The power set of natural numbers is uncountable
September 20, 2015
Theorem 0.1. The power set of the natural numbers, P(N), is not countable.
Proof. Assume that P(N) is countable, i.e. there exists a pairing between N and P(N) such
that (i) every natural number is paired with a different element of P(N) and (ii) every element
of P(N) is paired with some natural number.
∀n ∈ N, let An denote the element of P(N) that the natural number n is paired with. (Note
that An is a subset N). We construct a subset B of natural numbers that is not paired with
any natural number.
Consider a table whose rows correspond to elements of P(N) and whose columns correspond
to the natural numbers. The table entry in the row corresponding to Ai and column j tells us
whether j is a member of Ai .
Now let’s construct our set B. We put j into B if j ∈
/ Aj , and say that j ∈
/ B if j ∈ Aj .
Hence, we have B = {x ∈ N | x ∈
/ Ax }.
Remember that every subset of the natural numbers must be paired with some natural
number . This means that B = Ak for some k. But by the way we defined B, k ∈ B if and
only if k ∈
/ Ak . This means that exactly one of B and Ak contains the element k. Therefore,
B is actually not Ak . Since k was arbitrary, this means that B is not paired with any natural
number. This contradicts the assumption that there exists a pairing between N and P(N), so
the assumption must be false. Furthermore, since we chose an arbitrary pairing, this means
that there is no pairing between N and P(N). It follows that P(N) is not a countable set.
In the proof above, we showed that for every pairing between natural numbers and subsets
of natural numbers, there is a subset of the natural numbers that is not paired. Note that we
didn’t make any assumptions about the pairing.
To give you a better understanding of why our argument works, let’s illustrate how it
constructs B for one particular pairing between N and subsets of N. Suppose A0 = ∅, A1 = {0},
A2 = {1}, and A3 = {1, 2, 3} are the sets paired with natural numbers 0, 1, 2, 3 respectively.
Then since 0 ∈
/ A0 , we put 0 in B, which ensures that B 6= A0 . Next, since 1 ∈
/ A1 , we put 1 in
B, and ensure that B 6= A1 . This process continues on, preventing all paired sets from being
B. We show part of this process in Table 0.1. We see from Table 0.1 that in order to construct
B, we just “flip” the diagonal entries of that table. For this reason, the technique we used in
our proof by contradiction is called diagonalization.
1
a\b
A0
A1
A2
A3
..
.
B
0
N
1
N
2
N
N
Y
N
N
..
.
N
Y
Y
..
.
N
Y
Y
Y
Y
..
.
3
N
N
N
Y
..
.
N
···
···
···
···
···
..
.
···
Table 0.1: Example of a table used to prove that P(N) is uncountable. A Y in row corresponding
to Ai and column corresponding to j indicates that j ∈ Ai . An N indicates that j ∈
/ Ai .
2