Show equivalence between $\overline{\bigcup_{n=1}^{\infty} \operatorname{span}(\{x_j\}_{j=1}^n)}$ and $\operatorname{span}(\{x_j\}_{j=1}^{\infty})$.

Question

Let $\{x\}_{j=1}^{\infty}$ be a sequence in a Hilbert space, $\mathcal{M}_n = \operatorname{span}(\{x_j\}_{j=1}^n)$, $\mathcal{M}_{\infty} = \overline{\bigcup_{n=1}^{\infty} \mathcal{M}_n}$ and $\mathcal{N} = \operatorname{span}(\{x_j\}_{j=1}^\infty)$. Herman J. Bierens says they are equivalent for $\mathcal{M}_{\infty}$ and $\mathcal{N}$ in Definition 1 on page 2 at http://www.personal.psu.edu/hxb11/WOLD.PDF. Although Herman gives proofs for $\mathcal{M}_n$ being a Hilbert space for finite $n$, I can not work out the equivalence between $\mathcal{M}_{\infty}$ and $\mathcal{N}$. Any help?

In addition, Herman says $\sigma_n^2$ is non-increasing in $n$ after the formula (39) on page 18. But it seems no information for this claim. Any explanation? Thanks.

Answer 1

He is defining $\operatorname{span} \{x_j\}_{n=1}^\infty$ to be $\overline{\bigcup_{n=1}^\infty \operatorname{span} \{x_j\}_{j=1}^n}$. If you define the span of an infinite set to be a set of finite linear combinations, you will find that span isn't necessarily closed, and if you define the span to be a set of infinite linear combinations, well it's not necessarily well-defined (take any divergent sum)