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 | | From: | Stephen J. Herschkorn | | Subject: | Why 14 is an interesting number (topology) | | Date: | 23 Jan 2005 08:53:25 -0800 |
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 | And it's not just because all the natural numbers 0 through 13 are
interesting.
Let A be a subset of a topological space. Let C be the smallest
collection of subsets of the space which contains A and is closed
under the closure and the complementation operations. Then C has at
most fourteen elements (i.e., subsets of the space). Furthermore,
there exists a subset of a topological space where this maximum of
fourteen is attained.
Munrkes attributes this result to Kuratowski; Munkres gives it as an
exercise (which he marks as difficult). Since this is a popular
textbook, I will not post the proof here.
Pretty neat, huh?
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| | From: | George Cox | | Subject: | Re: Why 14 is an interesting number (topology) | | Date: | Sun, 23 Jan 2005 18:46:20 +0000 (UTC) |
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| "Stephen J. Herschkorn" wrote:
>
> And it's not just because all the natural numbers 0 through 13 are
> interesting.
>
> Let A be a subset of a topological space. Let C be the smallest
> collection of subsets of the space which contains A and is closed
> under the closure and the complementation operations. Then C has at
> most fourteen elements (i.e., subsets of the space). Furthermore,
> there exists a subset of a topological space where this maximum of
> fourteen is attained.
>
> Munrkes attributes this result to Kuratowski; Munkres gives it as an
> exercise (which he marks as difficult). Since this is a popular
> textbook, I will not post the proof here.
>
> Pretty neat, huh?
Let X be a set ordered by <= such that
x <= x
x <= y & y <= x implies x = y
x <= y & y <= z implies x <= z
for all x, y, z in X.
Let f:X -> X and g:X -> X be such that
x <= y implies f(x) <= f(y) & g(y) <= g(x)
and
x <= f(x), f(f(x)) = f(x), g(g(x)) = x.
Let Int = gofog then
x <= y implies Int(x) <= Int(y)
Int(x) <= x, Int(Int(x)) = Int(x)
for all x, y in X.
So Int is an interior operator.
The semigroup generated by f and g consists of 14 elements:
identity, g, f, fog, gof, Int, fogof, foInt, Intof,
gofoInt, foIntof, fogofoInt, gofoIntof and IntofoInt.
[Sorry that looks horrible, I tried "f o g o f o Int" for "fogofoInt"
(etc) but it looked no better.]
There are no more than 14 because
f(Int(f(Int(x)))) = f(Int(x)) & Int(f(Int(f(x)))) = Int(f(x)).
But how to produce an example were there are 14?
I don't have Munkres. Does his solution proceed like this?
A supplementary question: In the modal logic S4 there are 14 distinct
modalities, what's the connection?
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| | From: | Thomas Nordhaus | | Subject: | Re: Why 14 is an interesting number (topology) | | Date: | Sun, 23 Jan 2005 22:25:24 +0100 |
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| George Cox schrieb:
>
>[Sorry that looks horrible, I tried "f o g o f o Int" for "fogofoInt"
Yeah, real horrible - sounds like "fogg off!" (Cockney slang?)
Thomas [SCNR]
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| | From: | Keith Ramsay | | Subject: | Re: Why 14 is an interesting number (topology) | | Date: | 23 Jan 2005 20:16:04 -0800 |
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| Stephen J. Herschkorn wrote:
| Let A be a subset of a topological space. Let C be the smallest
| collection of subsets of the space which contains A and is closed
| under the closure and the complementation operations. Then C has
at
| most fourteen elements (i.e., subsets of the space). Furthermore,
| there exists a subset of a topological space where this maximum of
| fourteen is attained.
If I remember correctly, there is an example on the real line
where the maximum is attained.
I remember when I first read this result I found it curious
that the maximum would be such a seemingly arbitrary number
as 14. I think now I can demystify it a little bit. It's 14
basically because its 2*(1+2*3).
First, we only have to consider complementary pairs of sets.
If the original space is nonempty, a set and its complement
are always distinct. This means that we can ignore closed
sets in favor of their open complements. If instead of using
the _closure_ operator we use the _exterior_ operator E, the
complement of the closure, we get the same collection of sets.
But the exterior operator gives open sets instead of closed.
Second, the original set A and its complement are a special
case, because they're the only ones that aren't necessarily
open or closed. It's not so hard to adjust examples so that
A is neither open nor closed, which automatically makes it
and its complement distinct from the other sets.
Third, the exterior of A and its interior (the exterior of
its complement) are disjoint open sets, but otherwise don't
have any particularly close relationship to each other. I'm
being vague but I can elaborate if I need to. This is not
needed to prove the upper bound, just to motivate that it's
the best possible upper bound. Each open set is the interior
of some set (e.g. itself) and is the exterior of some set
(e.g. its complement), so separately they are not special
in any way.
Finally, we show that starting from an arbitrary open set U
and applying the exterior operator are only U, E(U), and
E(E(U)), since E(E(E(U)))=E(U) always holds. Moreover these
three can be distinct.
One way to look at this is that the open sets of a topological
space are a Heyting algebra. A Heyting algebra is a kind of
lattice used to model intuitionist logic, which is the logic
ordinarily used in constructive mathematics. Since it is a
lattice, any two elements have a least upper bound and a greatest
lower bound. There also is a greatest element 1 and a least
element 0. For any two elements x and y, there exists an
element u=(x->y) which is the greatest element such that the
join u&x of u and x is <=y. The open sets of a topological
space also satisfy the property that any family of elements
has a least upper bound (even if the family is infinite).
In a Heyting algebra, negation is represented by ~u = u->0.
In the case of a topological space, this is the exterior of
an open set u. Now in intuitionist logic, the double negation
rule ~~u=u does not necessarily hold. Classical logic is the
same as intuitionist logic plus the double negation elimination
rule. However, triple negation is equivalent to single
negation, ~~~u=~u.
I think it's maybe easier to understand this as a special case
of the following little abstract lemma. Given two partially
ordered sets A and B, and a relationship R that holds between
them satisfying
1. If a is in A, b is in B, aRb, a'<=a, and b'<=b,
then a'Rb' (downward closure).
2. For each a in A, there exists a maximal b=f(a) such
that aRb, and for each b in B there exists a maximal
a=g(b) such that aRb,
we get the identities f(g(f(a)))=f(a) and g(f(g(b)))=g(b).
Galois theory deals with the case where A consists of the
subsets S of a field, and B consists of subsets T of a
Galois group of the field over a subfield, and the relation
R is that the elements of T fix the elements of S. Starting
with the subset S of the field, we take the subgroup H
fixing it. The subfield F fixed by H may be larger than S.
But then the subgroup fixing F is still H. It can't be
smaller than H because by definition H fixes F. It also can't
be any larger, because anything not in H fails to fix some
element of the original set S.
The same argument can be used starting with a set of elements
T of a Galois group. We take the field F fixed by it. The
subgroup H of the Galois group fixing F contains T but may
be larger than T. The fixed field of H, however, is F again.
It has to contain at least F, since F was fixed by H. But it
can't contain anything outside of F, since F contains
everything fixed by the elements of T.
This argument is cute in that it's so very abstract and so
simple as to be almost tautological. But it's nontrivial
enough not quite to go without saying.
Essentially the same argument was presented to me again in
an exposition of elementary algebraic geometry. Here A is the
subsets of C^n where C is the complex numbers (we could take
k^n for a field k more generally, but it's a more complicated
theory (although less complex)), and B is the set of polynomials
in n variables with complex coefficients. Starting from a set
of polynomials, we can consider the set of points that are
zeros of all of them, known as an algebraic set. Conversely
given a set of points we can consider the set of polynomials
that are zero at all the points. Characterizing the sets of
polynomials of this form is a standard problem. This same little
lemma for this case is one of the standard little lemmas of
algebraic geometry.
The fact that E(E(E(U)))=E(U) is the special case where
A=B is the set of open subsets of a topological space where
<= is inclusion, and the relation R is disjointness.
The open sets that are the exteriors of open sets are called
regular open sets. Equivalently, an open set is regular if
it's the exterior of its exterior.
What I just wrote seems to overlap a bit with what George Cox
wrote, but I thought it was worth putting it a different way.
Keith Ramsay
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