# Show disjoint $\theta$-sets in the plane is countable.

Let $\theta$-set be the union of a circle with one of its chords. Show that any collection of disjoint $\theta$-sets in the plane is countable?

I cannot understand the question and I do not know where to start with, could anyone kindly help? Thanks.

• – Dave L. Renfro Apr 4 '17 at 16:18
• @DaveL.Renfro The "Y result" you link to is easier than showing every disjoint collection of Y's is countable, isn't it? – zhw. Apr 6 '17 at 19:36
• @zhw: You're correct. I commented too quickly (I think I googled-and-wrote this just before I had to leave for some reason) and didn't look carefully enough at what I linked to. Better are ... 8-signs ... AND ... letter T's ... AND ... 8 signs.... – Dave L. Renfro Apr 6 '17 at 20:23
• @zhw: I believe these problems originated from R. L. Moore's 1928 paper Concerning Triods in the Plane and the Junction Points of Plane Continua. Googling the phrase "triodic continua" turns up a lot of literature. Also of interest is Timothy Chow's 6 January 1990 sci.math question. No, not THAT Timothy Chow, but Herbert Enderton did contribute in that sci.math thread. – Dave L. Renfro Apr 6 '17 at 20:29

A chord is just a straight line segment between two points on the circle (like in the letter $\theta$!)
• Could you kindly tell the difference between the two situations (circle vs $\theta$)? I have thought on it but quite do not see the true reason. That has to be related to the fact that the chord is "straight" but I fail to see precisely the details. And what happens to "swallows" (i.e. degenerate Y's, like $\curlyvee$)? Thank so much for your help. – Romeo Apr 10 at 7:18