I think I have this but it seems a little too easy.
Let $S$ be a collection of points $\{y_j\}$. Denote closure as $\bar{S}$ and closed linear span as $cls(S)$. Suppose $x \in \bar{S}$. If $x$ is not a limit point then it's clearly in $cls(S)$ and there is nothing to prove. Suppose x is a limit point of $S$, then $x$ belongs to every closed linear subspace that contains $S$ so $x \in cls(S)$.
Now suppose $x \in cls(S)$, then $x \in S$ or is a limit point of $S$, either way it's a point in $\bar{S}$
Is this correct?
Best Answer
Here's my proof.
Let $A=\{A_j, j\in J\}$ be the collection of all closed linear subspace containing $S$, where $J$ is the index set. Let $B=\{B_i, i\in I\}$ be the collection of all linear subspace containing $S$, where $I$ is the index set. If $K$ is a set, we denote $\overline K$ the closure of $K$.
We know $\bigcap \limits_{i\in I} B_i$ is a linear subspace. Thus, $\overline {\bigcap \limits_{i\in I} B_i}$ is a closed linear subspace. Therefore, from the definition of $A$, we know $\overline{\bigcap \limits_{i\in I} B_i} \in A$. So, we know $\bigcap \limits_{j\in J} A_j \subseteq \overline{\bigcap \limits_{i\in I} B_i}$.
Next, we let $x\in \overline{\bigcap \limits_{i\in I} B_i}$. Then, there exists a convergent sequence $\{x_n\}$ such that $x_n\in \bigcap \limits_{i\in I} B_i, x_n\rightarrow x$. From the definitions of A and B, we know $\bigcap \limits_{i\in I} B_i \subseteq \bigcap \limits_{j\in J} A_j$. So, $x_n\in \bigcap \limits_{j\in J} A_j$. Moreover, because $\bigcap \limits_{j\in J} A_j$ is closed. Thus, $x\in\bigcap \limits_{j\in J} A_j$. Hence, $\overline{\bigcap \limits_{i\in I} B_i} \subseteq \bigcap \limits_{j\in J} A_j$.
Therefore, we have $\overline{\bigcap \limits_{i\in I} B_i} = \bigcap \limits_{j\in J} A_j$. Done.