say $\lbrace y_i\rbrace_i$. suppose the basis is infinite. then take a non-zero element $x\in H$ then $x=\sum_{i\geq1}a_iy_i$. but $x$ can be written as linear combination of finitely many elements, say $x=\sum_{k=1}^{n}b_ky_k$. Therefore, $0=\sum_{k=1}^{n}(a_k-b_k)y_k+\sum_{i\geq n+1}a_iy_i$. So that implies the tail of the series converges to $0\Longrightarrow x=0$ (Contradiction.)

If this prove is correct then where are we really using the inner product of Hilbert space and the fact that Hilbert spaces are complete.

Answer 1

Here is a proof, which looks like the one you had in mind. The contradiction is again in finite versus infinite sums.

Let $(y_i)$ be a countable basis of $H$. Then using the Gram-Schmidt orthonormalization process we can construct an orthonormal basis $(z_i)$ of $H$: $(z_i)$ is a basis and $\langle z_i,z_j\rangle = \delta_{i,j}$ for all $i,j\in \mathbb N$.

Define $u_n:=\sum_{k=1}^n \frac1k z_k$. This is a Cauchy sequence in $H$. Hence converging to $u$. One can verify $u=\sum_{k=1}^\infty \frac1kz_k$ by looking at $\langle u-u_n,z_k\rangle$. Let $v\in span(z_1\dots z_n)$ with $v=\sum_{k=1}^n v_kz_k$. Then $$ \|u-v\|_H^2 = \|\sum_{k=1}^n(\frac1k - v_k)z_k\|_H^2 + \|\sum_{k=n+1}^\infty \frac1kz_k\|_H^2 \ge \sum_{k=n+1}^\infty \frac1{k^2} >0. $$ Hence, $u$ cannot be a finite linear combination of the $(z_i)$'s. Contradiction.