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MA2224 (Lebesgue integral) Tutorial sheet 1

[January 21, 2010]

Name: Solutions

1. If S  = {s1, s2, . . .} is a countably infinite set and t /∈ S , show that S ∪{t} is also countably

infinite.

Solution: We can list the elements of  S ∪ {t} by inserting t at the beginnings, that is as

S ∪ {t} = {t, s1, s2, . . .}.

Would this still be true if t ∈ S ? (Why?)

Solution: Yes because if t ∈ S , then S ∪ {t} = S .

Would it be true if we replaced “countably infinite” by “countable”? (Why?)

Solution: Yes, because countable means countably infinite or finite. We have already

considered the case when S  is infinite. In the remaining case when S  is finite, so that

S  = {s1, s2, . . . , sn} has n elements for some n ≥ 0, then S ∪ {t} has n + 1 elements if 

t /∈ S (and just n elements if t ∈ S ).

2. Show thatN×N is countably infinite. (Recall N×Nmeans the set of ordered pairs (n, m)with n, m ∈ N. Hint: Make a picture of N×N.)

Solution:

We can list the elements (the grid points) by following a zig-zag pattern

N×N = {(1, 1), (2, 1), (1, 2), (1, 3), (2, 2), (3, 1), (4, 1), (3, 2), . . .}

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so that the points (n, m) ∈ N×N with smaller values of n + m are listed earlier than larger

values and all pairs are included in the list. This shows that N × N is countably infinite

(elements can be listed in an infinite list).

3. If S  and T  are countably infinite sets, show that S × T  is countable.

Solution: This is really almost the same again, at least if  S  and T  are countably infinite.

Start with lists of the elements of S  = {s1, s2, s3, . . .} and T  = {t1, t2, t3, . . .}, list all the

pairs (sn, tm) by following the same zig-zag pattern as we did for N×N.

S × T  = {(s1, t1), (s2, t1), (s1, t2), (s1, t3), (s2, t2), (s3, t1), (s4, t1), (s3, t2), . . .}

Aside: This is also true in the case S  and T  are both countable, not necessarily

countably infinite. The argument is maybe messy as there are a number of cases

to consider.

If S  or T  is empty (zero elements) then S × T  is also empty. (So finite and so

countable.)

If S and T  are both finite, then S × T  is again finite.

If S  = {s1, s2, . . . , sn} is finite, but T  = {t2, t2, . . .} is (counatbly) infinite then

S ×T  = {(s1, t1), (s2, t1), . . . , (sn, t1), (s1, t2), (s2, t2), . . . , ((sn, t2), (s1, t3), . . .}

is countably infinite. On the other hand if  T  = {t1, t2, . . . , tm} is finite, while

S  = {s1, s2, . . .} is infinite then

S ×T  = {(s1, t1), (s1, t2), . . . , (s1, tm), (s2, t1), (s2, t2), . . . , (s2, tm), (s3, t1), . . .}

It is a fact that if S  is an infinite set, then S  contains a countably infinite subset.

Idea of how this is proved: Certainly S = ∅ and so there is an element s1 ∈ S . Now S \{s1}can’t be empty either as then S  = {s1} would be finite. So there is some s2 ∈ S \{s1}. We

can continue inductively to find sn ∈ S \ {s1, s2, . . . , sn−1}. Finially we get a counatbly

infinite subset {s1, s2, . . .} ⊆ S .

4. If S  is an infinite set and t /∈ S , show that there is a bijection f : S → S ∪ {t} (so that the

sets S and S 

∪ {t

}have the ‘same number’ of elements).

Solution: Consider a countably infinite subset {s1, s2, . . .} ⊆ S . Define a bijection f : S →S  ∪ {t} by f (s1) = t, f (s2) = s1, f (s3) = s2, in general f (sn+1) = sn, and finally

f (x) = x for x ∈ S \ {s1, s2, . . .}.

So the idea is to match the lists {s1, s2, . . .} and {t, s1, s2, . . .}, while leaving the rest of 

the elements of S alone.

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If S is an infinite set and T  is a countable set disjoint from S , show that there is a bijection

f : S → S ∪ T .

Solution: The idea is similar in the case of countably infinite T  = {t1, t2, . . .}, but now tomatch {s1, s2, . . .} and {t1, s1, t2, s2, t3, s3, . . .}, while leaving the rest of the elements of 

S alone.

So define f (s2n−1) = tn, f (s2n) = sn, f (x) = x for x ∈ S \ {s1, s2, . . .}.

In the case T  is finite, say T  = {t1, t2, . . . , tm} we might more easily be able to match

{s1, s2, . . .} and {t1, t2, . . . , tm, s1, s2, s3, . . .} (while leaving the rest of the elements of  S alone again).

(Another way would be to use induction and the step of adding one extra element. In fact,

that would produce bijections S  → S  ∪ {t1} → (S  ∪ {t1}) ∪ {t2} = S  ∪ {t1, t2} →S 

∪ {t1, t2, t3

} → · · ·S 

∪T  after m steps. The composition of the bijections would be a

bijection from S  to S ∪ T .)

In both of these approaches to the case of finite T , the situation when T  = ∅ is maybe

being overlooked. It would correspond to m = 0 above, we have S ∪ T  = S ∪ ∅ = S  in

this case and the identity map is a bijection from S  to itself.

Show that there is a bijection f : R \Q → R (from the irrationals to the whole of R).

Solution: This follows from the previous point by taking S  = R\Q and T  = Q. We know

Q is countably infinite and the the R \ Q of irrationals is infinite. (Are we sure we know

that? Well √ 2 is irrational as is n√ 2 for n ∈ N. In fact R \Q is uncountable.)

Richard M. Timoney

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