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Math 1001: Real Analysis
The Baire Category Theorem and Some
Applications
Tom Potter
997597051
March 8, 2010
The Baire Category Theorem is an important theorem in functional
analysis which was proved by René-Louis Baire in his 1899 doctoral thesis.
It has many important consequences, some of which will be shown in this
paper. But let us begin with the statement and proof of the theorem.
∞
Theorem (Baire): Let X be a complete
T∞ metric space. If {Un }1 is a
sequence of open dense subsets of X, then 1 Un is dense in X. It follows
that X is not a countable union of nowhere dense sets (a set is said to be
nowhere dense if its closure has empty interior).
Proof. We show that {Un }∞
1 is dense in X by showing that it has nonempty
intersection with every nonempty open subset of X. Let W ⊂ X be nonempty
and open. Since U1 is dense in X, U1 ∩ W is nonempty and open, and so it
contains an open ball B(x0 , r0 ), where we can choose r0 so that 0 < r0 < 1.
Then since U2 is dense in X, B(x0 , r0 ) ∩ U2 is nonempty and open, so it
contains a ball B(x1 , r1 ) whose radius r1 can be chosen so that r1 < 2−1
and B(x1 , r1 ) ⊂ U1 ∩ B(x0 , r0 ). We proceed in a recursive manner using the
density of each Un to get a sequence of nested closed balls centered at points
xn , with ball n having radius less than 2−n (the reason for taking closed balls
will become apparent below). So we get a Cauchy sequence {xn }, which
converges to a limit x since X is complete. Note that the choice of the xn ’s
invokes the axiom of countable choice. Then given any N ,
x ∈ B(xN , rN ) ⊂ UN ∩ B(x0 , r0 ) ⊂ UN ∩ W,
which shows that the intersection of the Un ’s meets any nonempty open
set W ⊂ X, and is therefore dense. To see that X is not a countable
union of nowhere dense sets, consider any sequence {ET
n } of such sets in X.
c
c
Then {(E
S
Sn ) } is a sequence of open dense sets. Since (E n ) 6= ∅, we have
En ⊂ E n 6= X, which concludes the proof.
This theorem is called the Baire Category Theorem because of Baire’s
terminology for sets: given a topological space X, a set E ⊂ X is said to be
of the first category if it is a countable union of nowhere dense sets; otherwise
E is of the second category. The BCT, as we shall henceforth refer to the
Baire Category Theorem, asserts that every complete metric space X is of
the second category in itself. In some more modern texts the term meager
1
is used in place of “of the first category” and a set is called residual if it is
the complement of a meager set.
We now discuss some applications of the theorem. This next theorem
uses the fact that the property “nowhere dense” is purely topological, and
is therefore preserved under homeomorphisms (in fact, the conclusion of the
BCT is purely topological, so if X is homeomorphic to a complete metric
space then X is nonmeager). First we will introduce some terminology. A
linear map T : X → Y between two normed vector spaces is bounded if
there exists C ≥ 0 such that
kT xk ≤ Ckxk for all x ∈ X .
We denote the spaces of all such maps by L (X , Y ). Also, a map f : X → Y
between topological spaces X and Y is said to be open if f (U ) is open in
Y whenever U is open in X. If X and Y are normed linear spaces and f is
linear, then it’s easy to see that f commutes with translations and dilations.
In this case showing that f is open amounts to showing that if B is the unit
ball centered at the origin in X, then f (B) contains a ball centered at the
origin in Y . We will follow the proof presented in Folland’s Real Analysis,
filling in some of the details.
The Open Mapping Theorem: Let X and Y be Banach spaces. If
T ∈ L (X , Y ) is surjective, then T is open.
Proof. Let Br denote the open ball of radius r centered at the origin in
X . ItSsuffices to show that T (B1 ) contains a ball
S∞ about 0 in Y . Since
∞
X = 1 Bn and T is surjective, we have Y = 1 T (Bn ). If we consider
the map y 7→ ny, which is a homomorphism of Y which maps T (B1 ) to
T (Bn ), we see that T (B1 ) cannot be nowhere dense, for if it were Y would
be meager, which cannot be the case since Y is complete (and by the BCT
we know
thatcomplete metric spaces are nonmeager). Thus there exists
◦
y0 ∈ (T (B1 ))
⊂ (T (B1 )), and in particular there is some r > 0 such that
the ball B(y0 , 4r) is contained in (T (B1 )). We assert that for each ε > 0 there
exists a point pε ∈ T (B1 ) such that kpε − y0 k < ε. For suppose not. Then
c
there exists ε > 0 such that T (B1 ) ∩ B(y
0 , ε) = ∅, i.e.: B(y0 , ε) ⊂ T (B1 ) ,
c
which implies y0 ∈ (T (B1 )c )◦ = T (B1 ) , a contradiction. So we can pick
2
y1 = T x1 ∈ T (B1 ) such that ky1 − y0 k < 2r; then B(2r, y1 ) ⊂ B(4r, y0 ) ⊂
T (B1 ), so if kyk < 2r,
y = −T x1 +(y+y1 ) ∈ −T x1 +B(2r, y1 ) ⊂ −T x1 +T (B1 ) = T (−x1 + B1 ) ⊂ T (B2 ).
Dividing both sides of this by 2 and using the fact that T commutes with
dilations, we conclude that there exists r > 0 such that if kyk < r then
y ∈ T (B1 ). If we could replace T (B1 ) with T (B1 ) we would be done. Making
r smaller we can accomplish this, as we now show.
Since T commutes with dilations, it follows that if kyk < r2−n , then
y ∈ T (B2−n ). Suppose kyk < 2r . Then y ∈ T (B 1 ), so there exists a sequence
2
{λn } inT (B 1 ) such that kλn − yk → 0 as n → ∞. So we can find x1 ∈ B 1
2
2
such that ky − TP
x1 k < 4r , and proceeding inductively, we can find xn ∈ B2−n
such that ky − n1 T xj k < r2−n−1 . Since X isP
complete, every absolutely
convergent seriesP
in X converges, so the series ∞
1 xn converges, say to x.
∞ −n
But then kxk < 1 2 = 1 and y = T x. Therefore, T (B1 ) contains all y
with kyk < 2r , which completes the proof.
Corollary 1: If X and Y are Banach spaces and T ∈ L (X , Y ) is
bijective, then T −1 ∈ L (X , Y ), and hence T is an isomorphism. This is
because we know from the open mapping theorem that T is open, and T
being open means T = (T −1 )−1 carries open sets forwards, so T −1 carries
open sets backwards.
Another important corollary of the BCT is the closed graph theorem.
First we introduce a little bit more terminology. If T is a linear map between
normed vector spaces X and Y , we define the graph of T to be
Γ(T ) = {(x, y) ∈ X × Y : y = T x}.
This is a subspace of X × Y , and we say T is closed if Γ(T ) is a closed
subspace of X × Y . If T is continuous then it’s easy to see that T is
closed. Indeed, if we take a convergent sequence {(xn , T xn )} ⊂ Γ(T ), then
xn → x for some x ∈ X, and the continuity of T implies T xn → T x, so
{(xn , T xn )} converges to (x, T x) ∈ Γ(T ), and hence T is closed. The closed
graph theorem asserts that if X and Y are complete then the converse is
also true:
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The Closed Graph Theorem: If X and Y are Banach spaces and
T : X → Y is a closed linear map, then T is bounded.
Proof. Let π1 and π2 be the projections of Γ(T ) onto X and Y . Then
π1 ∈ L (Γ(T ), X ) and π2 ∈ L (Γ(T ), Y ), since they are easily seen to be
linear, and kπi (x, T x)k ≤ max(kxk, kT xk) = k(x, T x)k so they are bounded.
Then since X and Y are complete, their Cartesian product X × Y is too,
and hence so is Γ(T ) since T is closed. The map π1 is a bijection from Γ(T )
to X , so by Corollary 1 π1−1 is bounded. Thus T = π2 ◦ π1−1 is bounded.
Yet another powerful application of the BCT is the Uniform Boundedness
Principle. It allows us to obtain uniform estimates from pointwise estimates
when certain conditions are met.
The Uniform Boundedness Principle: Suppose that X and Y are
normed vector spaces and A is a subset of L (X , Y ).
a. If supT ∈A kT xk < ∞ for all x in some nonmeager subset E ⊂ X , then
supT ∈A kT k < ∞. b. If X is a Banach space and supT ∈A kT xk < ∞ for all
x ∈ X , then supT ∈A kT k < ∞.
Proof. Let
En = {x ∈ X : sup kT xk ≤ n} =
T ∈A
\
{x ∈ X : kT xk ≤ n}.
T ∈A
The En ’s are
S closed since arbitrary intersections of closed sets are closed.
Also, E = En , and since E is nonmeager at least one of the En must have
nonempty interior. Thus for this n, En◦ contains on open ball B(x0 , r), and
hence B(x0 , r) ⊂ En = En . Then for kxk ≤ r we have x + x0 ∈ En , and
hence
kT xk ≤ kT (x + x0 )k + kT x0 k ≤ 2n
for each T ∈ A . In other words, T (B(0, r)) ⊂ B(0, 2n) whenever T ∈ A .
Then since each such T commutes with dilations we have
T (B(0, 1)) = T (δ 1 B(0, r)) = δ 1 T (B(0, r)) ⊂ δ 1 B(0, 2n) = B(0, 2n
),
r
r
r
r
where δ 1 denotes dilation by 1r . We’ve shown that kxk ≤ 1 =⇒ kT xk ≤ 2n
,
r
r
2n
and therefore kT k = sup{kT xk : kxk ≤ 1} ≤ r , and this applies for all
T ∈ A . Thus we’ve proven (a), and (b) follows directly form the BCT, since
the BCT implies that any Banach space is nonmeager.
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The BCT and the notion of category that Baire introduced are especially
useful in the theory of functions on the real line. We will present a little
bit of this theory now. First we give a slightly more descriptive definition of
nowhere dense for subsets of R, and some other relevant terminology. A set
A ⊂ R is dense in the interval I if A has nonempty intersection with every
subinterval of I. A set A ⊂ R is nowhere dense if it is dense in no interval,
that is, if every interval has a subinterval contained in the complement of A.
Let f be a real-valued funcion on R. For any interval I, the quantity
ω(I) = sup f (x) − inf f (x)
x∈I
x∈I
is called the oscillation of f on I. For any fixed x, the function ω((x−δ, x+δ))
decreases with δ and approaches the limit
ω(x) = lim ω((x − δ, x + δ)),
δ→0
called the oscillation of f at x. ω is an extended real-valued function on R,
and we see that ω(x0 ) = 0 if and only if f is continuous at x0 . When it is
not 0, ω(x0 ) gives us a measure of the size of the discontinuity of f at x0 . If
we let Df denote the set of points at which f is discontinuous, then one can
show that
∞
[
{x : ω(x) ≥ n1 }.
Df =
n=1
A function f is said to be of the first class (of Baire) if it can be represented
as the pointwise limit of a sequence of continuous functions. Such functions
do not need to be continuous. For a counterexample, consider the functions
fn (x) = max(0, 1 − n|x|). These are continuous, but converge to the discontinuous function f (x) = χ{0} (x). However, we prove a theorem below which
shows that a function of the first class cannot be everywhere discontinuous.
Baire’s theorem on function of the first class: If f is of the first
class then f is continuous except at a meager set of points.
Proof. It suffices to show that, for each ε > 0, the set F = {x : ω(x) ≥ 5ε} is
nowhere dense, since then Df is meager. Let f (x) = lim fn (x), fn continuous,
and define
\
En =
{x : |fi (x) − fj (x)| ≤ ε} (n = 1, 2, ...).
i,j≥n
5
S
Then En is closed, En ⊂ En+1 , and En = R. Let I ⊂ R be any closed
interval. Then I itself is a complete metric space, and henceSby the BCT
cannot be a countable union of nowhere dense sets. Since I = ∞
i=1 (En ∩ I),
this implies that the sets (En ∩ I) cannot all be nowhere dense. Hence,
for some positive integer n, En ∩ I contains an open interval J. We have
|fi (x) − fj (x)| ≤ ε for all x ∈ J; i, j ≥ n. Putting j = n and letting i → ∞,
it follows that |f (x)−fn (x)| ≤ ε for all x ∈ J. Now since fn is continuous, for
any x0 ∈ J there is a neighbourhood I(x0 ) ⊂ J such that |fn (x)−fn (x0 )| ≤ ε
for all x ∈ I(x0 ). Hence |f (x) − fn (x0 )| ≤ 2ε for all x ∈ I(x0 ). Therefore,
ω(x0 ) ≤ ω(I(x0 ))
= sup f (x) − inf f (x)
x∈I(x0 )
x∈I(x0 )
≤ sup f (x) − fn (x0 ) + fn (x0 ) − inf f (x)
x∈I(x0 )
x∈I(x0 )
≤ 2ε + 2ε
= 4ε.
Therefore no point of J belongs to F . Thus for every closed interval I there
is an open interval J ⊂ I \ F . It easily follows that every interval I ⊂ R has
a subinterval in F c , which shows that F is nowhere dense.
The BCT is also useful in proving existence results, such as the existence
of nowhere differentiable functions (see Oxtoby pp.45-46). One thing which
I only briefly made reference to in my paper is that the proof of the BCT
for arbitrary complete metric spaces requires the axiom of countable choice
(ACω ). In the special case that the space is separable, however, we do not
need this (see Levy pp.212-213). Another thing of interest that I mentioned
in my proposal was the equivalence of the BCT to the axiom of dependent
choices (DC), which states that for any nonempty set X and any entire
binary relation R on X, there is a sequence {xn } in X such that xn Rxn+1 for
each n ∈ N (and by entire we mean for each a ∈ X ∃ b ∈ X such that aRb).
Since (ACω ) is strictly weaker than (DC), the latter implies the BCT. The
converse is true as well, and was proven by Charles Blair 1977. I was unable
to find the paper containing the proof (our library doesn’t have the necessary
volume of the journal), and it did not occur to me until last week to email
him and ask if he had an electronic copy. I did email him and he offered to
send me a paper copy by surface mail, but I knew I wouldn’t receive it on
6
time. He also mentioned that the result is presented in the second edition of
Oxtoby’s Measure and Category, but that’s only available at the Scarborough
library, and I knew if I placed a request for it it would not arrive until this
week. I only mention this because I think that theorem looked like one of the
more interesting things related to the BCT, and I wanted to at least mention
it in my paper.
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Bibliography
[1]. Folland, Gerald B. Real Analysis: Modern Techniques and Their
Applications. 2nd ed. New York: John Wiley & Sons, 1999.
[2]. Oxtoby, John C. Measure and Category: A Survey of the Analogies
between Topological and Measure Spaces. New York: Springer-Verlag, 1971.
[3]. Levy, Azriel. Basic Set Theory. Berlin: Springer-Verlag, 1979.
[4]. Charles E. Blair. The Baire category theorem implies the principle
of dependent choices. Bulletin de l’Academie Polonaise des Sciences, 25 :
933 − 934, 1977.
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