Equivalence of Definitions of Closed Linear Span

== Theorem ==
Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$, and let $A \subseteq H$ be a subset.
:$(1): \quad \ds \vee A = \bigcap \Bbb M$, where $\Bbb M$ consists of all closed linear subspaces $M$ of $H$ with $A \subseteq M$
:$(2): \quad \vee A$ is the smallest closed linear subspace $M$ of $H$ with $A \subseteq M$
:$(3): \quad \ds \vee A = \map \cl {\set {\sum_{k \mathop = 1}^n \alpha_k f_k: n \in \N_{\ge 1}, \alpha_i \in \Bbb F, f_i \in A} }$, where $\cl$ denotes closure
== Proof ==
Let the proposition $(1)$ hold:
Assume the closed linear subspace $M'$ contains the set $A$.
Then because $M' \in \Bbb M$:
:$\vee A \subseteq M'$
The intersection of arbitrary family of subspaces is a subspace.
For suppose $\CC$ is a family of subspaces.
Denote $\bigcap \CC = \set {f\in H: \forall V \in \CC: \exists f \in V}$
If $f \in \bigcap \CC$, then for any $V \in \CC$, $f \in V$ there exists $a f \in V$ for $a \in \Bbb F$.
If $f, g \in \bigcap \CC$, we have for any $V \in \CC$:
:$f + g \in V$
Therefore, $\vee A$ is a subspace.
It is closed, as intersection of arbitrary family of closed sets is closed.
The choice of $M'$ is arbitrary.
Hence, $(2)$ holds.
This is the proof $(1) \implies (2)$.
Next, assume $(2)$ holds.
Since $A \subseteq \vee A$:
:$\vee A \in \Bbb M$
$\vee A$ is the smallest one in $\Bbb M$.
Hence:
:$\ds \vee A = \bigcap \Bbb M$
Thus we have established the equivalence between $(1)$ and $(2)$.
Finally we come to $(3)$:
We prove that $\map \cl {\map \span A}$ is a subspace.
Let $f \in \map \cl {\map \span A}$. 
By :
:$H$ is a metric space.
From Point in Closure of Subset of Metric Space iff Limit of Sequence:
:there is a sequence $\sequence {f_i}$ in $\map \span A$ that its limit is $f$.
By the continuity of the function of the multiplication of numbers of the field $\Bbb F$ and points in $H$, the sequence $\sequence {af_i}$ with $a\in \Bbb F$ converges to $af$.
Because all the terms of the sequence $\sequence {af_i}$ are points in $\map \span A$, we have $af\in \map \cl {\map \span A}$.
The proof regards the addition of two vectors is by a similar manner.
If $f \in \map \cl {\map \span A}$ and $g\in \map \cl {\map \span A}$, then a sequence $\sequence {f_i}$, converges to $f$, and a sequence $\sequence {g_i}$, converges to $g$, of $\map \span A$ are given.
By the continuity of the function of the addition of points in $H$, $\sequence {f_i + g_i}$ converges to $f+g$. 
Because each terms of the sequence $f_i + g_i$ is in $\map \span A$, we have $f + g \in \map \cl {\map \span A}$.
Notice that in general, if $H'$ is a subspace, then $\map \cl {H'}$ is a subspace.
Finally we come to the proof of equivalence of $(2)$ and $(3)$:
$\map \cl {\map \span A}$, the closed linear subspace, contains $\map \span A$ and thus contains $A$.
For any closed linear subspace $M$ which contains $A$, $\map \span A \subseteq M$, since the linear span of $A$ is the smallest subspace that contains $A$.
Because $M$ is closed, $\map \cl {\map \span A} \subseteq M$.
$\map \cl {\map \span A}$ is the the smallest closed linear subspace $M$ of $H$ with $A \subseteq M$.
Because arbitrary intersection of closed sets is closed and arbitrary intersection of subspaces is a subspace, the smallestness is unique.
Hence $\vee A = \map \cl {\map \span A}$.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Exercise $4$
* P130, Lemma $21.2$
Category:Hilbert Spaces