Lecture 2
2. The Concept of Convergence: Ultrafilters and Nets
In this lecture we discuss two points of view on the notion of convergence. The
first one employs a set theoretical concept, which turns out to be technically useful.
Definition.
Suppose
X
is a fixed (nonempty) set. A
filter in
X
is a (non
empty) family
F
of nonempty subsets of
X
which has the property
2
:
(f)
Whenever
F
and
G
belong to
F
, it follows that
F
∩
G
also belongs to
F
.
What is important here is that all the sets in the filter are assumed to be
non
empty
. The set of all filters in
X
can be ordered by inclusion. A simple application
of Zorn’s Lemma yields:
•
For each filter
F
there exists at least one
maximal
filter
U
with
U
⊃
F
.
Maximal filters will be called
ultrafilters
.
An interesting feature of ultrafilters is given by the following:
Lemma 2.1.
Let
X
be a nonempty set, and let
U
be a filter on
X
.
The
following are equivalent:
(i)
U
is an ultrafilter.
(ii)
For any subsets
A
⊂
X
, it follows that either
A
or
X
A
belongs to
U
,
but not both!
Proof.
(
i
)
⇒
(
ii
).
Assume
U
is an ultrafilter.
First remark that
X
always
belongs to
U
. (Otherwise, if
X
does not belong to
U
, the family
U
∪ {
X
}
will be
obviously a new filter which will contradict the maximality of
U
).
Let us assume that
A
is nonempty and it does not belong to
U
. This means
that the family
M
=
U
∪ {
A
∩
U

U
∈
U
}
is no longer a filter (otherwise, the maximality of
U
will be contradicted). Note that
if
F
and
G
belong to
M
, then automatically
F
∩
G
belongs to
M
. This means that
the only thing that can prevent
M
from being a filter, must be the fact that one
of the sets in
M
is empty. That is, there is some set
V
∈
U
such that
A
∩
V
=
∅
.
In other words,
V
⊂
X
A
.
But then, it follows that for any
U
∈
U
we have
U
∩
(
X
A
)
⊃
U
∩
V
=
∅
and then the set
N
=
U
∪ {
U
∩
(
X
A
)

U
∈
U
}
will be a filter. By maximality, it follows that
N
=
U
, in particular,
X
A
belongs
to
U
. It is obvious that
A
and
X
A
cannot simultaneously belong to
U
, because
this will force
∅
=
A
∩
(
X
A
) to belong to
U
.
2
Some textbooks may use a slightly different definition.
9
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10
LECTURE 2
(
ii
)
⇒
(
i
).
Assume property (ii) holds, but
U
is not maximal, which means
that there exists some ultrafilter
V
with
V
U
. Pick then some set
A
∈
V
U
.
Since
A
∈
U
, by (ii) we must have
X
A
∈
U
. This would force both
A
and
X
A
to belong to
V
, which is impossible.
Exercise 1
.
Let
U
be an ultrafilter on
X
, and let
A
∈
U
.
Prove that the
collection
U
A
=
{
U
∩
A
:
U
∈
U
}
is an ultrafilter on
A
.
Remark 2.1.
If
U
is an ultrafilter on
X
, and
A
∈
U
, then
U
contains all sets
B
with
A
⊂
B
⊂
X
. Indeed, if we start with such a
B
, then by the above result,
either
B
∈
U
or
X
B
∈
U
. Notice however that in the case
X
B
∈
U
we would
get
U
(
X
B
)
∩
A
=
∅
,
which is impossible. Therefore
B
must belong to
U
.
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 Three '10
 Smith
 Empty set, Open set, Topological space, u., Ultrafilter

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