Let $X\subseteq\mathbb R^n$ be a convex set. Can we always find a sequence $(X_n)_{n\geq 0}$ of sets subject to the following conditions?
- The sequence $(X_n)_{n\geq 0}$ is increasing, that is, $X_0\subseteq X_1\subseteq X_2\subseteq\cdots$.
- The $X_n$ are convex and compact.
- The union $\bigcup_{n\geq 0}X_n$ equals $X$.
I already suggested a solution here, but with no success.
Best Answer
That's not true.
Take $X = \{(x, y) | x^2 + y^2 < 1\} \cup \{(\cos \alpha, \sin \alpha) | \alpha \in [0, 2\pi] \setminus \mathbb Q\}$ - open unit disk plus points with irrational argument on border.
Let $Y_n = \{\alpha | (\cos \alpha, \sin \alpha) \in X_n\}$. If $X_n$ is compact, then $Y_n$ is compact. If $\bigcup_n X_n = X$ then $\bigcup_n Y_n = [0, 2\pi] \setminus \mathbb Q$. But $[0, 2\pi] \setminus \mathbb Q$ isn't $F_\sigma$ set, thus not a countable union of compact sets.