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\[ \int_{-\infty}^b 2^\mathbb{N} \]    This is a contradiction because no number under 0 is natural.  \subsection{No one-to-one correspondence}  Cantor established a simple yet remarkable theorem that for all sets A, there cannot be a one-to-one correspondence from A to the set $\mathcal{P}(A)$ of all subsets of A. We examine the basic idea of Cantor’s theorem in the case $A=\mathbb{N}$, the set of natural numbers. Cantor’s idea is to see that no function $f:\mathbb{N}\to\mathcal{P}(\mathbb{N})$can be a one-to-one correspondence:  To this end, for a function $f:\mathbb{N}\to\mathcal{P}(\mathbb{N})$ call a natural number n ordinary if n\in f(n), and extraordinary if n \not\in f(n). Let \mathcal{E} be the set of all extraordinary natural numbers. Can there be a natural number k with $f(k)=\mathcal{E}?$ Can $f:\mathbb{N}\to\mathcal{P}(\mathbb{N})$ be a one-to-one correspondence?  Generalize Cantor’s argument to arbitrary sets A: that is, show that no function $f:A\to \mathcal{P}(A)$ can be a one-to-one correspondence.    $f(k)=\mathcal{E} = 0$