NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/math/17371

Text File | 1992-12-23 | 3.3 KB | 65 lines

Newsgroups: sci.math
Path: sparky!uunet!think.com!mintaka.lcs.mit.edu!zurich.ai.mit.edu!ara
From: [e-mail] (Allan Adler)
Subject: Re: more math puzzles
In-Reply-To: [e-mail]'s message of 18 Dec 92 10:14:09 GMT
Message-ID: <[e-mail]>
Sender: [e-mail]
Organization: M.I.T. Artificial Intelligence Lab.
References: <[e-mail]> <[e-mail]>
Date: Wed, 23 Dec 1992 16:53:12 GMT
Lines: 52



I'm not entirely convinced by Macolm Beattie's explanation that the tangent
bundle to L is nontrivial and there is another point he raises that I
would like to address.

Suppose the tangent bundle to L is nontrivial. Then as he points out
there is a nowhere vanishing vector field on L. I don't integrate
vector fields very often, but it seems to me that this implies that
there is an action of R on L, where R is the additive group of real
numbers, which I will refer to as translation on L. Translating by
a real number x will be a mapping from L to itself and any mapping
from L to itself fixes a closed cofinal subset. This is therefore
true for x=1, x= 1/2, x=1/4, and so on. The intersection of countably
many closed cofinal subsets of L is itself closed and cofinal,
so there is a closed cofinal (in particular nonempty) subset of L
every points of which is fixed by all translations. The vector field
vanishes at such a point.

It is not true that every continuous map from R to L is eventually constant.
R can be mapped homeomorphically onyo an open interval of R, hence of
one of the intervals used to make up L. It is, however, true that
any map from L to R is eventually constant. More generally, any
map from L to a metric space is eventually constant. To prove this,
note that L is sequentially compact and that any continuous map must
preserve sequential compactness. So if f: L--> M is continuous and
u is in L then, denoting by Lu the part of L above and including u,
f(L_u) is sequentially compact and hence compact, since M is a metric
space. Therefore, the intersection of f(L_u) is nonempty as u goes
to infinity in L. That  proves there is a u such that f(u) is
taken on cofinally as a value. Since any two cofinal closed subsets
of L meet, it follows that f is eventually constant and equal to f(u).

One application of this is that L is itself not metrizable. Here is another:
it is proved in Singer and Thorpe, e.g. that the product of continuum
many unit intervals is not metrizable. But using the long line, you can
prove that the product of aleph-1 unit intervals is not metrizable:
for each countable ordinal b, you can map L onto the b-th interval that
makes up the long half line by making the map the identity on the interval,
scrunching stuff before the interval into the left endpoint and scrunching
stuff after the interval into the right endpoint. That gives a family
of mappings into [0,1] indexed by the countable ordinals and these maps
separate points of the long half line(which is what I actually proved the
property for, i.e. that maps into metric spaces are eventually constant).
The product of this family of maps is a continuous injective map from the
long half line into the product of aleph-1 copies of [0,1]. Therefore,
that product is not a metric space.

This proves that the long line is good for something. :)

Allan Adler
[e-mail]