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Prove that $\mathcal{P}(\Bbb N)$ is not countable. Alternatively, prove that $|\mathcal{P}(\mathbb N)|\ne|\mathbb N|$.

(You may assume the following result from tutorial: If $S$ is infinite and countable, then there exists a bijection $f:S\rightarrow\mathbb N$.)

And I think when they say $\mathcal{P}(\mathbb N)$ they mean the power set.

I'm not sure about this questions, any tips would be great thanks!

user136088
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    You mean to say the power set of the natural numbers. Also if you aren't sure, could you state where you are having trouble? – IAmNoOne Mar 23 '14 at 07:15
  • This is very standard. Have you seen any results showing that some set is uncountable? – Andrés E. Caicedo Mar 23 '14 at 07:18
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    Cantor's theorem – Jose Antonio Mar 23 '14 at 07:29
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    Take a real number $r$ in $[0,1]$. Construct $S_r$ as follows: If $r=1$ set $S_r=\mathbb N$ If $r<1$, take binary expansion of $r$. If the $nth$ decimal place is 1, add the element $n\in\mathbb N$ to $S_r$ else don't add it. This way get a 1-1 correspondence from $r\mapsto S_r$. Example $r=\frac12$ implies $S_r={1}$ – Karthik C Mar 23 '14 at 07:36

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For any set $S$, if $|\mathcal P(S)|=|S|$, there is a bijective function $\varphi$ from $\mathcal P(S)$ to $S$.

Let $E$, be the set of all $x\in S$ such that $x\notin\varphi^{-1}(x)$. Let $y=\varphi(E)$, hence $E=\varphi^{-1}(y)$.

  1. If $y\in E$, by definition of $E$, $y\notin\varphi^{-1}(y)=E$, a contradiction.
  2. If $y\notin E$, by definition of $E$, $y\in\varphi^{-1}(y)=E$, a contradiction.

Hence $\varphi$ does not exists, and for any set $S$, $$|\mathcal P(S)|\neq|S|$$

Xoff
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