This is a follow-up question to a previous one.
Assume that $K$ is a compact convex set in a Hausdorff locally convex space, and $K$ is metrizable with the induced topology. Then the set $\textrm{ex}(K)$ of extreme points of $K$ is a $G_\delta$ set (countable intersection of open sets).
In the proof of the above lemma, a subset of $K$ is defined as $$ F_n=\{x\in K:\textrm{there exist }\ y,z\in K\textrm{ such that }x=(y+z)/2\textrm{ and }d(x,y)\geq 1/n\}. $$ It is said that $F_n$ is closed, but I don't know why.
Let $d$ be a metric on $K$ defining the induced topology and let $(x_m)$ be a sequence in $F_n$ such that $x_m\to x$ in the metric. We want to show that $x\in F_n$. By the definition of $F_n$, there exist $y_m,z_m\in K$ with $$ x_m=(y_m+z_m)/2\quad \textrm{ and } \quad d(x_m,y_m)\geq 1/n. $$ By the compactness of $K$, there exist $y,z\in K$ as limit points of $(y_m)$ and $(z_m)$ respectively. I want to show that $x=(y+z)/2$. If $(y_m)$ and $(z_m)$ have subsequences $(y_{m_k})$ and $(z_{m_k})$ (with the same index) converging to $y$ and $z$ respectively, then I'm done. When it is not the case, I don't see how to go on.
Here are my questions:
- How can I show that $F_n$ is closed?
- If $K$ is not compact, is $F_n$ still closed?
Answer to the first question.
Since $K\times K$ is compact, $(y_m,z_m)$ has a convergent subsequence $(y_{m_k},z_{m_k})$ so that $y_{m_k}\to y$ and $z_{m_k}\to z$. Now we have $$ x=\frac{y+z}{2} $$ and also $$d(x,y)=\lim_{k\to\infty} d(x_{m_k},y_{m_k})\geq \frac1n.$$