Path Connectedness and fixed points

Question

We have the following given to us,

Let $α, β \colon [0, 1] \to [0, 1]$ be (not necessarily continuous) functions such that

• $α(x) ≤ β(x)$, for all $x ∈ [0, 1]$.
• The set $K = \{\,(x, y); α(x) ≤ y ≤ β(x)\,\}$ is closed in $\mathbb R^2$ .

Show that there exists $t ∈ [0, 1]$ such that $(t, t) ∈ K$.

We think that if $K$ is closed, then it must be path-connected. In which case, we have a continuous function from $[0,1]$ to $[0,1]$ that is contained in $K$. Therefore can use the intermediate value theorem to show that this function must have a fixed point, which would conclude our argument.

Our difficulty is that we are unable to prove that $K$ is actually path-connected, even though every example of $α$ and $β$ that we can think of satisfying the required properties make $K$ path connected.

Answer 1

Let $A=\{\,x\in[0,1]:\alpha(x)>x\,\}$. Then $A$ is an open subset of $[0,1]$: If $\alpha(x_0)>x_0$, then $K$ is disjoint to an open neighbourhood of the compact set $\{x_0\}\times[0,x_0]$, hence $\alpha(x)>x$ in a neighbourhood of $x_0$. Similarly, $B=\{\,x\in[0,1]:\beta(x)<x\,\}$ is open. Since $A\cap B=\emptyset$ and $[0,1]$ is connected, we conclude that $A\cup B\ne [0,1]$. Let $t\in[0,1]\setminus (A\cup B)$. Then $\alpha(t)\le t\le\beta(t)$, i.e. $(t,t)\in K$.

Answer 2

You only need that $K$ is connected, so suppose $K$ can be divided into $2$ nonempty disjoint closed sets $A,B$.

$K$ is composed of vertical segments and since they are connected, every such segment is entirely in $A$ or in $B$. Therefore the projection of $A$ and $B$ onto the $x$ axis divides the interval $[0,1]$ into $2$ nonempty disjoint closed sets, which is not possible.