82
LECTURE
NOTES
FOR
18.102,
SPRING
2009
Lecture
13.
Thursday,
Mar
19:
Baire’s
theorem
Note
from
Lecture
9,
modiFed
and
considerably
extended.
Theorem
8
(Baire)
.
If
M
is
a
non-empty
complete
metric
space
and
C
n
⊂
M,
n
∈
N
,
are
closed
subsets
such
that
[
(13.1)
M
=
C
n
n
then
at
least
one
of
the
C
n
’s
has
an
interior
point.
Proof.
So,
choose
p
1
∈
/ C
1
,
which
must
exist
since
otherwise
C
1
contains
an
open
ball.
Since
C
1
is
closed
there
exists
1
>
0
such
that
B
(
p
1
,
1
)
∩
C
1
=
∅
.
Next
choose
p
2
∈
B
(
p
1
,
1
/
3)
which
is
not
in
C
2
,
which
is
possible
since
otherwise
B
(
p
1
,
1
/
3)
⊂
C
2
,
and
2
>
0
,
2
<
1
/
3
such
that
B
(
p
2
,
2
)
∩
C
2
=
∅
.
So
we
have
used
both
the
fact
that
C
2
has
empty
interior
and
the
fact
that
it
is
closed.
Now,
proceed,
inductively.
Assume
that
there
is
a
Fnite
sequence
p
i
, i
= 1
,...,k
and
positive
numbers
0
<
k
<
k
−
1
/
3
<
k
−
2
/
3
2
<
<
1
/
3
k
−
1
<
3
−
k
such
that
p
j
∈
B
(
p
j
−
1
,
j
−
1
/
3)
and
···
B
(
p
j
,
j
)
∩
C
j
=
∅
.
Then
we
can
add
another
p
k
+1
by
using
the
properties
of
C
k
–
it
has
non-empty
interior
so
there
is
some
point
in
B
(
p
k
,
k
/
3)
which
is
not
in
C
k
+1
and
then
B
(
p
k
+1
,
k
+1
)
∩
C
k
+1
=
∅
where
k
+1
>
0
but
k
+1
<
k
/
3
.
Thus,
we
have
constructe
and
inFnite
sequence
{
p
k
}
in
M.
Since
d
(
p
k
+1
,p
k
)
<
k
/
3
this
is
a
Cauchy
sequence.
In
fact
(13.2)
d
(
p
k
,p
k
+
l
)
<
k
/
3 +
+
k
+
l
−
1
/
3
<
3
−
k
<
2
k
/
3
···
for
all
l >
0
,
and
this
tends
to
zero
as
k
→ ∞
.
Since
M
is
complete
this
sequence
converges.
±rom
(13.2)
the
limit,
q
∈
M
must
lie
in
the
closure
of
B
(
p
k
,
2
k
/
3)
for
every
k.
Hence
q /
∈
C
k
for
any
k
which
contradicts
(13.1).
Thus,
at
least
one
of
the
C
n
must
have
non-empty
interior.
One
application
of
this
is
often
called
the
uniform
boundedness
principle,
I
will
just
call
it:
Theorem
9
(Uniform
boundedness)
.
Let
B
be
a
Banach
space
and
suppose
that
T
n
is
a
sequence
of
bounded
(i.e.
continuous)
linear
operators
T
n
:
B
−→
V
where
V
is
a
normed
space.
Suppose
that
for
each
b
∈
B
the
set
{
T
n
(
b
)
} ⊂
V
is
bounded
(in
norm
of
course)
then
sup
n
T
n
<
∞
.
Proof.