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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.

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  • $\begingroup$ $K_n:=V_n\cap K $? $\endgroup$ – Bananach Sep 4 '17 at 7:13
  • $\begingroup$ @Bananach: This does not work. Let $V = \mathbb R$, $\{x_i\}$ be the rational numbers and $K = \{\pi\}$. $\endgroup$ – gerw Sep 4 '17 at 9:05
  • $\begingroup$ @gerw I didn't check if it works, but I don't understand your example: wouldnt $ V_n=\mathbb{R}$ and $ K_n=\{\pi\}$? $\endgroup$ – Bananach Sep 4 '17 at 14:16
  • $\begingroup$ @Bananach: Of course. I missed the span... $\endgroup$ – gerw Sep 4 '17 at 19:46
  • $\begingroup$ Please specify what you mean by "dense in $K$" here. If you expect $K_n\subset K$, the answer is "can't be done". If $K_n\subset K$ is not required, then let $K_n=V_n$. $\endgroup$ – user357151 Sep 5 '17 at 22:35
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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.

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    $\begingroup$ This does not work. Let $V = \mathbb R$, $\{x_i\}$ be the rational numbers and $K = \{\pi\}$. $\endgroup$ – gerw Sep 4 '17 at 9:06
  • $\begingroup$ Are thinking of the span of $x_i$'s as the set of rational linear combinations? If $V=R$ then each $V_n$ is V. $\endgroup$ – Kabo Murphy Sep 15 '17 at 11:27
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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.

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  • $\begingroup$ Could you be more specific please? $\endgroup$ – Tien Kha Pham Sep 4 '17 at 10:34
  • $\begingroup$ Did you try to follow my hint? At which step are you struggling? I will not give you a step-by-step solution. $\endgroup$ – gerw Sep 4 '17 at 11:12
  • $\begingroup$ I'm considering whether the set $\{y_n\}$ is necessarily dense in $K$? Following your hint, I approximate each $y_i$ by a sequence $x_{i, n}$ such that $x_{i, n}\in V_n$ and $x_{i, n}\to y_i$. For each $n$, I define $S_n$ is the convex hull of $\{x_{i, n}: i = 1, 2, \ldots\}$, and set $K_n = S_n \cap K$. Howerver, it seems that defining $K_n$ in this way does not work. $\endgroup$ – Tien Kha Pham Sep 4 '17 at 12:04
  • $\begingroup$ Oh, you want to have $K_n \subset K \cap V_n$? Then it cannot work since $K \cap V_n$ might be empty. $\endgroup$ – gerw Sep 4 '17 at 19:48
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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 $K = \{x\}.$

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