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Topic:
Connected?
Replies:
5
Last Post:
May 12, 2013 8:52 PM




Connected?
Posted:
May 12, 2013 3:44 AM


On Fri, 10 May 2013, Leon Aigret wrote: > 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.
For a ball of radius r < 1, the projected image is (pr, p+r)\Q /\ [0,1].
> 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
A is closed within Sz which isn't complete. A is not closed within [,1]x[1,1]. Thus the limit (q,0) is not in A.
> 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.
I'll believe it when I see the details.



