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.


 * $(1): \quad \displaystyle \vee A = \bigcap \Bbb M$, where $\Bbb M$ consists of all closed linear subspaces $M$ of $H$ with $A \subseteq M$
 * $(2): \quad \displaystyle \vee A$ is the smallest closed linear subspace $M$ of $H$ with $A \subseteq M$
 * $(3): \quad \displaystyle \vee A = \map \cl {\set {\sum_{k \mathop = 1}^n \alpha_k f_k: n \in \N_{\ge 1}, \alpha_i \in \Bbb F, f_i \in A} }$, where $\cl$ denotes closure

Proof
Let the proposition (1) holds; assume the closed linear subspace $M'$ contains the set $A$, then because $M' \in \Bbb M$, we have


 * $ \qquad \vee A \subseteq M' $

The intersection of arbitrary family of subspaces is a subspace. For suppose $\mathcal C$ is a family of subspaces.

Denotes $\bigcap \mathcal C =\{f\in H|\text{ for any }V\in \mathcal C \text{, there }f\in V \}$

If $f\in \bigcap \mathcal C$, then for any $V\in \mathcal C $, $f\in V$ there $af\in V$ for $a\in \Bbb F$. If $f\text{, }g\in \bigcap \mathcal C$, we have for any $V\in \mathcal C $, $f+g\in V$.

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)=>(2).

Next if (2) holds. Since $A\subseteq \vee A$, $\vee A \in \Bbb M$.

$\vee A$ is the smallest one in $\Bbb M$; hence


 * $\qquad \displaystyle \vee A = \bigcap \Bbb M$

We have established the equivalence between (1) and (2).

Finally we come to (3):

We prove that $\cl(\text{span}(A))$ is a subspace.

Let $f\in \cl(\text{span}(A))$.

$H$, the Hilbert space, is a metric space, which satisfies the first countability axiom. For reason then one shall consider the collection of open balls which all are centred at a point and all have their length of radiuses be rational.

We then need "the sequence lemma," which is:


 * Let $A$ be a subset of a topological space $X$. If there is a sequence of points of $A$ converging to $x$, then $x\in \cl(A)$;
 * the converse holds if $X$ is first-countable.

So there is a sequence $\{f_i\}$ in $\text{span}(A)$ 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 $\{af_i\}$ with $a\in \Bbb F$ converges to $af$. Because all the terms of the sequence $\{af_i\}$ are points in $\text{span}(A)$, we have $af\in \cl(\text{span}(A))$.

The proof regards the addition of two vectors is by a similar manner. If $f\in \cl(\text{span}(A))$ and $g\in \cl(\text{span}(A))$, then a sequence $\{f_i\}$, converges to $f$, and a sequence $\{g_i\}$, converges to $g$, of $\text{span}(A)$ are given. By the continuity of the function of the addtion of points in $H$, $\{f_i+g_i\}$ converges to $f+g$. Because each terms of the sequence $f_i+g_i$ is in $\text{span}(A)$, we have $f+g\in \cl(\text {span}(A))$.

Notice that in general, if $H'$ is a subspace, then $\cl(H')$ is a subspace.

Finally we come to the proof of equivalence of (2) and (3):

$\cl(\text{span}(A))$, the closed linear subspace, contains $\text{span}(A)$ and thus contains $A$. For any closed linear subspace $M$ which contains $A$, $\text{span}(A)\subseteq M$, since the linear span of $A$ is the smallest subspace that contains $A$. Because $M$ is closed, $\cl(\text{span}(A))\subseteq M$. $\cl(\text{span}(A))$ 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 $\vee A=\cl(\text{span}(A))$.