Weakly closed and stable for convex combinations with rational coefficients implies strongly closed

Question

It is a standard result that for a convex subset of a Banach space weak-closure and strong-closure coincide. Is this result still true for a subset which is convex only for combinations with rational coefficients?

Answer 1

Let the set $M\subset X$, where $X$ is a Banach space.

1) Let first have that $M$ is convex for rational coefficients and that $M$ is norm-closed. This means that for each $\lambda\in\mathbb Q:\quad \lambda x_1+(1-\lambda) x_2\in M$ for $x_1,x_2\in M$. Let $r\in\mathbb R$ and $\{\lambda_n\}\subset\mathbb Q: \lambda_n\rightarrow r\in \mathbb R$. For each $n$: $\lambda_n x_1+(1-\lambda_n)x_2 \in M$ and after letting $n\rightarrow \infty$ we see $\lambda_n x_1+(1-\lambda_n)x_2\rightarrow rx_1+(1-r)x_2\in M$ because $M$ is norm-closed. Therefore $M$ is convex also with real coefficients.

2) If we have that $M$ is convex with rational coefficients and also $M$ is weakly closed, then it is also norm-closed. Therefore applying the above argument we see that again $M$ is convex for real coefficients.