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

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William Elliot

Posts: 2,637
Registered: 1/8/12
Posted: May 12, 2013 3:44 AM
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On Fri, 10 May 2013, Leon Aigret wrote:
> On Sat, 4 May 2013, William Elliot <> 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 non-empty 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 (p-r, 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 non-empty complement.
> Disproof of pathwise-connectedness 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.

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