Closure operators
Today in class, I said that a topology on X was a collection
of subsets of X (called the open sets), containing X and the
empty set, which is closed under finite intersection and
arbitrary union. Somebody asked if the phrase "closed under"
indicated a closure operator somewhere.
There are two answers to this question. The technically correct
answer is "yes." After class I began thinking that the question
meant "is there a strange connection between THIS definition
of a topology and the definition given in Danny's notes ..
specifically, does the closure operator do the same thing in
both cases?" In this case, the answer is no. There ARE
closure operators floating behind the phrase "closed under,"
but they apply to COLLECTIONS of subsets of X, as opposed to
subsets of X. (In Danny's notes, a topology on the space X
is defined from a closure operator on subsets of X).
I could have answered "well, to think about closure operators
right now is a little misleading. The important concept for
today is that of a topology as defined by a collection
of open sets." You could definitely think through the
proof of today's lemma without ever once thinking about
closure operators.
But back to the "yes" answer. Mathematicians often use closure
operators even when they aren't explicitly thinking about topology.
That is, even when their primary objective isn't defining the
closed sets in some topology as the sets that are fixed by
the closure operator in question.
Consider the closure operator K associated to the phrase
"closed under arbitary union." We assume there is an underlying
space X. As mentioned above, K acts on collection os subsets
of X. Specifically, if S is a collection of subsets, then K(S)
is another collection of subsets -- the ones that are arbitary
unions of subsets belonging to S.
The lemma proved in class today can be stated as follows. If
B is a basis, then closing B under arbitrary union and adding
the empty set gives a topology.
(WARNING: do not read if your brain is full)
Of course, from the point of view of set theory, the set of subsets
of X -- called the power set P(X) of X -- can be the underlying space
for a topology just as easily as X can be. If you define a closed
set on P(X) as any collection of subsets of X closed under the
operator K above, then you certainly get a topology on P(X).
Moral: there is always a topology generated by a closure operator,
but it may be on a somewhat abstract space.
Jessica Young
Last modified: Tue Sep 30 13:01:53 EDT 2003