3 point compactification??
Paul R. Chernoff
I am told that, via the theory of "ends", one can show that the
real line R has no 3 point (Hausdorff) compactification -- that is,
there does not exist a compact Hausdorff space X, containing a homeomorphic
image of R as a dense subspace, such that X\R has exactly 3 points.
(The theory of ends shows much, much more, in fact.)
Since the theory of ends is rather exotic, at least to this humble
functional analyst, I ask if anyone can supply a direct, fairly short,
and reasonably intuitive proof that R has no 3 point compactification.
***************************************************
Paul R. Chernoff
Department of Mathematics
University of California
Berkeley, CA 94720
cher...@math.berkeley.edu
***************************************************
--
# Paul R. Chernoff cher...@math.berkeley.edu #
# Department of Mathematics 3840 #
# University of California "Against stupidity, the gods themselves #
# Berkeley, CA 94720-3840 struggle in vain." -- Schiller #
artfl...@my-deja.com
cher...@math.berkeley.edu (Paul R. Chernoff) wrote:
>
> I am told that, via the theory of "ends", one can show that the
> real line R has no 3 point (Hausdorff) compactification -- that is,
> there does not exist a compact Hausdorff space X, containing a homeomorphic
> image of R as a dense subspace, such that X\R has exactly 3 points.
> (The theory of ends shows much, much more, in fact.)
>
> Since the theory of ends is rather exotic, at least to this humble
> functional analyst, I ask if anyone can supply a direct, fairly short,
> and reasonably intuitive proof that R has no 3 point compactification.
Let R denote the real line and let T be a three-point compactification of
R. Let a,b,c denote the three points of T-R. Let A,B,C be open
neighborhoods (in T) of a,b,c, respectively; because T is Hausdorff, we
can and do assume that A,B,C are pairwise disjoint. For each point x of
R let I(x) be the open interval of length 1 centered at x. The
collection {A,B,C,I(x) (x in R)} is an open cover of T. By compactness
of T there is a finite subset of R such that {A,B,C,I(x) (x in F)} is a
(finite) cover of T. The union of the I(x) (x in F) is contained in some
interval J of the form [-M,M]. Each of the sets A,B,C is unbounded (in
R), and either (i) there must be two of these sets whose intersection
with (-infinity,M) is non-empty, or (ii) there must be two of these sets
whose intersection with (M,+infinity) is non-empty. Without loss of
generality we assume the latter to be the case, and relabeling if
necessary, that the sets in question are A and B. Then
A* := [A intersect (M,+infinity)] and B* := [B intersect (M,+infinity)]
are open subsets of R with union equal to (M,+infinity). This
contradicts the fact that (M,+infinity) is connected, so T cannot exist
after all.
--
A.
Sent via Deja.com http://www.deja.com/
Before you buy.
Stephen Montgomery-Smith
>
> I am told that, via the theory of "ends", one can show that the
> real line R has no 3 point (Hausdorff) compactification -- that is,
> there does not exist a compact Hausdorff space X, containing a homeomorphic
> image of R as a dense subspace, such that X\R has exactly 3 points.
> (The theory of ends shows much, much more, in fact.)
>
> Since the theory of ends is rather exotic, at least to this humble
> functional analyst, I ask if anyone can supply a direct, fairly short,
> and reasonably intuitive proof that R has no 3 point compactification.
>
I am but a humble functional analyst myself, so I can only provide this
humble attempt. I hope it is not messed up.
So suppose that R union {a,b,c} is the 3 point compactification.
Let A,B and C be open neighborhoods of a,b,c respectively that are
disjoint. Then K = R-(A union B union C) is compact. Let
x = inf K, y = sup K. Then A-[x,y], B-[x,y] and C-[x,y] are
also open neighborhoods of a, b and c respectively. Now
R intersect (A union B union C - [x,y]) is R-[x,y], and splits
into three componants R intersect (A - [x,y]), R intersect (B - [x,y])
and R intersect (C - [x,y]). But R - [x,y] has only two componants.
QED.
--
Stephen Montgomery-Smith
ste...@math.missouri.edu
http://www.math.missouri.edu/~stephen
L. Andrew Campbell
compact Hausdorff space X, with X\R consisting of 3 points.
Let U_i (i=1,3) be disjoint open neighborhoods of the 3 points.
Let V_i = U_i intersect R. V_i is a countable union of open
intervals, of which at least one has +/- oo as an endpoint
(for otherwise the closure would not contain a point at oo
or, if it did, it could not be a full neighborhood of such
a point, since one could find a sequence of points in the gaps
between the open intervals converging to some point at oo -
necessarily unique by the choice of U_i). Choose such an
interval for each i = 1,3. This yields 3 disjoint open intervals
with at least one of +/-oo as an endpoint. Consideration of
cases easily yields a contradiction.
--
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
| L. Andrew Campbell internet: camp...@aero.org |
| M1-102 PO Box 92957 organization: The Aerospace Corp. |
| Los Angeles CA 90009 telephone: (310) 336-8642 |
David C. Ullrich
cher...@math.berkeley.edu (Paul R. Chernoff) wrote:
>
> real line R has no 3 point (Hausdorff) compactification -- that is,
> there does not exist a compact Hausdorff space X, containing a homeomorphic
> image of R as a dense subspace, such that X\R has exactly 3 points.
> (The theory of ends shows much, much more, in fact.)
>
> Since the theory of ends is rather exotic, at least to this humble
> functional analyst, I ask if anyone can supply a direct, fairly short,
> and reasonably intuitive proof that R has no 3 point compactification.
Let's pretend R is literally a subset of a 3-point compactification X.
Say A, B, and C are disjoint neighborhoods of the three points at
infinity. Now X\(A union B union C) is a compact subset of R, so
wlog it's contained in [-1,1].
Note that (A intersect R) cannot be contained in [-1,1]
since [-1,1] is compact and hence closed, while A must
intersect any neighborhood of one of the three points
at infinity. Similarly for B and C.
Let R' = R \ [-1,1] = (-infinty,-1] union [1, +infinity);
note that R' is a subset of A union B union C. One
of the two intervals must intersect at least two of the sets
A, B and C; wlog [1,infinity) intersects both A and B, and
possibly C. Now [1, infinity) is the union of two or three
(relatively) open subsets, contradicting the connectedness of
[1,infinity).
>
> ***************************************************
> Paul R. Chernoff
> Department of Mathematics
> University of California
> Berkeley, CA 94720
>
> cher...@math.berkeley.edu
> ***************************************************
>
> --
> # Paul R. Chernoff cher...@math.berkeley.edu #
> # Department of Mathematics 3840 #
> # University of California "Against stupidity, the gods themselves #
> # Berkeley, CA 94720-3840 struggle in vain." -- Schiller #
--
Oh, dejanews lets you add a sig - that's useful...
Robert Israel
Paul R. Chernoff <cher...@math.berkeley.edu> wrote:
>I am told that, via the theory of "ends", one can show that the
>real line R has no 3 point (Hausdorff) compactification -- that is,
>there does not exist a compact Hausdorff space X, containing a
>homeomorphic
>image of R as a dense subspace, such that X\R has exactly 3 points.
>(The theory of ends shows much, much more, in fact.)
>Since the theory of ends is rather exotic, at least to this humble
>functional analyst, I ask if anyone can supply a direct, fairly short,
>and reasonably intuitive proof that R has no 3 point compactification.
Suppose X = R union {a,b,c} is a three-point compactification of R.
Then a, b and c have open neighbourhoods A, B and C which are
disjoint. Let Ar, Br and Cr be the intersections of these with R.
X is covered by the open sets A, B, C and (-N,N) for all positive
integers N, so by compactness there must be some N such that
A, B, C and (-N,N) cover K. Now [N,infinity) is connected and
contained in the union of the disjoint open sets Ar, Br and Cr,
so at most one of Ar, Br and Cr can intersect [N,infinity).
Similarly for (-infinity,-N]. We conclude that at least one
of Ar \ [-N,N], Br \ [-N,N] and Cr \ [-N,N] is empty. But then
A \ [-N,N], B \ [-N,N] or C \ [-N,N] is a single point {a}, {b} or
{c} which is an open set, contradicting the assumption that R is
dense in X.
Of course this generalizes to an n-point compactification for
any finite n > 2.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Toby Bartels
>Paul R. Chernoff wrote:
>>I am told that, via the theory of "ends", one can show that the
>>real line R has no 3 point (Hausdorff) compactification -- ...
>>Since the theory of ends is rather exotic, at least to this humble
>>functional analyst, I ask if anyone can supply a direct, fairly short,
>>and reasonably intuitive proof that R has no 3 point compactification.
>So suppose that R union {a,b,c} is the 3 point compactification.
>Let A,B and C be open neighborhoods of a,b,c respectively that are
>disjoint. Then K = R-(A union B union C) is compact. Let
>x = inf K, y = sup K. Then A-[x,y], B-[x,y] and C-[x,y] are
>also open neighborhoods of a, b and c respectively. Now
>R intersect (A union B union C - [x,y]) is R-[x,y], and splits
>into three componants R intersect (A - [x,y]), R intersect (B - [x,y])
>and R intersect (C - [x,y]). But R - [x,y] has only two componants.
This is just the theory of ends in disguise, as one might expect.
The 2 components of R - [x,y] are the representatives for K
of the 2 ends of R.
-- Toby
to...@math.ucr.edu