
Re: Connected?
Posted:
May 10, 2013 11:47 AM


On Sat, 4 May 2013, William Elliot <marsh@panix.com> wondered:
>Let Sz = ([0,1] /\ Q)x[1,0] \/ ([0,1] /\ R\Q)x(0,1]
> Sz appears to be connected but not path connected. > Is that correct? How would you proof it?
It is. Sketch of proof of connectedness:
Let A be a nonempty subset of Sz that is both open and closed in Sz. Every stripe {p}x[1,0] or {p}x(0,1] in Sz is connected, so its intersection with A must be the whole stripe or the empty set, and the projection from Sz to [0,1] maps A and its complement to complementary subsets of [0,1]. Proving that these projection images are open will show, because of the connectedness of [0,1], that one of them fills the whole interval, which can only happen if A = Sz. That would prove the connectedness of Sz.
So let p in [0,1] lie in the projection image C of A. If p is irrational then (p, 0) must lie in A and, since A is open, so does an open disk around (p, 0) and its projected image is an open interval around p that lies inside C. If p is rational then an open disk around around (p, 1) must lie in A and its projection contains all rational points in an open interval around p. Each irrational point q in that interval is the limit of a sequence of rational points rn in the interval. All points (rn, 1/n) lie in A, which is closed, so their limit (q, 0) also lies in A and its projected image q is part of C, so once again there is an open interval around p that is part of C. So C is open. Same story for its hypothetically maybe nonempty complement.
Disproof of pathwiseconnectedness is trivial: a path that starts in the upper part of Sz must stay on the same stripe {p}x(0,1] and could only leave through (p, 0), which isn't there.
Leon

