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Re: Connected?
Posted:
May 12, 2013 8:52 PM


On Sun, 12 May 2013 00:44:00 0700, William Elliot <marsh@panix.com> wrote:
>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]
Relying on my eyes instead of my memory this time, I see that I interchanged the rationals and the irrationals throughout the following argument, with obvious consequences for its correctness. However, since the problem stays essentially the same for any division of [0,1] into two dense subsets, I'd rather correct for the difference than rewrite the whole thing.
>> > 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].
For P _rational_ that argument is OK. the lower half of the ball is projected onto the rationals in an interval and the upper half to the irrationals.
>> 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.
With p and the rn _irrational_ and q _rational_ (q, 0) lies in Sz and is the accumulation point of {(rn, 1/n)  n in N} and therefore lies in the closure of A, which is 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.
Some extra details: Assume the existence of a path f=(f1, f2): [a,b] > Sz such that f2(a) > 0 and f2(b) <= 0. Since the topology of Sz is the relativization of the topology of R^2, f is also continuous as a mapping from [a,b] to R^2, so f1 and f2 are continuous real functions. That makes f2^(1)(0) a nonempty closed set with minimum t0 > a. So for t in [a, t0) one has f2(t) > 0 and f1(t) must have a constant irrational value q, since it can not have two different values without assuming all intermediate values, which would include rationals. But now it follows from the continuity of f that f(t0) = (q, 0), which lies outside Sz.
Contradicton => bad assumption => no pathwiseconnectedness.
Leon



