Is the power set of the natural numbers countable?

Question

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?

Answer 1

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?)

Answer 2

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.

Answer 3

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

Proof

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

Answer 4

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 $x_0\in f(x_0)$, so $x_0\in B$; contradiction. Therefore, $B$ is not in the range of $f$, and the proof is complete.

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*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 $\mathcal P(\mathbb N)$.

Answer 5

No the power set of the Natural numbers is not countable and there is a very nice argument to show it is indeed uncountable.

Assume the contrary and we can list out all possible subsets of the naturals.

Draw a table listing out all the subsets and next to each subset have an infinite binary number where if the $i^{th}$ position being a $0$ means that $i\in \mathbb{N}$ does not belong to the subset and if it is 1 then it belongs to the subset.

Now go along the list where in the $j^{th}$ subset flip the $j^{th}$ digit and putting it all together we get a binary sequence that has not been listed out as it differs from every subset in atleast one position by construction. Thus this is a contradiction hence we would not be able to list out all subsets of naturals hence it is uncountable.

In fact it has the same cardinality as Reals as in some sense we are trying to count all the possible binary numbers in the above argument. If we look at Reals in the binary form the above argument follows to show uncountability and each binary representation of a real number would also represent the subset of a Natural number.

Answer 6

The easiest proof that the subsets of naturals has the same cardinality as the reals, is:

Associate a subset with the characteristic function: 1, if the number is in a subset, 0 otherwise. Write these as dyadic fractions, starting at zero, so:

0.1010000.. . represents the set {1,3}

The correspondence is not unique, because 0.1111... = 1, but - modulus a countable set (of rational numbers) the interval [0,1] maps to the subsets, "almost" one-to-one.