Prove that $n\not=n^+$ for all $n \in \omega$

Question

Prove that $n\not=n^+$ for all $n \in \omega$.

Where $\omega$ is the set of natural numbers, and $n^+ = n ∪ \{n\}$.

Attempt: Assume $n = n^+$ for some $n \in \omega$. Then $n=n∪\{n\}$. This is only possible if $n=\{n\}$, and this is not true for any of the natural numbers.

This is just really hand-wavy and I'm not sure how else to go about it. Any way to show that this isn't true for any natural number?

Answer 1

Prove by induction on $n$ that $n \notin n$. Since $n \in n+$, this establishes that the two sets aren’t equal. Trivially true for $0=\emptyset$. Can you do the inductive step?

Answer 2

Proof by induction. Define a set $I=\{n∈ω|n≠n^+\}$. We know that for any $n∈ω$, $n^+≠0$. We can then say that $0≠0^+$, and hence, $0∈I$. Since for any $n$, $m∈ω$, $n^+=m^+$ implies $n=m$, we can say that if $n∈I$, then $n≠n^+$, and $n^+≠(n^+)^+$. So if $n∈I$, $n^+∈I$. Therefore $I=ω$.