TITLE: Union of closed convex sets
QUESTION [8 upvotes]: Let $(K_n)_{n=1}^\infty$ be a sequence of convex, compact sets in the space $\mathbb{R}^k$ such that
$$K_1\subsetneqq K_2\cdots \subsetneqq K_n\cdots.$$
Put $K=\bigcup_{n=1}^\infty K_n$. Assume that $K$ is bounded. Can we deduce that $K$ is always not closed?
This is true for $k=1$. It is also true if the sequence is such that $K_n$ is contained in the interior of $K_{n+1}$. Is it possible to prove this in the general case or is there a counterexample?

REPLY [12 votes]: The claim is obviously true for $k=1$.
It is not true for $k\ge2$:
Define the sets
$$
S:= \{ x\in \mathbb R^k: \|x\|_2\le 1, \ x_1=0\},
$$
which is a closed, $k-1$-dimensional ball,
and
$$
B_n = \{x\in \mathbb R^k: \|x\|_2\le 1, \ x_k \le 1- \frac1n\},
$$
which is a ball that is cut.
Then
$$
K_n = conv(S \cup B_n)
$$
is an increasing sequence of convex and compact sets.
And its union $K=\cup K_n$ is the closed unit ball:
Let $\|x\|_2\le 1$. If $x_k=1$ then $x\in S \subset K_n$, if $x_k<1$ then $x\in B_n$ for all $n$ such that $1-\frac1n\ge x_k$, or equivalently for all $n$ with $n\ge \frac1{1-x_k}$.