Homework 3 Solutions
Problem 7.8.
Proposition. Let X = {0, 1}, and let B be the set of all countable subsets of X ω . Then
|B| = |X ω |.
Proof. Let M be the set of all functions Z+ × Z+ → X, i.e. the set of all infinite matrices


 x11 x12 x13


x
 21 x22 x23
{xij } = 

x
 31 x32 x33

 .
..
..
..
.
.
··· 


··· 



··· 


... 
where each xij ∈ X. Note that such a matrix has countably many entries (since Z+ × Z+ is
countable), so |M| = |X ω |.
Now, consider the function f : M → B that maps each matrix to its set of rows, i.e.
f ({xij }) =
(x11 , x12 , x13 , . . .), (x21 , x22 , x23 , . . .), (x31 , x32 , x33 , . . .), . . .
The image of f is B − {∅}. Indeed, since a matrix can have repeated rows, each element of
B − {∅} has infinitely many preimages. If we modify the function f slightly to map one of
the matrices to ∅, we obtain a surjection f : M → B, and therefore |M| ≥ |B|.
Finally, we can define an injection g : X ω → B by g(x) = {x}, so |X ω | ≤ |B|. Then
|X ω | ≤ |B| ≤ |M| = |X ω |,
which proves that |X ω | = |B|.
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Problem 10.4.
Let Z− denote the set of negative integers in the usual order.
Proposition. A simply ordered set A fails to be well-ordered if and only if it has a subset
with the same order type as Z− .
Proof. The backward direction is obvious, since a subset with the order type of Z− does not
have a minimum element. For the forward direction, suppose that A is not well-ordered.
Then there must exist a nonempty subset S of A that has no minimum element. Fix an
element b1 ∈ S. Since b1 is not the minimum element of S, there must exist a b2 ∈ S
for which b2 < b1 . Since b2 is not the minimum element of S, there must exist a b3 ∈ S
for which b3 < b2 . Continuing in this way, we can choose an infinite decreasing sequence
b1 > b2 > b3 > · · · of elements of S. Then {. . . , b3 , b2 , b1 } is a subset of A with the same
order type as Z− .
Proposition. Let A be a simply ordered set. If every countable subset of A is well-ordered,
then A is well-ordered.
Proof. By the previous proposition, if A is not well-ordered then A has a countable subset
that is not well-ordered.
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Problem 10.8.
Proposition. Let A1 and A2 be disjoint sets, well-ordered by <1 and <2 , respectively. Define
an order relation < on A1 ∪ A2 by letting a < b if either a, b ∈ A1 and a <1 b, or if a, b ∈ A2
and a <2 b, or if a ∈ A1 and b ∈ A2 . Then < is a well-ordering.
Proof. Let S ⊂ A1 ∪ A2 be nonempty. If S has any elements from A1 , then the minimum
element of S ∩ A1 is a minimum element for S. If S has no elements from A1 , then S ⊂ A2 ,
which implies that S has a minimum element.
Proposition. Let {Aα }α∈I be a disjoint family of sets, well-ordered by relations {<α }α∈I ,
and let <I be a well-ordering of the index set I. Define an order relation < on
S
α∈I
Aα by
letting a < b if either a, b ∈ Aα and a <α b for some α ∈ I, or if a ∈ Aα and b ∈ Aβ for
some α, β ∈ I with α <I β. Then < is a well-ordering.
Proof. Let S ⊂
S
α∈I
Aα be nonempty. Since <I is a well-ordering, there exists a minimum
element α ∈ I for which S ∩ Aα is nonempty. Then the minimum element of S ∩ Aα under
<α will also be the minimum element of S under <.
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