2Baire category theorem
II Linear Analysis
2.1 The Baire category theorem
When we first write the Baire category theorem down, it might seem a bit
pointless. However, it turns out to be a really useful result, and we will be able
to prove surprisingly many results from it.
In fact, the Baire category theorem itself does not involve normed vector
spaces. It works on any metric space. However, most of the applications we have
here are about normed vector spaces.
To specify the theorem, we will need some terminology.
Definition
(Nowhere dense set)
.
Let
X
be a topological space. A subset
E ⊆ X
is nowhere dense if
¯
E has empty interior.
Usually, we will pick
E
to be closed so that the definition just says that
E
has an empty interior.
Definition
(First/second category, meagre and residual)
.
Let
X
be a topological
space. We say that
Z ⊆ X
is of first category or meagre if it is a countable union
of nowhere dense sets.
A subset is of second category or non-meagre if it is not of first category.
A subset is residual if X \ Z is meagre.
Theorem
(Baire category theorem)
.
Let
X
be a complete metric space. Then
X is of second category.
This, by definition, means that it is not a countable union of nowhere dense
sets. This is equivalent to saying that if we can write
X =
∞
[
n=1
C
n
,
where each C
n
are closed, then C
n
has a non-empty interior for some n.
Alternatively, we can say that if
U
n
is a countable collection of open dense
sets, then
T
∞
n=1
U
n
6
=
∅
(for if
U
n
is open dense, then
X \ X
n
is closed with
empty interior).
Proof.
We will prove that the intersection of a countable collection of open dense
sets is non-empty. Let U
n
be a countable collection of open dense set.
The key to proving this is completeness, since that is the only information
we have. The idea is to construct a sequence, show that it is Cauchy, and prove
that the limit is in the intersection.
Construct a sequence
x
n
∈ X
and
ε
n
>
0 as follows: let
x
1
, ε
1
be defined
such that
B(x
1
, ε
1
) ⊆ U
1
. This exists
U
1
is open and dense. By density, there is
some x
1
∈ U
1
, and ε
1
exists by openness.
We define the
x
n
iteratively. Suppose we already have
x
n
and
ε
n
. Define
x
n+1
, ε
n+1
such that
B(x
n+1
, ε
n+1
) ⊆ B(x
n
, ε
n
) ∩ U
n+1
. Again, this is possible
because
U
n+1
is open and dense. Moreover, we choose our
ε
n+1
such that
ε
n+1
<
1
n
so that ε
n
→ 0.
Since
ε
n
→
0, we know that
x
n
is a Cauchy sequence. By completeness of
X
, we can find an
x ∈ X
such that
x
n
→ x
. Since
x
is the limit of
x
n
, we know
that x ∈ B(x
n
, ε
n
) for all n. In particular, x ∈ U
n
for all n. So done.