$\{x_n\} \to x$ iff $\bigcap_{n=1}^\infty K_n = \{x\}$

Question

Let $\{x_n\}$ be a sequence in $\mathbb{R}^k$ and let $K_n$ be the intersection of all closed convex sets that contain $x_m$ for all $m \ge n$. How do I show that $\{x_n\}$ converges to $x$ if and only $\bigcap_{n=1}^\infty K_n = \{x\}$?

Answer 1

We first demonstrate that if $\{x_n\}$ converges to $x$, then $\bigcap_{n=1}^\infty K_n = \{x\}$. For $\epsilon > 0$, let $N$ be large enough such that $n > N$ implies $|x_n - x| < \epsilon$. Then for all $n > N$, we have that the set $T_\epsilon = \{(y_1, \dots, y_n) \text{ }|\text{ }|x_i - y_i| \le \epsilon \text{ }\forall i\}$ is clearly a closed convex set containing $x_n$ for all $n > N$. It follows then, that $$\bigcap_{n=1}^\infty K_n \subset \bigcap_{\epsilon \in \mathbb{R}^+} T_\epsilon = \{x\}.$$We also note that since $x_n \to x$, and $K_n$ contains $x_m$ for $m \ge n$, $K_n$ has a sequence converging to $x$, so $x \in K_n$ for all $n$, and the result follows.

We now demonstrate if $\bigcap_{n=1}^\infty K_n = \{x\}$, then $\{x_n\} \to x$. We note that for any $\epsilon > 0$ the sphere $S_\epsilon(x) := \{y \text{ }|\text{ }d(x, y) = \epsilon\}$ is covered by finitely many of the sets $K_n^c$, and since $K_{n+1} \subset K_n$, $K_{n+1}^c \supset K_n^c$, we see that in fact $S_\epsilon(x)$ is contained in $K_n^c$ for some $n$. Fix such an $n$, and from now on assume $m > n$. $S_\epsilon(x) \subset K_m^c$. Thus $K_m$ is covered by two disjoint open sets, namely $\{y \text{ }|\text{ }|x - y| < \epsilon\}$ and $\{y\text{ }|\text{ }|x-y| > \epsilon\}$. Since $K_m$ is convex, $K_m$ is connected, it cannot intersect both of these open sets. Since $x \in K_m$, we conclude $K_m \subset N_\epsilon(x)$ for $m > n$, whence $|x_m - x| < \epsilon$ for $m > n$, and the result follows.

Answer 2

Let ${\cal H_x}$ be the set of closed halfspaces containing $x$ in their interior. Hahn Banach shows $\cap_{H \in {\cal H_x}} = \{x\}$. Suppose $x_n \to x$, and select $H \in {\cal H_x}$. Then there is some $N$ such that $x_n \in H$ for all $n \ge N$ and so we have $K_n \subset H$ for all $n \ge N$. Consequently, $\cap_n K_n \subset H$. Hence we have $\cap_n K_n \subset \cap_{H \in {\cal H_x}} = \{x\}$. Since the $K_n$ are closed, we have $x \in K_n$ for all $n$, hence $\{x\} \subset K_n$, from which we have $\cap_n K_n = \{x\}$.

Note that if we have points $y_n \in K_n$ and $y_n$ has some accumulation point $y$, then we must have $y \in K_n$ since the $K_n$ are closed and nested, and hence $y \in \cap_n K_n$.

Now suppose $\cap_n K_n = \{x\}$. Let $\epsilon>0$ and let $C= \bar{B}(x,\epsilon)$. Then I claim that $K_n \subset C$ for some $n$. If not, then choose $c_n \in C \setminus K_n$. By convexity, $y_n = x+ {1 \over \|c_n-x\|} (c_n-x) \in K_n$ (note that $\|c_n-x\| >1$) and $\|y_n-x\|=1$. Hence $y_n$ has an accumulation point $y$, and we have $y \in \cap_n K_n$. Since $\|y-x\| =1$, this is a contradiction. Hence there is some $n$ such that $K_n \subset C$. It follows that $x_n \to x$.

To see why the latter doesn't work in infinite dimensions, consider the sequence $0,e_2,0,e_3,0,e_4,...$ in $l^2$. Let $K_n = \overline{\operatorname{co}} \{ x_k \}_{k \ge n}$, and note that $\{ x_k \}_{k \ge 2n} \subset \overline{\operatorname{sp}}\{e_n, e_{n+1},... \}$. It follows that $\cap_n K_n = \{0\}$, but $x_n \not\to 0$.