2

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.

Zuy
  • 4,656

1 Answers1

1

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.

mihaild
  • 15,368
  • What's an $F_\sigma$ set? – Zuy Oct 11 '22 at 15:53
  • Set that is countable union of closed sets https://en.wikipedia.org/wiki/F%CF%83_set. See proof that irrational numbers are not such in, for example, https://math.stackexchange.com/a/1321896/659499. – mihaild Oct 11 '22 at 15:54
  • Fantastic, thank you a lot. Never would I have guessed that this is not possible... – Zuy Oct 11 '22 at 16:04
  • I’m having trouble believing this, although your argument makes sense. What’s wrong with (up to translation) taking the sets $X_n = (1-1/(n+1))X$? – While I Am Oct 11 '22 at 16:12
  • @William these sets don't care about border. I think if $X$ contains an inner point, then $\cup X_n$ will be interior of $X$ (but I am not sure, although in this case it will be). Also, you need to take closure of scaled $X$, as just scaled $X$ don't have to be closed. – mihaild Oct 11 '22 at 16:35
  • @William The $X_n$ you suggest will never reach the border points of $X$, will they? – Zuy Oct 11 '22 at 19:30
  • Right you are! Thanks for the clarification. – While I Am Oct 11 '22 at 19:44