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Let $V$ be a closed subspace of $\mathcal H$ and if $y \in \mathcal H$ but $y \notin V$ prove that $W = \text{Span}(V \cup\{y\})$ is a closed subspace.

Here's my attempted proof.

Proof:

Note first that theorem 4.11 in Big Rudin says that $\mathcal{H} = V \oplus V^\perp$. Therefore, since $y \in \mathcal{H}$ but $y \notin V$, $y = z+z^\perp$ with $z \in V$ and $z^\perp \in V^\perp$. Now, since span is the set of all finite linear combinations, $$\text{Span}(V \cup\{y\}) = \text{Span}(V \cup\{z+z^\perp\}) = \text{Span}(V \cup\{z^\perp\})$$ because $z \in V$ already. Thus, it suffices to show the rhs is closed. Now, take a sequence in this subspace, $\{x_n\}_{n \in \mathbb{N}} \to x \in \mathcal{H}$. We need to show $x \in \text{Span}(V \cup\{z^\perp\})$. Let $\epsilon >0$; since $x_n \to x$, there exists $n$ such that

$$\|x_n - x\|^2 < \epsilon.$$ We know there exists linear maps (since $V$ is closed) $P: \mathcal{H} \to V$ and $P: \mathcal{H} \to V^\perp$ giving unique orthogonal decomopsitions. We have $P(x_n) \in V$, so that $\beta_nz^\perp=Q(x_n)$, and let $x = P(x)+Q(x)$.

Then, using the fact that $P(x_n) - P(x) \perp Q(x_n)-Q(x)$, \begin{eqnarray*} \|x_n-x\|^2 &=& \|(P(x_n)+\beta_nz^\perp)-(P(x)+Q(x))\|^2\\ &=& \|P(x_n) - P(x)\|^2 + \|\beta_nz^\perp - Q(x)\|^2 \\ &<& \epsilon \end{eqnarray*}

and the above says precisely that $P(x) = \lim \limits_{n \to \infty} P(x_n)$ and $Q(x) = \lim \limits_{n \to \infty} Q(x_n) = \beta z^\perp$ in norm for some $\beta \in \mathbb{C}$. \ Thus, since $V$ is closed, $P(x) \in V$, so thus $x \in \text{Span}(V \cup\{z^\perp\})$ since it is of the form $P(x) + \beta z^\perp$.

Are there any issues with this proof? I've EDITED this post substantially

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