From: Wade Ramey <wrameyxiii@home.remove13.com>
Subject: Re: Stories About Sets: Q10
Date: Sun, 30 Sep 2001 02:48:46 GMT
Newsgroups: sci.math
Summary: Triods (homeomorphic copies of a T-shape in the plane): how many?

In article <3BB57028.BEAD1902@chariot.net.au>,
 Raymond Kennington <raymondk@chariot.net.au> wrote:

> Show that it is not possible to find in the plane a larger than countable set 
> of curves
> having the form of the letter T.

Suppose F is an uncountable pairwise disjoint family of letter T's in the 
standard orientation, all assumed to be similar. Let Fn be the set of those 
T's in F whose diameters are > 1/n, n = 1,2, ... . F is the union of all 
the Fn's, so some Fn is uncountable. We can thus assume the diameter of 
every member of F is, say, > 1.

Let's call the point where the two segments of a given T the "cross point" 
of T. The cross points of the T's in F form an uncounbable set in the 
plane, hence some bounded subset of them is infinite. This implies we have 
a sequence T1, T2, ... in F whose crosspoints converge to some point in the 
plane. Recalling that the diameter of each Tn is > 1, it's pretty easy to 
see now that these Tn's cannot all be disjoint, -><-.

You can jazz this up to cover T's that are are 1-1 affine linear images of 
a standard T, but I'm not sure how you'd deal with curves merely 
homeomorphic to a T.

--W.
==============================================================================

From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: Stories About Sets: Q10
Date: Sat, 29 Sep 2001 23:32:01 -0500
Newsgroups: sci.math

On Sun, 30 Sep 2001, Neo 1061 wrote:

> On Sun, 30 Sep 2001 02:48:46 GMT, Wade Ramey
> <wrameyxiii@home.remove13.com> jacked into the Matrix and remarked
> thus:
> 
> >You can jazz this up to cover T's that are are 1-1 affine linear images of 
> >a standard T, but I'm not sure how you'd deal with curves merely 
> >homeomorphic to a T.
> 
> My proof covers that case. :) It assumes each is two connected curves
> of positive length with one ending on a point belonging to the other,
> and the two curves are not tangent where they join. That's basically
> it.

So the two curves can be joined at a common *endpoint*? Are you saying
that you can't have uncountably many disjoint translates of the letter
T?

But your statement does not cover all homeomorphs of the letter T,
because the curves in such a homeomorph *could* be tangent where they
join. Anyway, it is true that a collection of disjoint homeomorphs of
the letter T in the plane must be countable. Let me repost the
reference, which I posted in February 2000:

R. L. Moore, Concerning triods in the plane and the junction points of
plane continua, Proc. Nat. Acad. Sci. U.S.A. 14 (1928), 85-88.

The result is probably also in Moore's book on Point Set Theory, but I
haven't looked that up. Here is Moore's definition of "triod", which
is more general than "homeomorph of the letter T":

"DEFINITION.--If O, A_1, A_2 and A_3 are four distinct points, and for
each n (1 <= n <=3), r_n is an irreducible continuum from A_n to O and
no two of the continua r_1, r_2 and r_3 have, in common, any point
except O then the continuum r_1+r_2+r_3 is a _triod_, the point O is
an _emanation_point_, and the continua r_1, r_2 and r_3 are _rays_ of
this triod."

Here is the theorem:

"THEOREM 1.--If, in space of two dimensions, G is an uncountable set
of triods, there exists an uncountable subset of G such that every two
triods of this subset have a point in common."

For simpler proofs and generalizations of Moore's triod theorem, do a
literature search on the keyword "triod".