An anonymous question from the 20-questions seminar:
Can you explicitly write $\mathbb{R}^2$ as a disjoint union of two totally path disconnected sets?
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Sign up to join this communityAn anonymous question from the 20-questions seminar:
Can you explicitly write $\mathbb{R}^2$ as a disjoint union of two totally path disconnected sets?
Let S be a subset of the reals such that S∩[a,b] and S^{c}∩[a,b] cannot be written as a countable union of closed sets for any a<b. This can be done (this explicit example of a non-Borel set achieves this). Let ℚ be the rationals. Then, A=(Sxℚ)U(S^{c}xℚ^{c}) and B=(Sxℚ^{c})U(S^{c}xℚ) should do it.
The proof is as follows. Suppose that the curve t→(f(t),g(t)) lies in A, and consider a closed bounded interval I. As the curve lies in A, f(I)∩S = f(I∩g^{-1}(ℚ))=∪_{x∈ℚ}f(I∩g^{-1}(x)) is a union of countably many closed sets. By the choice of S, f(I) must be a single point. Hence, f is constant. Then, g is a continuous function mapping into either ℚ or ℚ^{c}, so is also constant. So A is totally path disconnected. The argument for B follows in the same way by exchanging S and S^{c}
I we omit "path" in the formulation, then it cannot be done. I guess that is why it is formulated this way.
Suppose the plane could be written as the union of two totally disconnected sets. Intersect with a closed square to write the square as the union of two totally disconnected sets. But the square is compact, so these two sets are zero dimensional. Finally, the union of two zero dimensional sets has dimension at most 1. So that union cannot be the square.
So, in the construction, your two sets are totally path disconnected, but not totally disconnected!
I'll apply the following simple result:
THEOREM Let $f : I\rightarrow X$ be an arbitrary non-constant continuous function (a path) of interval $I:=[0;1]$ into an arbitrary topological space $X$. Then there exist continuous maps $\alpha:I\rightarrow I$ and $g:I\rightarrow X$ such that $f = g\circ \alpha$, and $g$ is not constant on any non-empty open subinterval of $I$.
Here is a simple positive solution for the question of this thread, and proof:
Let $\mathbb C := \mathbb R^2$ be the complex plane. Let $K \subseteq \mathbb C$ be a Knaster pseudo-arc. Let $$L := i\cdot K := \{i\cdot z : z \in K\}$$
where $i^2=-1$. Let $D$ be a dense countable subset of $\mathbb C$. Define $$A := \left(\bigcup_{d\in D}\ \left(d+K\right)\right)\cup\left(\bigcup_{d\in D}\ \left(d+L\right)\right)$$
where $d+X := \{d+x:x\in X\}$. Finally, let $$B := \mathbb C\setminus A$$
Then $\dim(B) = 0$, and $B$ does not contain any non-constant path.
Also, there does not exist any non-constant continuous map $f : I \rightarrow A$ --indeed, if there was one then we may assume that it is not constant on any open subinterval of $I$. Then the inverse images:
$(\bigcirc^{-1}f)(d+K)\quad$ and $\quad(\bigcirc^{-1}f)(d+L)$
would be 0-dimensional closed subsets of $I$, for every $d\in D$. Thus $I$ would be a countable union of 0-dimensional closed subsets, which is a contradiction. It means that $A$ does not contain any image of any non-constant path.
This completes a positive answer to the Question of this thread.
Let me try. QxQ and its complement in R^2. Does this work? Edit: this does not work, :(. Let me try something else then. Take S = (R\Q)x(R\Q). QxR U RxQ would be S complement. Let p be a prime number and let q_p be in Q. For each q_p, let J_p be the set of all rational numbers times the square root of p. Since Q is countable, there is a bijection between the set of primes and {q x R : q is rational} union {R x q : q is rational}. So let A be the union of all q_p X J_p and J_p X q_p. Take A union S and the complement of that in R^2 to be your disjoint union. Dunno if this works but at least it seems like a better attempt.