# Find a set $A$ such that $A \notin \mathrm{Rng}{f}$

Let $f:\mathbb N\to \mathbb P(\mathbb N)$ be given by the following expression: $$f(n)=\{m\in\mathbb N\mid 3m-10>n\}$$ Find a set $A$ such that $A \notin \mathrm{Rng}{f}$.

I came up with $A=\{n\in\mathbb N\mid n\notin f(n)\}$, but I don't believe that it works because $6\in f(6)$.

• Where's the contradiction? Does that mean $A=f(6)$? No. It means that $6\notin A$ and therefore that $A\neq f(6)$. – Thomas Andrews Apr 19 '15 at 22:50
• Of course, you could just take $A=\{1\}$. Since $1\notin f(n)$ for any $n$, so this would mean that $A\cap f(n)=\emptyset$ for all $A$. Or you could take $A$ any finite set, since $f(n)$ is always infinite. Or you could take $A$ to be the set of all primes, since if $m\in f(n)$ then $m+1\in f(n)$, so $f(n)$ contains all numbers larger than its least element. – Thomas Andrews Apr 19 '15 at 22:54

The set $A=\{n\in\mathbb N\mid n\notin f(n)\}$ should work for every application $f$, for instance in your application it's clear that $n\in f(n)$ if and only if $3n-10> n$ which means that: $$A=\{0,1,2,3,4,5\} \tag 1$$ now assume that there exists some $x$ such that $f(x)=A$ then $A$ is infinte because always $f(x)$ is infinite for any $x$, this contradicts $(1)$.
Note that the set $A$ always works without any restriction on $f$, and it's a part of Cantor's argument that $\mathbb N$ and $\mathbb P(\mathbb N)$ don't have the same cardinal.