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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 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 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].

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 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.

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 non-empty 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.

Leon

Date Subject Author
5/4/13 William Elliot
5/4/13 fom
5/4/13 William Elliot
5/10/13 Leon Aigret
5/12/13 William Elliot
5/12/13 Leon Aigret