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

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Leon Aigret

Posts: 31
Registered: 12/2/12
Re: Connected?
Posted: May 10, 2013 11:47 AM
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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 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.
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 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.

Leon



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