How to prove following set is dense?

Question

Let $\{e_n\}$ be orthonormal basis of infinite dimensional Hilbert space $X$. Let $X_n = \operatorname{span} \{e_1, e_2, \ldots, e_n\}$ for all $n$.

Then how to prove that union of $X_n$ is dense in $X$?

I know that $X_n$ is closed subspace. But how to union is dense?

Answer 1

Take an element $x \in X$, and let $x_n$ be the orthogonal projection of $x$ onto $X_n$. What can you say about $\|x-x_n\|$ as $n$ increases? What conclusion can you draw about any open ball of positive radius around $x$?

Answer 2

That is the definition of a Hilbert basis: $\;\operatorname{span}(e_1,\dots e_n,\dots)$, i.e. the set of all linear combinations $\;\displaystyle\sum_{k\in\mathbf N}\lambda_k e_k\;$ with finite support, is dense in $X$.

Now note that, if each $X_n$ is closed, their union is not necessarily closed. A simple counter-example is the subsets $\;\biggl[\dfrac1n,1-\dfrac1n\biggr]\subset\mathbf R$: their union is the open interval $(0,1)$.