Special Counterexample to Kakutani's Fixed-Point Theorem

Question

For reference, here is the statement of Kakutani's fixed point theorem.

Let $X$ be a compact, convex subset of $\mathbb{R}^n$ and let $f:X\to \mathcal{P}(X)$ be a set-valued function such that $f(x)$ is convex and non-empty for every $x,$ and so that the graph of $f$ is closed. Then $f$ has a fixed point, i.e. there exists $x\in X$ for which $x\in f(x).$

There are obvious counterexamples when each of the conditions are not satisfied. My question is whether or not there exists $f$ such that $f(x) = \emptyset$ for exactly one $x$ such that $f$ satisfies all other conditions but does not have a fixed point.

Answer 1

If there is exactly one $x_0 \in X$ with $f(x_0) = \varnothing$, then the graph

$$\Gamma(f) = \{ (x,y) : x \in X, y \in f(x)\}$$

can only be closed if $X = \{x_0\}$. For otherwise, we'd have a sequence $(x_n)$ in $X\setminus \{x_0\}$ converging to $x_0$, and choosing any $y_n \in f(x_n)$, since $X$ is compact, there'd be an accumulation point $y_0$ of the sequence $y_n$. Then

$$(x_0,y_0) \in \overline{\Gamma(f)}\setminus \Gamma(f),$$

so the graph wouldn't be closed.