Date: Jan 16, 2013 6:07 AM Author: William Elliot Subject: Property related to denseness By definition, D is a dense subset of a space S when cl D = S.

Exercise. D is dense subset of S iff for all not empty open U,

U /\ D is not empty.

Exerise. If D is a dense subset of S and U an open set,

then cl U = cl U/\D

> Not just Willard but "indiscrete" is the standard term I would think.

> I think that a trivial object would be one that is embedded in all

> objects of the same type.

Yes, the indiscrete or trivial topology of S is { empty set, S }.

The indiscrete topology of S is the coarsest (smallest) topology

for S of all possible topologies.

The discrete topology of S is P(S).

The discrete topology for S is the finest (largest) topology

for S of all possible topologies for S

Exercise. The identily function id:X -> Y is continuous

iff the topolgy of X is finer than the topology of Y

iff the topolgy of Y is coarser than the topology of X.

Exercise. Assume f:X -> Y.

Show if Y is indiscrete or if X is discrete, then f is continuous.

Generalize to show X is discrete iff for all Y and f:X -> Y, f is continuous

and Y is indiscrete iff for all f:X -> Y all X and f:X -> Y, f is continuous.

> For example, the trivial group is the group

> with one element. If there is such a thing as "the trivial topology

> [without mentioning the underlying set]" then that might be the topology

> on the empty set where the only open set is the empty set.

Exercise. Show there's a unique space with exactly one open set.

Exercise. Give some examples of spaces that are both discrete

and indiscrete. Show they're all homeomophic except for one.

> If the underlying set is X, then I would think "The trivial topology on

> X" is a fine way of describing the indiscrete topology, since the open

> sets in that topology on X are exactly the sets that are open in every

> topology on X. So it's analogous to the "trivial group". Is "trivial

> topology on X" a standard way of referring to the indiscrete topology on

> X? If not, I think it should be.&

If S is given the indiscrete topology, that can be unually notated

(S,indiscrete) or for short, indiscrete S.

The lattice L, of topologies for a set S is a complete, complemented, not

distributive lattice with top element discrete S, and bottom element

indiscrete S. It's complete because the intersection of every collection

of topologies for S, is another topology for S. It suffices to use a few

finite topology spaces to show L is not distributive. That it's

complemented is a long complicated proof.