Is the power set of the natural numbers countable?

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

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?)

Attribution
Source : Link , Question Author : Martin Thoma , Answer Author : hmakholm left over Monica

Leave a Comment