Assume the theorem is false. Then there must be at least one number
\in \N that is boring.
Let\mathbb{B} \subset
\N be the set of boring numbers.
By the assumption, we have:
\mathbb{B} \ne
\phi\qquad{(1)}
Thus, because \mathbb{B} is
non-empty \mathbb{B} \subset \N, the Well
Ordering Principle tells us that:
\exists b \in \mathbb{B}
\| b = \mathrm{inf}(\mathbb{B})\qquad{(2)}
Thus this b is the smallest boring
natural number in existence; which makes b interesting. A contradiction! ∎
Note that the same doesn’t trivially hold for the set of real numbers
since you’d first have to prove that the set of boring real numbers is
either finite or bounded, and that the infimum is a member of the set.
Neither does this hold for \Z since a
non-finite subset in \Z doesn’t
necessarily have a minimum.
