Induction, set theory drama

Question

Let $P$ be a subset of $\def\N{\mathbf N}\N$ so that for all $n \in \N$, if forall $m \in \N$ with $m<n$ we have $m\in P$, then $n \in P$. Then $P=\N$ (prove this with induction)

Now what if I take the set $P = \N\setminus\{1\}$, i'll never get $P=\N$...? Or $P$ as the empty set? (I may do this as the empty set is defined to be a subset of all sets)

This excercise is supposed to give insight on the principle of induction, but as you can see, I'm in trouble...

Help please, and sorry for the bad format (typing this on my phone, while riding my bike)

Answer 1

Note, that your condition on $P$, namely

(1) If forall $m <n $ we have $m \in P$, then $n \in P$.

does imply that $1\in P$ (hence $P = \mathbf \N \setminus \{1\}$ or $P= \emptyset$ do not fulfull the condition). Why? Lets apply (1) for $n=1$. The statement reads:

($1_1$) If for all $m < 1$ we have $m \in P$, then $1 \in P$.

As there are no $m\in \N$ with $m < 1$, the implication $m < 1 \implies m \in P$ is (trivially) true, hence $1 \in P$.