Problem 4

Suppose that $x$ and $y$ are two points in the unit disk $D=\{\vert z\vert<1;z\in\mathbb{R}^2\}$ in $\mathbb{R}^2.$ Using results from class (or otherwise) show that the set $\{tx+(1-t)y;t\in[0,1]\}$ is connected. Using this, or otherwise, show that if $D$ has a decomposition $D=A\cup B$ where $\bar A\cap B=\emptyset$ and $A\cap\bar B=\emptyset$ (closure in $D)$ then one of them must be empty. What does this say about $D?$

Proof. [Solution] For two fixed points $x,y\in D,$ the map $f:[0,1]\longrightarrow tx+(1-t)y$ is continuous since each component is a linear function of $t,$ hence continuous. By theorems from class, the interval $[0,1]$ is connected and the image of a connected set under a continuous map is connected. Thus the line $L_{x,y}=\{tx+(1-t)y;t\in[0,1]\}$ is connected. It lies in $D$ by the triangle inequality

$$\vert tx+(1-t)y\vert\le t\vert x\vert+(1-t)\vert y\vert<1.$$

Now, suppose that $D=A\cup B$ is a decomposition as described above and that neither $A$ nor $B$ is empty. Thus we can choose $x\in A$ and $y\in B.$ Now consider the decomposition of $L_{x,y}=A'\cup B'$ where $A'=A\cap L_{x,y}$ and $B'=B\cap L_{x,y}.$ Since $L_{x,y}$ is the continuous image of a compact set it is also compact, and hence closed. Thus the closures in $L_{x,y}$ satisfy $\overline{A'}\subset \bar A\cap L_{x,y}$ and $\overline{B'}\subset \bar B\cap L_{x,y}.$ Hence we deduce that $\overline{A'}\cap B'=A'\cap\overline{B'}=\emptyset.$ So using the connectedness of $L_{x,y}$ we deduced that one of $A'$ of $B'$ must be empty, but this contradicts the assumption that both $A$ and $B$ are non-empty. Thus one of $A$ or $B$ must in fact be empty and this means that $D$ itself must be connected. $\qedsymbol$

Remark: What you are showing here is that `pathwise connectedness implies connectedness'.