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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    $\begingroup$ I'd be happy with a non-explicit way to make such a partition. $\endgroup$ – Anton Geraschenko Oct 6 '09 at 23:57
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    $\begingroup$ Here's a nonexplicit construction. Enumerate (well-order) all possible paths with order-type the continuum. By induction, put one point from each path in each of the sets of the partition. This is possible since each path contains continuum many points and at any stage of the induction, you've only chosen where less than continuum points go. $\endgroup$ – Eric Wofsey Oct 9 '09 at 2:32
  • $\begingroup$ @Eric: that's awesome! What else could you possibly do? $\endgroup$ – Anton Geraschenko Oct 9 '09 at 20:17
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    $\begingroup$ What does "well-order with order-type the continuum"? Maybe "biject the set of paths with the first uncountable ordinal" so that all initial segments of the ordinal are numerable? $\endgroup$ – Mariano Suárez-Álvarez Nov 18 '09 at 6:52
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    $\begingroup$ Mariano: No, you are assuming CH there. It means "biject with the first ordinal in bijection with the reals." $\endgroup$ – Andrés E. Caicedo Oct 31 '10 at 15:59

Let S be a subset of the reals such that S∩[a,b] and Sc∩[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(Scxℚc) and B=(Sxℚc)U(Scxℚ) 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∈ℚf(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 ℚc, so is also constant. So A is totally path disconnected. The argument for B follows in the same way by exchanging S and Sc

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    $\begingroup$ +1 Wow, I really thought that this question was destined to never get a good answer, but that's a pretty explicit, and the proof looks good to me. $\endgroup$ – Anton Geraschenko Oct 10 '09 at 3:00

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!

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  • $\begingroup$ @Gerald: The two sets in my construction are connected and locally connected, yet totally path disconnected. This came up in a recent MO question (mathoverflow.net/questions/46748/…). $\endgroup$ – George Lowther Nov 20 '10 at 21:14
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    $\begingroup$ @Gerald: What is your definition of "totally disconnected"? A subset of square which does not contain any connected subsets but the empty set or 1-element sets, can still have topological dimension = 1. Together with a 0-dimensional set they can cover the whole square. $\endgroup$ – Włodzimierz Holsztyński Feb 24 '13 at 20:42
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    $\begingroup$ Remedial question: what's a reference for the statement "the union of two zero dimensional sets has dimension at most 1"? $\endgroup$ – Greg Martin Mar 13 '13 at 7:25
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    $\begingroup$ "the union of two zero dimensional sets has dimension at most 1" ... Decomposition Theorem $\dim X \le n$ if and only if $X$ can be written as a union of $n+1$ sets, each of dimension $\le 0$. Engleking, Dimension Theory (somewhere on pages 257 to 260). $\endgroup$ – Gerald Edgar Mar 13 '13 at 13:22
  • $\begingroup$ This answer above, as I've pointed out in a previous comment, is not correct. The correct one is given in a new thread/question which has a similar title but without word "path". $\endgroup$ – Włodzimierz Holsztyński Feb 26 '15 at 5:09

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.

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  • $\begingroup$ Images $f(I)$ of paths $f\rightarrow X$ in arbitrary metric spaces $X$ are the same as the connected and locally connected compact subsets of $X$ (Hahn-Mazurkiewicz Theorem). Each of them can be decomposed into a union of two smaller continua, if it contains more than one point. On the other hand the Knaster pseudo-arc $K$ is hereditarily indecomposable (all subcontinua of $K$ are homeomorphic to $K$, how nice!) hence it does not contain any image of a non-constant path. Probably I could use some of it to make the proof above a bit nicer. $\endgroup$ – Włodzimierz Holsztyński Feb 25 '13 at 4:57

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.

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    $\begingroup$ okay, I now see why this doesn't work, sorry. $\endgroup$ – Patrick Tam Oct 31 '10 at 12:17
  • $\begingroup$ @Patrick, see wonderful Hurewicz and Wallman, Dimension Theory :-) $\endgroup$ – Włodzimierz Holsztyński Feb 24 '13 at 21:03

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