1. The Baire category theorem - Aarhus Universitet

1. The Baire category theorem 

Klaus Thomsen matkt@imf.au.dk 

Institut for Matematiske Fag 

Det Naturvidenskabelige Fakultet 

Aarhus Universitet 

September 2005 

Klaus Thomsen 1. The Baire category theorem

We read in W. Rudin: Functional Analysis 


We read in W. Rudin: Functional Analysis 

Chapter 2, Starting on page 41 of my edition 


Definition of first and second category 



Let S be a topological space, i.e. 




S is a set with a specified collection τ of subsets, the open sets, 

such that 




S is a set with a specified collection τ of subsets, the open sets, 

such that 

i) S, ∅ ∈ τ, 

ii) when U1,U2,... ,Un is a finite number of elements of τ the 

intersection 

n 

is also in τ. 

i=1 

iii) for any collection Uα,α ∈ I, of sets from τ the union 

is also in τ. 

 

α∈I 

Ui 

Uα 



Example: A metric space X with metric d is a topological space: 



Example: A metric space X with metric d is a topological space: 

The open sets, τ, consists of the subsets U of X which satisfies the 

following condition: 

∀x ∈ U ∃δ > 0 : {y ∈ X : d(y,x) < δ} ⊆ U. 


A subset A of S is nowhere dense when the closure A has empty 

interior. 



interior. 

Reminder: The closure A of A is by definition the intersection of 

the closed sets containing A. The interior of any subset B ⊆ X is 

the union of the open sets contained in B. 



interior. 




Thus a subset A of S is nowhere dense when the closure A does 

not contain any open subset. 



interior. 




Thus a subset A of S is nowhere dense when the closure A does 

not contain any open subset. 

Examples: The integers Z is nowhere dense in R. 

The set { 1 

n : n ∈ N} is nowhere dense [0,1]. 



First category 

A set A ⊆ S is of first category if it is the union of a countable 

collection of nowhere dense sets, i.e. if A = ∞ i=1 Ai, where Ai have 

empty interior. 







Example: The rational numbers Q form a set of first category in 

the real line R. Note that Q itself is not nowhere dense; but it is 

the union of a countable collection of sets that are! 







Example: The rational numbers Q form a set of first category in 

the real line R. Note that Q itself is not nowhere dense; but it is 

the union of a countable collection of sets that are! 

A set of first category is ’small’ from the topological viewpoint. 



Second category 

A set A ⊆ S is of second category when it is not of the first 

category. 






Via the Baire category theorem we will become able to give many 

examples - for example show that R is of the second category as a 

subset of itself. (As well as [0,1] ⊆ R). 






Via the Baire category theorem we will become able to give many 

examples - for example show that R is of the second category as a 

subset of itself. (As well as [0,1] ⊆ R). 

Subsequently we shall then see why the notion is usefull ! 

Philosophically a set of second category is a large subset of S from 

the topological point of view. 


Basic properties the notions of category 

(a) If A ⊆ B and B is of the first category, then so is A. 




Indeed if B = ∞ 

i=1 Bi, then A = ∞ 

i=1 Bi ∩ A and Bi ∩ A ⊆ Bi. 







(b) Any countable union of sets of the first category is of the first 










Indeed, if B = ∞ i=1 Bi and Bi = ∞ j=1 B(i,j), for all i, then 

B(i,j), i,j ∈ N is a countable collection of sets whose union is B. 











(c) Any closed set with empty interior is of the first category. 











(c) Any closed set with empty interior is of the first category. 

Indeed, if E is closed, E = E ∪ ∅ ∪ ∅ ∪ ∅ ∪ ... is a union of a 

countable collection of nowhere dense sets. 



(d) If h : S → S is a homeomorphism, E ⊆ S and h(E) are of the 

same category. 





It suffices to show that h(E) is of the first category if and only if E. 






Furthermore, since E = h −1 (h(E)), it suffice to show that h(E) is 

of the first category when E is. 






Furthermore, since E = h−1 (h(E)), it suffice to show that h(E) is 


So assume that E = ∞ i=1 Ei, where each Ei has empty interior. 

Then h(E) = ∞ i=1 h (Ei) and h(Ei) = h 

Ei . 










Ei . 

If U ⊆ h 

Ei is open, h−1 (U) ⊆ Ei is also open. 










Ei . 

If U ⊆ h 

Ei is open, h−1 (U) ⊆ Ei is also open. 

Since Ei has empty interior, this implies that h−1 (U) = ∅, i.e. 

U = h−1 (h(U)) = ∅. 


The Baire’s category theorem 

The Baire’s category theorem. 

If S is either 

(a) a complete metric space, or 

(b) a locally compact Hausdorff space, 

then the intersection of every countable collection of dense open 

sets in S is dense is S. 









Thus if Ei,i = 1,2,3,... , are dense and open in S, then ∞ 

i=1 Ei is 

dense in S. 









Thus if Ei,i = 1,2,3,... , are dense and open in S, then ∞ 

i=1 Ei is 

dense in S. 

Corollary. 

A complete metric space or a locally compact Hausdorff space is of 

the second category; thus they are not the union of a countable 

collection of nowhere dense subsets. 


Proof of the corollary 

Reminder ?: A subset E of S is dense when E ∩ U = ∅ for all open 

and non-empty subsets U. 





Examples: 1) Q is dense in R. 






The polynomials are dense in C[0,1]. 







Nullermændene er tætte i min ældste søns værelse. 








Proof of corollary using the theorem: 








Proof of corollary using the theorem: 

If S = ∞ i=1 Ei, where each Ei is nowhere dense, observe that 

∅ = ∞ i=1 Ec 

i ⊇ ∞ i=1 Ei 

c c 

and each Ei is dense and open. This 

contradicts the theorem. 


Proof of Baire’s category theorem 

Let V1,V2,V3,... be open and dense subsets of S. We must show 

that ∞ i=1 Vi is dense in S when S is either a complete metric 

space or a locally compact Hausdorff space. 




that ∞ 

i=1 Vi is dense in S when S is either a complete metric 


Consider an open non-empty subset B0 of S. We want to show 

that ∞ 

i=1 Vi ∩ B0 = ∅. Since V1 is dense in S, B0 ∩ V1 = ∅. 







that ∞ i=1 Vi ∩ B0 = ∅. Since V1 is dense in S, B0 ∩ V1 = ∅. 

We claim that there is an open set B1 such that 

B1 ⊆ V1 ∩ B0 

Klaus Thomsen 1. The Baire category theorem 

(1)








B1 ⊆ V1 ∩ B0 

When S is a complete metric space we will require that d(y,x) ≤ 1 

when x,y ∈ B1 and when S is locally compact Hausdorff we require 

that B1 is compact. 


(1)








B1 ⊆ V1 ∩ B0 




The argument for this differ in the two cases. Assume first that S 

is a complete metric space. Take x ∈ V1 ∩ B0. 


(1)








B1 ⊆ V1 ∩ B0 




The argument for this differ in the two cases. Assume first that S 

is a complete metric space. Take x ∈ V1 ∩ B0. 

Since V1 ∩ B0 is open there is a 1 > δ > 0 such that 

{y ∈ S : d(y,x) < δ} ⊆ V1 ∩ B0. Then 

B1 = y ∈ S : d(y,x) < δ 

 

2 has the desired property. 


(1)

On locally compact Hausdorff spaces 

Before we continue the proof of Baire’s category theorem we 

develop a little of the general theory of locally compact Hausdorff 

spaces. 





spaces. 

Recall that a topological space is locally compact when every point 

has an open neighborhood with compact closure. 





spaces. 

Recall that a topological space is locally compact when every point 

has an open neighborhood with compact closure. 

Lemma 

Let X be a Hausdorff space, K a compact subset of X and p a 

point in the complement K c = X \K of K. There are then open 

sets U and W such that p ∈ U,K ⊆ W and U ∩ W = ∅. 



Proof. 

Consider an element k ∈ K. Since X is Hausdorff there are open 

sets, Wk ⊆ X and Uk ⊆ X, such that k ∈ Wk,p ∈ Uk and 

Wk ∩ Uk = ∅. 



Proof. 



Wk ∩ Uk = ∅. 

Note that the collection {Wk : k ∈ K} is an open cover of K. 

Since K is compact there is a finite collection k1,k2,... ,kN such 

that 

K ⊆ Wk1 ∪ Wk2 ∪ · · · ∪ WkN . 



Proof. 



Wk ∩ Uk = ∅. 



that 

K ⊆ Wk1 ∪ Wk2 ∪ · · · ∪ WkN . 

Set W = Wk1 ∪ Wk2 ∪ · · · ∪ WkN and U = Uk1 ∩ Uk2 ∩ · · · ∩ UkN . 



Proof. 



Wk ∩ Uk = ∅. 



that 

K ⊆ Wk1 ∪ Wk2 ∪ · · · ∪ WkN . 


Corollary: In a Hausdorff space, any compact subset is closed. 



Proof. 



Wk ∩ Uk = ∅. 



that 

K ⊆ Wk1 ∪ Wk2 ∪ · · · ∪ WkN . 


Corollary: In a Hausdorff space, any compact subset is closed. 

Proof: By Lemma1 the complement of a compact set is open. 



Lemma 

Let X be a Hausdorff space and {Kα : α ∈ A} a collection of 

compact subsets of X. If 

α∈A Kα = ∅, there is a finite subset 

F ⊆ A such that 

α∈F Kα = ∅. 



Lemma 





α∈F Kα = ∅. 

Proof. 

 

Fix one of the Kα’s; say Kα1 . Since α∈A Kα = ∅, every element of 

Kα1 is in the complement Kc α of Kα, for some α. This means that 

{Kc α : α ∈ A} covers Kα1 . 



Lemma 





α∈F Kα = ∅. 

Proof. 

 

Fix one of the Kα’s; say Kα1 . Since α∈A Kα = ∅, every element of 

Kα1 is in the complement Kc α of Kα, for some α. This means that 

{Kc α : α ∈ A} covers Kα1 . 

By Corollary ?? each Kc α is open, so by compactness of Kα1 there is 

a finite subset F0 ⊆ A such that 

 

⊆ K c α. 

Kα1 

α∈F0 

Set F = {α1} ∪ F0. Then 

α∈F Kα = ∅. 



Lemma 

Let X be a topological space. 

a) The union of a finite number of compact sets in X is itself 

compact. 

b) If K is a compact subset of X and B ⊆ K is a closed subset of 

K, then B is compact. 



Lemma 



compact. 



Proof. 

a) is left to the reader. 



Lemma 



compact. 



Proof. 

a) is left to the reader. 

To prove b), consider an open cover {Uα : α ∈ A} of B. Then 

{Bc } ∪ {Uα : α ∈ A} is an open cover of K and hence 

K ⊆ B c ∪ 

for some finite collection F ⊆ A. 

α∈F 

Uα 



Proof. 

It follows that B ⊆ 

α∈F Uα, proving that B is compact. 



Proof. 

It follows that B ⊆ 

α∈F Uα, proving that B is compact. 

Lemma 

Let X be a locally compact Hausdorff space, K a compact subset 

of X and U an open subset of X such that K ⊆ U. There is then 

an open set V in X with compact closure V such that 

K ⊆ V ⊆ V ⊆ U. 



Proof. 

Since X is locally compact we can, for each k ∈ K, find an open 

set Wk such that k ∈ Wk and Wk is compact. 



Proof. 



Since {Wk : k ∈ K} is an open cover of K, and K is compact, 

there is a finite collection k1,k2,...,kN such that 

K ⊆ Wk1 ∪ Wk2 ∪ · · · ∪ WkN . Set W = Wk1 ∪ Wk2 ∪ · · · ∪ WkN and 

note that W ⊆ Wk1 ∪ Wk2 ∪ · · · ∪ WkN . 



Proof. 







It follows from Lemma 3 that W is compact. This completes the 

proof in the case where U = X, since we can then take V = W. 



Proof. 









When U = X, let C = Uc be the complement of U in X. For each 

p ∈ C there is an open set W ′ p such that K ⊆ W ′ p and p /∈ W ′ p . 

This follows from Lemma 1. 



Proof. 









When U = X, let C = Uc be the complement of U in X. For each 

p ∈ C there is an open set W ′ p such that K ⊆ W ′ p and p /∈ W ′ p . 

This follows from Lemma 1. 

Note that 

C ∩ W ∩ W ′ p = ∅, 

p∈C 

and that each set C ∩ W ∩ W ′ p is compact. 



Proof. 

It follows from Lemma 2 that there is a finite set, p1,p2,... ,pN, 

such that 

N 

C ∩ W ∩ W ′ pi = ∅. (2) 

i=1 



Proof. 

It follows from Lemma 2 that there is a finite set, p1,p2,... ,pN, 

such that 

N 

C ∩ W ∩ W ′ pi = ∅. (2) 

i=1 

Set V = W ∩ W ′ p1 ∩ W ′ p2 ∩ · · · ∩ W ′ pN . Then 

K ⊆ V ⊆ V ⊆ W ∩ W ′ p1 ∩ W ′ p2 ∩ · · · ∩ W ′ pN . It follows from (2) 

that W ∩ W ′ p1 ∩ W ′ p2 ∩ · · · ∩ W ′ pN ⊆ U and from Lemma 3 that V 

is compact. 


Proof of Baire’s category theorem - continued 







B1 ⊆ V1 ∩ B0 


(3)








B1 ⊆ V1 ∩ B0 





(3)








B1 ⊆ V1 ∩ B0 




This is now straightforward in case of a locally compact Hausdorff 

space; thanks to the last lemma. 


(3)


Next we find an open set B2 such that 

B2 ⊆ V2 ∩ B1 

such that d(y,x) ≤ 1 

2 when S is complete metric space and such 

that B2 is compact when S is a locally compact Hausdorff space. 




B2 ⊆ V2 ∩ B1 




This is done as above: Since V2 is dense, V2 ∩ B1 is not empty. 

Repeat then the previous arguments. 




B2 ⊆ V2 ∩ B1 




This is done as above: Since V2 is dense, V2 ∩ B1 is not empty. 

Repeat then the previous arguments. 

Thus we construct recursively a sequence B0,B1,B2,... of open 

sets such that 

Bn ⊆ Vn ∩ Bn−1 

for all n ≥ 1, d(x,y) ≤ 1 

n when x,y ∈ Bn in the case where S is a 

complete metric space, and such that Bn is compact in the case 

where S is a locally compact Hausdorff space. 



The proof is now completed as follows; first the case when S is a 

locally compact Hausdorff space: 





B0 ⊇ B1 ⊇ B2 ⊇ B3 ⊇ ... and each Bk,k ≥ 1, is compact and 

non-empty. 






non-empty. 

It follows that ∞ k=1 Bk = ∅. Since Bk ⊆ Vk−1 for all k, we find 

that 

∞ ∞ 

Vk. 

k=1 

Bk ⊆ 

k=0 






non-empty. 

It follows that ∞ k=1 Bk = ∅. Since Bk ⊆ Vk−1 for all k, we find 

that 

∞ ∞ 

Vk. 

k=1 

Bk ⊆ 

But ∞ 

k=1 Bk ⊆ B0, so we see that 

∞ 

k=1 

Bk ⊆ 

k=0 

∞ 

Vk ∩ B0, 

k=1 

which was what we wanted to prove. 



Finally the completion of the proof in the case of a complete metric 

space: 




space: 

As above B0 ⊇ B1 ⊇ B2 ⊇ B3 ⊇ ... and ∞ 

k=1 Bk ⊆ ∞ 

k=0 Vk ∩B0. 




space: 

As above B0 ⊇ B1 ⊇ B2 ⊇ B3 ⊇ ... and ∞ 

It remains therefore only to show that ∞ 

k=1 Bk = ∅. 

k=1 Bk ⊆ ∞ k=0 Vk ∩B0. 




space: 

As above B0 ⊇ B1 ⊇ B2 ⊇ B3 ⊇ ... and ∞ k=1 Bk ⊆ ∞ k=0 Vk ∩B0. 

It remains therefore only to show that ∞ Choose xk ∈ Bk. We claim that {xk} ∞ k=1 

k=1 Bk = ∅. 

is a Cauchy sequence in S. 




space: 


It remains therefore only to show that ∞ k=1 Bk = ∅. 

Choose xk ∈ Bk. We claim that {xk} ∞ k=1 is a Cauchy sequence in S. 

To see this, let ǫ > 0 be given. Choose N so large that 1 

N ≤ ǫ. 




space: 





N ≤ ǫ. 

Let n,m ≥ N. Then xn,xm ∈ BN, and hence d (xn,xm) ≤ 1 

N ≤ ǫ. 




space: 





N ≤ ǫ. 


N ≤ ǫ. 

Thus {xk} ∞ k=1 is a Cauchy sequence in S and we set 

x = limk→∞ xk. 




space: 





N ≤ ǫ. 


N ≤ ǫ. 



Since xn ∈ Bk when n ≥ k, we conclude that x ∈ Bk. 




space: 





N ≤ ǫ. 


N ≤ ǫ. 




(Indeed, if x /∈ Bk there is a δ > 0 such that 

c 

{y ∈ S : d(y,x) < δ} ⊆ Bk . Since xn ∈ {y ∈ S : d(y,x) < δ} 

for all large n, this is a contradiction.) 




space: 





N ≤ ǫ. 


N ≤ ǫ. 




(Indeed, if x /∈ Bk there is a δ > 0 such that 

c 

{y ∈ S : d(y,x) < δ} ⊆ Bk . Since xn ∈ {y ∈ S : d(y,x) < δ} 

for all large n, this is a contradiction.) 

Hence x ∈ ∞ k=1 Bk and the proof is complete!

1. The Baire category theorem - Aarhus Universitet

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