Math 300: Final Exam Practice Solutions

For surjectivity, suppose that (x, y)E × O, so x is even and y is odd. Since x is even x 2
is an
integer. Hence if y < 0 we get f( x 2 , −y) = (x, y), while if y > 0 we have f(x 2
, y + 1) = (x, y). Thus
either way there is something in Z × Z + mapping to (x, y), so f is surjective. Thus f is bijective
so Z × Z + and E × O have the same cardinality as claimed.
5. (Possibly Tricky) The Cantor set is defined at the end of Section 2.3 in the book, under the
heading “Mathematical Perspective: An Unusual Set”. Show that the Cantor set is uncountable.
Remark. This is tricky, especially if you’ve never seen the Cantor set before. My purpose for
including this here is to show how one question can be converted into another, which you already
know the answer to. This is vague, but the point here is that we will show that an element of the
Cantor set can be uniquely described by a sequence of 0’s and 1’s, and the set of such sequences
can be shown to be uncountable using Cantor’s diagonalization argument. As for preparing for the
final, don’t worry so much about the part of this solution having to do with the Cantor set, but
you should at least be able to show that the set of sequences of 0’s and 1’s is uncountable.
Proof. We will use the notation the book uses when defining the Cantor set C. Let x ∈ C. We
construct an element of {0, 1} ∞ associated to this as follows. Since C = ⋂ A n , x ∈ A n for all n. In
particular, x ∈ A 1 so x is in one of the two intervals making up A 1 ; take the first element in our
sequence to be 0 if x is in the “left” interval [0, 1/3] and take the first element in our sequence to
be 1 if x is in the “right” interval [2/3, 1].
Now, whichever of these intervals x is in will itself split into two smaller intervals in the construction
of A 2 . Since x ∈ A 1 , x will be in one of these smaller intervals; take the next element
in our sequence to be 0 if it is the “left” interval x is in and take it to be 1 if x is in the “right”
interval. For instance, the interval [0, 1/3] splits into [0, 1/9] and [2/9, 1/3]. If x ∈ [0, 1/9] the first
two terms in the sequence we are constructing will be 0, 0, while if x ∈ [2/9, 1/3] we have 0, 1 as
the beginning of our sequence. Continuing in this manner, whichever interval making up A 2 that
x is in will split into two smaller pieces; take 0 as the third term in our sequence if x is in the left
piece and 1 if x is in the right piece, and so on. By keeping track of which interval x is in at each
step in the construction of the Cantor set in this manner we get a sequence of 0’s and 1’s.
For instance, if we get the sequence 0, 1, 1, 1, 0, 0, 0, . . ., x is in the “left” interval of A 1 , then in
the “right” smaller interval which this interval splits into, then in the “right” smaller interval this
splits into, then “right” again, then in the “left” smaller interval that this splits into, and so on.
(This is easier to imagine if you draw a picture of this splitting into smaller and smaller intervals
as we did during the review.)
This assignment of a sequence of 0’s and 1’s to an element x ∈ C defines a function C → {0, 1} ∞ .
It is injective since different elements in the Cantor set produces different sequences (at some point
in the construction, two different numbers in the Cantor set will belong to two different “smaller”
intervals), and it is surjective since given any sequence we can use it to single out an element of
the Cantor set. Thus C and {0, 1} ∞ have the same cardinality.
We claim that {0, 1} ∞ is uncountable, which then shows that the Cantor set is uncountable.
Given a listing of elements of {0, 1} ∞ :
n 11 , n 12 , n 13 , . . .
n 21 , n 22 , n 23 , . . .
n 31 , n 32 , n 33 , . . .
.
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