Approximation of closed convex set in separable Banach space

Question

Let $V$ be a separable Banach space with $\{x_1, x_2, \ldots\}$ as its countable dense subset and $K\subset V$ be a closed, convex set. We define $$V_n = \text{span}\{x_1, x_2, \ldots, x_n\}, \quad \forall n\ge 1.$$

It is easy to prove that $\displaystyle \bigcup_{n=1}^\infty V_n$ is dense in $V$.

How can we construct the sets $K_n$ such that $$K_n\subset V_n, K_n \text{ is convex and } \bigcup_{n=1}^\infty K_n \text{ is dense in } K?$$

Thank you very much.

Answer 1

Have you tried answering your questions? The first one is obvious because each x_n is in the union. For the second, take K_n to be the set of those points of K intersected with V_n whose norms do not exceed n.

Answer 2

Hint: Take a countable subset $\{y_i\}$ of $K$, approximate each of these points $y_i$ by points in $V_n$ and take the convex hull.

Answer 3

It is not always possible to construct such $K_n.$

For example consider the case where $\displaystyle \bigcup_{n=1}^\infty V_n \neq V$ then pick $ x \in V \setminus \displaystyle \bigcup_{n=1}^\infty V_n $ and define $K = \{x\}.$