If $M$ is the span of a finite orthonormal set in a Hilbert space then $M$ is closed

Question

Suppose that $\{e_1,e_2,\dotsc,e_n\}$ is a finite orthonormal set in a Hilbert space. I want to prove that $M= \operatorname{span}(\{e_1,e_2,\dotsc,e_n\})$ is closed. So what I was thinking is if I can show that $M$ is finite then I have that it is a finite subspace of my Hilbert space therefore it is complete and therefore closed.

But I'm wondering if it's true that if my set is finite then the span of my set is finite. And if thats true how do I go about proving it?

Answer 1

$\mathbb{R}$ is the span of $\{1\}$, so your attempt won't work. But write that $i:M\to H$ given by $i(e_j)=e_j$ is an isometry, and so send the complete space $M$ (because finite dimensional) to the complete and so closed set $M\subset H$, and the end of your proof works well.

Answer 2

Let $X$ be a normed linear space over $S$ where $S=\Bbb R$ or $S=\Bbb C$.

THEOREM. If $Y$ is a closed linear sub-space of $X$ and $v\in X\setminus Y$ then $Z=\{y+sv:y\in Y\land s\in S\}$ is closed.

Proof: Suppose a sequence $(y_n+s_nv)_n$ converges in norm to $z,$ with $y_n\in Y$ and $s_n\in S.$

Case (i). If $(s_n)_n$ is not a bounded sequence then it has a sub-sequence $(s_{n_j})_j$ with $|s_{n_j}| \to \infty$ as $j\to \infty,$ and with each $s_{n_j}\ne 0.$ Then $$0=\lim_{j\to \infty} \|(s_{n_j})^{-1}y_{n_j}+v-(s_{n_j})^{-1}z\|$$ $$\text { and }\quad 0=\lim_{j\to \infty}\| (s_{n_j})^{-1}z\|.$$ $$ \text {But these imply that }\quad \lim_{j\to \infty}\|(s_{n_j})^{-1}y_{n_j}+v\|$$ which in turn implies that $(-v)\in \overline Y=Y,$ and hence $v\in Y$ (because $Y$ is a vector space), contradicting $v\not\in Y.$ So this case is impossible.

Case (ii). If $(s_n)_n$ is a bounded sequence then it has a sub-sequence $(s_{n_i})_i$ converging to some $s\in S.$ Then as $i\to \infty$ we have $y_{n_i}+(s_{n_i}-s)v \to z-sv\;$ and we have $(s_{n_i}-s)v\to 0. \;$ So $y_{n_i} \to z-sv.\;$ So with $y=z-sv$ we have $y\in \overline Y=Y,$ and $$z=(z-sv)+sv=y+sz\in Z.$$

COROLLARY: If $Z$ is a finite dimensional vector sub-space of $X$ then $Z$ is closed.

Proof: If not, then take a counter-example $Z$ with the least possible dimension, $m$ . Clearly $m\ne 0.$ (Note: $\dim Z=0$ iff $Z=\{0\}.$) So let $Z$ be the linear span of $B=(\{y_j: 1\leq j\leq m\},$ with $m>0.$ By the minimality of $m,$ the linear span $Y$ of $B\setminus \{y_m\}$ is closed. (Note: In case $m=1,$ the linear span of $\emptyset$ is defined to be $\{0\}.$) And $y_m\in X\setminus Y.$ By the theorem, $Z=\{y+sy_m:y\in Y\land s\in S\}$ is closed, contrary to $Z$ being a counter-example.