NetNews Usenet Archive 1992 #26/NN_1992_26.iso/spool/sci/math/14580

Text File | 1992-11-08 | 9.4 KB | 170 lines

Newsgroups: sci.math
Path: sparky!uunet!stanford.edu!CSD-NewsHost.Stanford.EDU!Sunburn.Stanford.EDU!pratt
From: [e-mail] (Vaughan R. Pratt)
Subject: Re: definition of topological space
Message-ID: <[e-mail]>
Sender: [e-mail]
Organization: Computer Science Department,  Stanford University.
References: <[e-mail]> <[e-mail]> <[e-mail]>
Date: Sat, 7 Nov 1992 23:55:21 GMT
Lines: 158

In article <[e-mail]> [e-mail] (I am a terminator.) writes:
>What elements of the power set are excluded from the collection?

Your choice, depending on how crumbly or sticky you want the space to
be.  If you want the space to be capable of crumbling into total dust
(every point in its own open set) then you keep all subsets.  This is
the DISCRETE topology, which should be thought as a set of isolated
points with no glue to hold them together.  If you want it to be
impossible to partition the space at all then keep just the empty
subset and the whole space (you *have* to always have those).  This is
the COARSE topology, which should be thought of as a blob of titanium
which you have no way of chopping up.

Spaces encountered in practice are in between.  For example you can
slice and dice the real line R, you just can't puree it.

The basic slice of R into an open set and its complementary closed set
is the open interval (a,b), having as complement (-oo,a]u[b,oo).  These
are called the BASIC open sets.  This is called a BASIS for the
topology.  Dicing is parallel slicing: you can chop out as many open
intervals (basic open sets) as you want at the one time, and the union
of these is considered to be an open set.  The open sets formed in this
way can be seen to be a topology, called the topology GENERATED BY that
basis.  The topology on R generated by the finite open intervals is the
STANDARD TOPOLOGY on R.

Slicing out no intervals at all is permitted, making the empty set
open.  Slicing them all out (overlap is ok) makes R open.  Slicing out
all unit open intervals (i,i+1) for all integers i yields R-Z, the set
of noninteger reals, whence Z is a closed subset of R, as are all
subsets of Z.  Slicing out all the open intervals below a yields
(-oo,a), and above a, (a,oo).

Why this particular topology on R?  Well, the *motivation* comes from
more concrete information about R than present in its topology, namely
its arithmetic.  Let's say that point x on R is NEAR subset Y of R when
x is not in Y ("near" being the irreflexive version of Y as a
neighborhood of x) and for every real \epsilon > 0 there exists a point
y in Y for which |x-y| < \epsilon (notice the appeal to the arithmetic
of R).  Connect this notion with abstract topology by defining a closed
set to be one with no points near it, and an open set to be the
complement of a closed set (and conversely define near(Y), the set of
points near Y, to be cl(Y)-Y).  It is easy to see that every open
interval satisfies this definition.  It is a nice exercise to verify
that the remaining sets that satisfy this definition are exactly the
unions of open intervals.  Thus the definitions of open and closed in
this paragraph are exactly equivalent to those in the preceding
paragraphs.

And this makes it clear why the rationals do not form an open set.  The
rational 0 is near the set of irrationals, making that set not closed
and hence its complement not open.  (Indeed *every* rational is near
the irrationals.)  The converse situation (rationals and irrationals
interchanged) is identical, with the substitution of sqrt(2) for 0.  So
neither the rationals nor the irrationals is either open or closed.
This shows that you cannot puree R: you can't chop it up completely
arbitrarily, especially not into fine dust like the rationals and the
irrationals.

Puzzle: is the set of fractions 1/i for integer i>0 open?  Closed?
Solve it both ways, appealing first to the arithmetic notion of
nearness to make sure you get the right answers, then see what you have
to do to your argument so that it appeals only to the
union-of-open-intervals definition of "open".  What difference does
adding 0 to this set make?

So all the information in the standard topology on the reals is present
in real arithmetic.  What about the converse?   Can we recover the
arithmetic on the reals from the information in the standard topology
on R?

Answer: no.  From just the open sets we can't even tell where 0 is.
Even if we could, we can't tell the scale: 1 could be any point other
than 0 (kx is a homeomorphism of R for any real nonzero k).  But even
after we've chosen 1, we can't find 1/2 or 2, since topological spaces
needn't stretch linearly under homeomorphism (example: x^3).  On the
other hand, after choosing 0 and 1 we *are* able to distinguish the
positive reals, namely as the block containing 1 in the unique (see
below) partition of R-{0} into two nonempty open sets.

So haven't we thrown out the baby with the bathwater in discarding
arithmetic in this way?  No, because topology doesn't care about
arithmethic.  What topology cares about are things like nearness and
connectedness.

So have we thrown out connectedness?  In particular have we lost the
information about the real line that it is a connected space?  No, we
still have that information.  Consider "punctured" R, R-{0}.  This is
an open set consisting of two disjoint open sets, the negative reals
and the positive, but none of these three sets are closed.  Now suppose
we take R-{0} to be a space in its own right, taking as its open sets
those open sets in the standard topology on R that do not contain 0.
We are then obliged to declare R-{0} to be closed.  This in turn makes
the positive reals and the negative reals closed, each now being the
complement of an open set.  Thus R-{0} as a space in its own right
contains exactly four sets that are both closed and open, or CLOPEN:
the empty set, the whole space, the negative reals, and the positive
reals.  Every space contains at least the empty set and the whole space
as clopen sets.  Any other complementary pair of clopen sets is called
a SEPARATION.  A CONNECTED space is one with no separations.  According
to this definition R is connected, since the only open sets of R with
no points near them are the empty set and R.  But we have just seen a
separation of R-{0} (its only one), making that space disconnected.

So without appealing to arithmetic we were able to tell from the
standard topology on R that deleting 0 from R disconnects it.  Topology
forgets certain details of arithmetic but remembers those details to do
with nearness, connectedness, and other such notions characteristic of
topology.  The standard topology on the real line is an ABSTRACTION of
real arithmetic.

==================
Idiosyncratic footnote.

To make connectedness seem more concrete, let's say that sets are
NAILED together at point x when they have x in common, and GLUED
together at point x when one set is nailed to the closure of the other
at x.  Disjoint closed sets are disconnected from each other because
they are neither nailed nor glued together, the situation with the two
halves of R-{0}.  In R the halves are still neither nailed nor glued
together, but in R they are not closed, and it is possible for a closed
set, namely {0}, to form a bridge connecting them.

Conversely disjoint sets that are glued together must include at least
one nonclosed set.  Think of the nonclosed set X as being STICKY AT (or
having glue at) those points in cl(X)-X.  Stickiness of Y at x can
arise in two ways.  Either there is some point y in Y for which x < y
(i.e. cl({x}) subset-of cl({y})), in which case we may visualize the
stickiness concretely as a rubber band from x to y.  This is
necessarily the situation when Y is finite.  Or there is no such y in Y
(necessarily the case for T1 topologies, which forbid x<y), in which
case Y must be infinite.  In that case we may picture a cloud of dust
in Y aggregating around x, which is called an ACCUMULATION POINT for Y
(not idiosyncratic).  The dust can be thought of as collectively
generating a potential field attracting x, the infinite analog of the
rubber band.  Annihilating finitely many of these dust particles
(removing them from the whole space) cannot alter this imagined field,
in the sense that what is left of Y continues to be sticky at x.

An extreme case is given by the set of rationals, which is sticky at
every irrational, and vice versa, assuming the standard topology on R.
Hence the rationals and the irrationals are glued together at every
point of R.  Now it is impossible for an open-closed pair (O,C) to be
glued together at every point of a nonempty space.  This is because the
glue points must all be in C, making C the whole space.  But the whole
space is not glued to the empty space at any point, a contradiction.
So this gives another way to see that the set of rationals is neither
open nor closed in the standard topology on R.  It is however both open
and closed in the discrete topology on R.

So we see that connectedness is a mild form of togetherness: it is
possible to tear a connected space into an open and a closed set, both
nonempty, but there must be some resulting stickiness at the frontier
of the open set.  In contrast, a space with the coarse topology is a
much stronger form of togetherness:  no tearing into two nonempty sets
is permitted at all.
--
Vaughan Pratt                There's no truth in logic, son.