Which of the following lists of subsets of the set
form topologies.
-
- Yes
-
- No
not in the list.
-
Yes
-
No , closed under union and intersection and
is
in the list, but
is
not.
Prove that a function
is continuous if and only if the inverse image of every closed set is closed.
Proof:
We use the following equations from the Boolean algebra of sets.
If
is any subset. then
Suppose the inverse image of every closed set is closed. Let
be an open set and
Then
is closed and so
is closed hence
is
open.
Suppose that
is
continuous and
is closed then
is
open since
is
open
but then
is closed.
Let
be the floor function.
for
Is
continuous?
Answer:
No because the supspace topology on
is descrete so points are open sets but
not an open set.
Let
be a set and
a collection of subsets closed under finite intersection, then
the set of arbitrary unions of sets of
form a topology.on
[ Added :
and
are
in
]
Proof:
Again, the solution amounts to recalling some properties of Boolean Algebras. In this case
More generally, if
and are
indexed
set of sets then
and similarly for any arbitrary collection of indexed sets.
And if
and are
indexed
set of sets then
where
is
either in the index set
or in the index set
and
similarly for an arbitrary collections of index sets.
We are given that
and
are
in
.
We are also given that arbitrary unions of sets of
is in
.
This is because a union of unions of sets in
is
a union of sets. That a finit intersection of sets of
is in
is because an intersection of unions of sets in
is
a union of intersections of sets.