Landon Kavlie
Math 711
Homework #1
14-September-2010
1. Let X be a set and A the set of all subsets A ⊂ X such that either A or
Ac is countable. Show that A is a σ-algebra.
Proof. We must show that
a) X ∈ A
b) If A ∈ A , then Ac ∈ A
c) If Ai ∈ A for i = 1, 2, . . . , then
S∞
i =1
Ai ∈ A .
a) X c = X − X = ∅ is countable. Hence Ac ∈ A .
b) Suppose A ∈ A . Then, either A or Ac is countable. If Ac is countable,
then certainly Ac ∈ A . If A is countable, then A = ( Ac )c is countable.
Hence Ac ∈ A .
c) Suppose Ai ∈ A for i = 1, 2, . . . Then, if each Ai is countable, certainly
S∞
i =1 Ai is countable. So suppose there is some j ∈ N so that A j is not
countable. Note that A j ∈ A ⇒ Acj is countable. Now, by DeMorgan’s
Laws,
(
∞
[
Ai )c =
i =1
T∞
c
i =1 A i
Acj
∞
\
Aic .
i =1
Acj
Since
⊂
and
is countable, it follows that (
T∞
S∞
c
i =1 Ai is countable. That is, i =1 Ai ∈ A .
S∞
i =1
Ai )c =
2. Find the σ-algebra on R that is generated by the one-element subsets.
Claim. The σ-algebra described above is the σ-algebra from Problem 1. That is, it
is the set of all A ⊂ R such that either A or Ac is countable.
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Proof. Suppose A is the σ-algebra generated by the single point sets of R.
Call the set of all A ⊂ R such that A or Ac is contable B. Then, since
σ-algebras are closed under countable unions, all countable subsets of R
are in A . Also, since σ-algebras are closed under complementation, all
subsets whose complements are countable are in A . Hence, A ⊂ B. Now,
take B ∈ B. Then, either B is countable or Bc is countable. Then, B or Bc
can be written as the countable union of single elements of R. That is, B
or Bc ∈ A . Since A is closed under complementation, B ∈ A . Hence,
B ⊂ A . Therefore, A = B.
3. Show that the σ-algebra of Borel sets B (R) is generated by the intervals (−∞, t] where t ∈ Q.
Proof. Let the σ-algebra generated by the intervals (−∞, t] for t ∈ Q be
given by B1 . Let the σ-algebra generated by (−∞, b) for b ∈ R be given by
B2 . Let the σ-algebra generated by [ a, b) for a, b ∈ R and a < b be given by
B3 . We will show that B (R) ⊂ B3 ⊂ B2 ⊂ B1 ⊂ B (R).
First off, as we showed in Proposition 1.1.3, B (R) is generated by the intervals (−∞, b] for b ∈ R. Since, Q ⊂ R, B1 ⊂ B (R).
Now, recall that any real number can be written as an infinite decimal
expansion. That is, for x ∈ R,
x = xn xn−1 . . . x2 x1 .d1 d2 . . .
For x1 , . . . xn , d1 , · · · ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Now, for each k ∈ N, xn . . . x1 .d1 . . . dk
is a rational number. Hence, each real number can be written as a nondecreasing sequence of rational numbers. So, for each b ∈ R
(−∞, b) =
∞
[
(−∞, qn − 1/n]
n =1
for the sequence qn of rationals converging to b. We add the 1/n fudge
factor for the case that b itself is rational. Thus, we have that B2 ⊂ B1 .
Now, consider the intervals [ a, b) for a, b ∈ R and a < b. Since, [ a, b) =
(−∞, b) ∩ (−∞, a)c , B3 ⊂ B2 .
Finally, notice that, for a, b ∈ R,
2
( a, b) =
∞
[
[ a + 1/n, b)
n= N
for large enough N. Hence, B (R) ⊂ B3 .
In conclusion, B (R) ⊂ B3 ⊂ B2 ⊂ B1 ⊂ B (R) ⇒ B1 = B (R).
4. Show that the irrational numbers R − Q area a Gδ set in R. Use this
and the Baire Category Theorem to show that the rational numbers Q are
not a Gδ set in R.
Proof. First, notice that, for each q ∈ Q, the sets R − {q} are open. Also,
since Q is countable, the collection of sets Q := {R − {q} : q ∈ Q} is also
T
countable. Therefore, q∈Q (R − {q}) ∈ Gδ . By DeMorgan’s Laws,
\
(R − {q}) = R −
[
{q} = R − Q
q ∈Q
q ∈Q
R − Q = the irrational numbers are a Gδ set.
To prove that Q 6∈ Gδ , suppose to the contrary that Q ∈ Gδ . Then, Q =
n=1 An for An open in R. Notice that, since Q is dense in R, each An must
also be dense in R. Also, R − q for q ∈ Q is also dense in R. Then, since R
is a complete metric space, by the Baire Category Theorem, it is also a Baire
space. Hence, for any countable collection of dense open sets in R, their
intersection must also be dense in R. But,
T∞
(
∞
\
n =1
An ) ∩ (
\
(R − {q})) = Q ∩ (R − Q) = ∅
q ∈Q
This is a contradiction. Hence, Q 6∈ Gδ .
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