$\cap_{n}{ (C_n + D_n)} = \cap_{n}{C_n} + \cap_{n}{D_n}.$

Question

Let $X$ be a separable Banach space. By $w$ we shall indicate the weak topology on $X$.

Let $\{C_n\}$ consist of $w$-compact and $\{D_n\}$ of $w$-closed subsets of $X$. Assume that $\{C_n\}$ and $\{D_n\}$ are both nonincreasing.

Then: $$\cap_{n}{ (C_n + D_n)} = \cap_{n}{C_n} + \cap_{n}{D_n}.$$

Proof:

Evidently, the right-hand side is contained in the left side. The proof of the converse inclusion is straightforward:

let $x$ be an element of the intersection on the left, i.e., for every $n$ there exist $c_n \in C_n$ and $d_n \in D_n$ with $x = c_n+d_n$. Now by $w$-compactness of $C_1$ and the Eberlein-Smulian theorem, a subsequence $\{c_{n_j}\}$ of $\{c_{n}\}$ will converge to some $c \in X$. Of course, then $\{d_{n_j}\}$ converges to $x-c$. It follows immediately from the monotone inclusions that $c\in \cap_n C_n$ and $x -c \in \cap_nD_n$.

My problem:

Why: "By $w$-compactness of $C_1$ and the Eberlein-Smulian theorem, a subsequence $\{c_{n_j}\}$ of $\{c_{n}\}$ will converge to some $c \in X$"?

Answer 1

Compactness of $C_1$ is suffcient because $C_n\supseteq C_{n+1}$ for all $n$ is part of the hypothesis and therefore $c_n\in C_1$ for all $n$.