Is the powerset of a countably infinite set countable?

Question

I remember having heard that if $S$ is countable infinite, then $\mathcal P(S)$ is uncountably infinite. My intuition, however, tells me it should not be. Since $S$ is countable, you can enumerate it. Let's define:

$$S=\{x_1,x_2,\dots,x_n,\dots\}$$

One could define an injective function $f : N \to \mathcal P(S)$.

$f$ takes a natural number and converts it into binary form. Let's say $6$ is converted into $110$. Starting from the right, defining the rightmost digit to be digit number $1$, we can convert this into a subset of $S$ by including $x_n$ if and only if digit $n$ is $1$. What am I missing here?

An injection from the naturals is not enough to be countable. You need it in the other direction. — Tobias Kildetoft, Jun 25 '14 at 07:07
Whoops, messed up the order. Well isn't my $f$ bijective? For any subset you pick, there seems to exist a natural number that gives it. — user160013, Jun 25 '14 at 07:08
@user160013 The problem with $f$ is that in order to label a subset of $R\subseteq S$ in the way you indicated, you need to specify an infinitely long string of $0$'s and $1$'s that encodes what elements are included in $E$. Such infinite strings cannot be represented by natural numbers in a “bijective fashion.” — triple_sec, Jun 25 '14 at 07:19

score 0 · Answer 1 · answered Jun 25 '14 at 07:08

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You are correct that there is such a map. But $f: N \to P(S)$ is not surjective. Actually all infinite subsets of $S$ are not in the image.

answered Jun 25 '14 at 07:08

Wow, I forgot about the infinite subsets. So I take it the set of all finite subsets is countable, still? – user160013 Jun 25 '14 at 07:09
@user160013 Yes, that is correct. – Tobias Kildetoft Jun 25 '14 at 07:09
Yes, and the you just wrote down a proof. – Jun 25 '14 at 07:10
Am I also correct in that the set of all finite subsets of an uncountable set is not countable? I thought about applying my proof to that, but then I realized that there can be no order to enumerate the set in. – user160013 Jun 25 '14 at 07:15
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@user160013 In any case, you can write down an injection from $X$ to $P(X)$, sending $x\mapsto {x}$. – Jun 25 '14 at 07:16
@John: That would not give all finite subsets. – user160013 Jun 25 '14 at 07:17
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@user160013: Yes, but that shows "the set of all finite subset of $X$" is not smaller then $X$. – Jun 25 '14 at 07:18

score 0 · Answer 2 · answered Jun 25 '14 at 07:13

The existence of an injection $f:A\to B$ shows that “$B$ is larger than $A$,” in the sense that there are “sufficiently many” elements in $B$ such that you can assign each $a\in A$ a corresponding element $f(a)\in B$ with no repetition.

However, to show that “$B$ is smaller than $A$,” you must show that there exists a surjection $g:A\to B$. The existence of a surjection shows that there are “sufficiently many” elements in $A$ such that any $b\in B$ can be generated as $g(a)$ for some $a\in A$.

It turns out there is no surjective function $g:\mathbb N\to\mathcal P(S)$ for any countably infinite set $S$. There are “not enough” elements in $\mathbb N$ to represent every subset of $S$ as the image of a natural number.

Is the powerset of a countably infinite set countable?

2 Answers2