THEOREM There do not exist two disjoint sets $\ A\ B\subseteq\mathbb R^2\ $ which are totally disconected, and which cover the plane: $\ A\cup B=\mathbb R^2$.
PROOF Let $\ A\ B\ $ form a cover of $\ \mathbb R^2\ $ while they are totally disconnected (a proof by contradiction). Thus both are dense hence they have more than one point. Then, since A is disconnected, $\ A=E_0\cup F_0\ $ where the two summands are non-empty, closed with respect to $\ A,\ $ and $\ E_0\cap F_0=\emptyset.\ $ There exist $\ E\ F\ $ closed in $\ \mathbb R^2\ $ such that $ E\cap A = E_0\ $ and $\ F\cap A = F_0.\ $ Then
$$X_0:= E\cap F$$
is a closed subset of $\ \mathbb R^2\ $ which is disjoint with $\ A:\ X_0\cap A=\emptyset.\ $ Consider a continuos function $\ f : E\cup F\rightarrow [-1;1]\ $ such that
- $\ f^{-1}(0)\ =\ X_0$
- $\ f^{-1}(\,[-1;0)\,)\ =\ E\setminus X_0$
- $\ f^{-1}(\, (0;1]\, )\,\ =\ F\setminus X_0$
and let $\ \phi:\mathbb R^2\rightarrow\mathbb R\ $be a continuous extension of $\ f.\ $ We see that $\ X:=\phi^{-1}(0)\ $ is a closed subset of $\ \mathbb R^2\ $ which disconnects sets $\ E_0\ F_0,\ $ thus $\ X\ $ is not totally disconnected. Since $\ X\subseteq B\ $ it follows that $\ B\ $ is not totally disconnected. END of Proof
- Sets $\ A\ B\ $ in the theorem are arbitrary (not closed, nor nothing :-).
Sets $\ A\ B\ $ in the theorem are arbitrary (not closed, nor nothing :-).
- It doesn't matter what $\ X\ $ disconnects--only that disconnects whatever (and that $\ X\ $ is disjoint from $\ A$).
It doesn't matter what $\ X\ $ disconnects--only that disconnects whatever (and that $\ X\ $ is disjoint from $\ A$).
- I've answered the question exactly. Essentially the same proof allows for a more general formulation.
I've answered the question exactly. Essentially the same proof allows for a more general formulation.
