Prove by induction If A $\in$ n and n $\in$ ω then A$\in$ ω
Problem is from Pinter’s a book of set theory
6.1 Definition By the set of the natural numbers we mean the intersection of all the successor sets. The set of the natural numbers is designated by the symbol ω; every element of ω is called a natural number.
6.3 Theorem (Mathematical Induction). Let X be a subset of ω; suppose X has the following properties:
i) $0\in X$.
ii) If $ n\in X$,then $ n^{+} \in X$
Therefore X = ω.
The Peano axioms for the natural numbers are:
P1 0∈ω.
P2 Ifn∈ω,then n+ ∈ω.
P3 For each n∈ω,n+ ≠0.r
P4. If X is a subset of ω such that
i) 0∈X,and
ii)if n∈X,then n+ ∈X,thenX=ω.
P5Ifn,m∈ωandn+ =m+,then n=m.
Attempted proof
P(0):
$0\in A$and $ 0\subseteq$ ω so $0\in$ω
P(n): LetA$\in n$,A$\subseteq n$ ,by P2 $n^{+}∈ω.$ and since $n\in \omega$ ;$n\subseteq\omega$ then n is transitive, So A$\in. ω $ .
We conclude ω is a transitive set
I tried induction on n? Help
(Note that I originally thought you were trying to prove mathematical induction. This the proof here. I put the proof of your actual question below but leaving this here for completeness).
You are trying to prove Mathematical Induction using induction. You can't do that. One way you can do it is via proof by contradiction.
Assume $X \neq \omega$. So then, since $X\subseteq\omega$, there must be some element(s) in $\omega-X$.
Let $n'=min(\omega-X)$ (which must exist since the natural numbers are well-ordered). By definition, $n'\neq0$ so $n'>0$. Since $n'$ is the smallest natural number not in $X$ this means that $n'-1$ (defined as the number whose successor in $n'$) is in $X$.
But, by definition, since $n'-1 \in X$ then $n' \in X$ which contradicts our assumption. So, $X=\omega$.
(It seems I misunderstood what you were asking. I'll put the proof of your question below here).
Note that $A\in\omega$ means $A$ is a natural number (by definition). So, let's do this by induction.
For $n=0$ this is vacuously true since $0=\emptyset$ so there is no $A\in0$.
Assume its true for $n$, let's prove it for $n+1$. But $n+1=n\cup\{n\}$ so we know, by the induction hypothesis, that it contains only natural numbers and the number $n$. But $n=(n-1)^+$ i.e. its the successor to a natural number, so $n$ is also a natural number so $n\in\omega$.