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The set of all subsets of $\mathbb N$ is $ \mathcal{P}{(\mathbb N)} $ right?

So how can I prove that $ \mathcal{P}{(\mathbb N)} $ is countable?

Seen
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    $P(\Bbb N)$ is not countable. But the set you describe in your title is very different from $P(\Bbb N)$. – David Nov 03 '14 at 00:17
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    The set of all $\textit{finite}$ subsets of $\mathbb{N}$ is different from $\mathcal{P}(\mathbb{N})$. Notice for example that the set of positive even integers is not in the former, but is in the latter. – Peter Woolfitt Nov 03 '14 at 00:17
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    $\mathscr P(\mathbb N)$ is, in fact, uncountable. But that's not the set you describe in the title for this question. – MPW Nov 03 '14 at 00:17
  • If you mean finite subsets, how many subsets are there of cardinality 1? How many of cardinality 2? Dot dot dot ... – user4894 Nov 03 '14 at 00:19
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    Glad we all agree on that. . . even if we don't agree on what kind of $P$ to use ;-) – David Nov 03 '14 at 00:21
  • And you still don't have Cantor-Bernstein, right? That's going to make it more tricky to prove. (But there is in fact a nice and clever solution that doesn't need Bernstein. Hint: Binary!) – hmakholm left over Monica Nov 03 '14 at 00:22
  • Sorry I'm still new to settheory… Why is the set of positive even integers not in the set described but in $ \mathcal{P}{(\mathbb N)} $ ? – Seen Nov 03 '14 at 00:23
  • @Seen : The set of all subsets of $\mathbb N$ is not countable, but the set of all finite subsets of $\mathbb N$ is countable. ${}\qquad{}$ – Michael Hardy Nov 03 '14 at 00:23
  • ok and when is a subset finite? How do I tell the difference? – Seen Nov 03 '14 at 00:25
  • @Seen $\mathcal{P}(\mathbb{N})$ refers to all subsets of $\mathbb{N}$. The positive even integers are certainly a subset of $\mathbb{N}$ (every positive even integer is a positive integer), but it is certainly not a finite subset - you can't write all of the elements down (more technically, if this set were finite, then there would be a maximum element, but no such element exists). – Peter Woolfitt Nov 03 '14 at 00:30

4 Answers4

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Let $A$ be a finite subset of $\mathbb N$. Write $$ A = \{ a_1, a_2, \dots , a_n\}. $$

Let $B$ denote the set of all such finite subsets and define $f : B \rightarrow \mathbb N$ according to : $$ \text{for } A = \{a_1, a_2, \dots , a_n\} : f(A) = 2^{a_1} + 2^{a_2} + \dots + 2^{a_n}$$

Then it is easy to see that $f$ is an injection of $B$ into $\mathbb N$, so $B$ must be countable.

gamma
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  • I dont get this part: $$ f(A) = 2^{a_1} + 2^{a_2} + \dots + 2^{a_n}$$. What do you mean by the different degrees of the twos? – Seen Nov 03 '14 at 00:37
  • @Seen Every natural number can be expressed uniquely in binary form. This is just the binary representation of a natural number. The exponents (the set members) uniquely determine that natural number, and conversely, a natural number is uniquely determined by its exponents. – gamma Nov 03 '14 at 00:43
  • So every Set is defined by a binary representation of its set members and therefor unique? For countability only a injection is needed not a bijection? – Seen Nov 03 '14 at 00:46
  • @Seen Yes. Any binary number defines one and only one natural number, and any natural number is representable by one and only one binary number. Here, because we do not have $ 0 \in \mathbb N $, we cannot map onto odd numbers using our map. Therefore the image of $ B $ is an infinite, proper subset of $ \mathbb N $, so it must be countable. – gamma Nov 03 '14 at 00:57
  • Thanks! I'm really impressed by that idea :) – Seen Nov 03 '14 at 01:16
  • @Seen Ya, set theory has all sort of nifty ideas which can be really impressive. Good luck with your studies. – gamma Nov 03 '14 at 01:18
  • @Seen I should also add that I'm really impressed by computers. A year ago I would never have believed that machines could be intelligent. Today, I'm very open to this idea. Computation is so pervasive in the world around us - or more precisely, the world around us may be entirely computable - with emphasis on "may be". – gamma Nov 03 '14 at 01:23
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Hint. There are only finitely many subsets of $\Bbb N$ whose elements add up to a given sum $n$.

David
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$$\aleph_0\le|\{A\subseteq\mathbb N: |A|<\aleph_0\}|=\Big|\bigcup_{n\in\mathbb N}\{A\subseteq\mathbb N: |A|=n\}\Big|\le\aleph_0\cdot\sup_{n\in\mathbb N}\aleph_0^n=\aleph_0\cdot\sup_{n\in\mathbb N}\aleph_0=\aleph_0$$

user2345215
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\begin{align} \text{All nonempty subsets of } \{1\}: & \{1\} \\ \text{All nonempty subsets of } \{1,2\}: & \{1\}, \{2\}, \{1,2\} \\ \text{All nonempty subsets of } \{1,2,3\}: & \{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\} \\ \vdots \qquad \vdots \qquad \vdots \qquad{} & {} \qquad \vdots \qquad \vdots \qquad \vdots \end{align}

Now count them, starting on the first row above: one, two, three, four, etc., skipping duplicates.

Michael Hardy
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