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Some explanations:

A set S is countable if there exists an injective function $f$ from $S$ to the natural numbers ($f:S \rightarrow \mathbb{N}$).

$\{1,2,3,4\}, \mathbb{N},\mathbb{Z}, \mathbb{Q}$ are all countable.

$\mathbb{R}$ is not countable.

The power set $\mathcal P(A) $ is defined as a set of all possible subsets of A, including the empty set and the whole set.

$\mathcal P (\{\})=\{\{\}\}, \mathcal P (\mathcal P(\{\}))=\{\{\}, \{\{\}\}\} $

$\mathcal P(\{1,2\})=\{\{\}, \{1\},\{2\},\{1,2\}\}$

My question is:

Is $\mathcal P(\mathbb{N})$ countable? How would an injective function $f:S \rightarrow \mathbb{N}$ look like?

Mark Cidade
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Martin Thoma
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    It isn't countable. To prove this, you can use diagonalization directly, or use the fact, which presumably has been proved by now, that the reals are uncountable. – André Nicolas Oct 31 '11 at 23:16
  • This question has been asked before. – Asaf Karagila Oct 31 '11 at 23:19
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    The cardinality of the power $\mathcal P (A)$ of any set A is always higher than the cardinality of a set A (Source: "Lineare Algebra", ISBN 978-3-528-66508-1, page 14) – Martin Thoma Apr 04 '12 at 09:53
  • As a footnote to the answers already given, you should also see a useful result known variously as the Schroeder-Bernstein, Cantor-Bernstein, or Cantor-Schroeder-Bernstein theorem. Some books present the easy proof; some others have the hard proof of it. – DanielWainfleet Aug 24 '16 at 18:36
  • It's not countable, as provable by diagonal argument, but the set of all FINITE subsets, and even ordered sequences, of natural numbers, or even integers or rational numbers, is, which I first realized by using extended definitions of prime factorization as ordered sequences of exponents to first however-many primes, though there may be an easier proof. – Mr. Nichan Feb 17 '24 at 21:28

4 Answers4

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Cantor's Theorem states that for any set $A$ there is no surjective function $A\to\mathcal P(A)$. With $A=\mathbb N$ this implies that $\mathcal P(\mathbb N)$ is not countable.

(But where on earth did you find those nice explanations of countability and power sets that didn't also tell you this?)

hmakholm left over Monica
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    Those nice explanations of countability and power sets were written by myself. Thanks for the compliment ;-) I just wanted to make sure that everybody who reads this question knows what I'm talking about – Martin Thoma Oct 31 '11 at 23:58
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    Today I thought about this again and I am a little bit confused. The powerset $\mathcal{P}(A_n)$ with $A_n := {1, ..., n}$ is finite for every $n$. So can't I simply count up for every $A_n$ (so first $\mathcal{P}(A_1)$, then $\mathcal{P}(A_2)$, $\mathcal{P}(A_3)$, ...? Where does it break? – Martin Thoma Nov 08 '14 at 21:41
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    @moose: You're going to miss every subset of $\mathbb N$ that is not finite, such as the set of all even numbers. – hmakholm left over Monica Nov 08 '14 at 21:45
  • Thank you! It's so obvious when you read it ... I ask because I have an application where I need (theoretically; it is a generator) every finite subset of $\mathbb{N}$ and this is something that bothered me. – Martin Thoma Nov 08 '14 at 21:47
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    @moose: Bonus hint: The binary representation of a natural number provides a nice enumeration of the finite subsets of $\mathbb N$. – hmakholm left over Monica Nov 08 '14 at 21:50
  • would this be true for the rationals and integers as well? – Jared Mar 17 '15 at 22:55
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Power set of natural numbers has the same cardinality with the real numbers. So, it is uncountable.

In order to be rigorous, here's a proof of this.

Harrison Grodin
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Karatuğ Ozan Bircan
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Here is my favorite proof that $$|\mathcal{P}(M)| > |M|, \quad\forall M.$$

Proof

$$\big[\exists\phi:M \stackrel{\mathrm{on}}{\to}\mathcal{P}(M)\big]$$ $$\Downarrow$$ $$\Big[\big[A_{\not\phi} := \{m|m\in M, m\notin\phi(m)\}\big]\Rightarrow A_{\not\phi}\in\mathcal{P}(M)\Rightarrow\exists p\in M:\phi(p) = A_{\not\phi}\Big]$$ $$\Downarrow$$ $$\Big[\big[p\in A_{\not\phi}\Rightarrow p\notin A_{\not\phi}\big]\quad \mathrm{and}\quad\big[p\notin A_{\not\phi}\Rightarrow p\in A_{\not\phi}\big]\Big]$$ $$Q.E.D.$$

LRDPRDX
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Cantor's Theorem tells us that for every set $A$, there is no surjection from $A$ to $\mathcal P(A)$. In particular, there is no surjection from $\mathbb N$ to $\mathcal P(\mathbb N)$, and so $\mathcal P(\mathbb N)$ is not countable.* For completeness, I will give the standard proof of Cantor's Theorem here.

Suppose $f:A\to\mathcal P(A)$ is a function, and consider the set $B=\{x\in A:x\not\in f(x)\}$. We shall show that $B$ is not in the range of $f$, and so $f$ is not surjective. Suppose, to the contrary, that $B$ is in the range of $f$. Then, there is an $x_0\in A$ such that $f(x_0)=B$.

Now consider the question of whether $x_0$ is an element of $B$. If $x_0\in B$, then $x_0\not\in f(x_0)$, so $x_0\notin B$; contradiction. On the other hand, if $x_0\not\in B$, then $x\in f(x_0)$, so $x_0\in B$; contradiction. Therefore, $B$ is not in the range of $f$, and the proof is complete.

*This means that there is no injection from $\mathcal P(\mathbb N)$ to $\mathbb N$ either, for if $f:\mathcal P(\mathbb N)\to\mathbb N$ were an injection, then $f$ would have a left inverse, and such a left inverse would be a surjection from $\mathbb N$ to $\mathcal P(\mathbb N)$.

Joe
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