Prove: if $m \in n$ then $m^+ \subseteq n$.

Question

The following is exercise 5(d), section 6.2, from A book of set theory, by Charles Pinter (pg. 122).

5. Prove the following, where $m, n, p \in \omega$.

d) If $m \in n$, then $m^+ \subseteq n$.

These are the relevant definition and results, quoted from the same book:

6.2 Theorem For each $n \in \omega$, $n^+ \neq 0$.

6.4 Lemma Let $m$ and $n$ be natural numbers; if $m \in n^+$, then $m \in n$ or $m = n$.

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 $\omega$; every element of $\omega$ is called a natural number.

Attempted proof:

Suppose $m^{+} \nsubseteq n$ then $n \in m$ and so $n \subseteq m$ ,but $m \subseteq m^{+}$ and so $n \subseteq m^{+}$.

But by 6.4 $m \in n$, then (because $n$ is transitive) $m \subseteq n$; and then $m \subseteq n^{+}$.

So we have two successors for $m$ by 6.4 (contradiction, since we must have $m \in n$).

Hence $S(m) \subseteq n$.

Answer 1

Recall that $m^+=m\cup\{m\}$.

Assuming $m\in n$, you want to show that $m^+\subseteq n$, which means that for any $x$ you have $$x\in m^+ \qquad\implies x\subseteq n.$$

If $x\in m^+=m\cup\{m\}$, then we have two possibilities:

• If $x=m$ we immediately get $x\in $.
• If $x\in m$ and $m\in n$, we get that $x\in n$ from the fact that $n$ is transitive (Lemma 6.6).

Notice that there are two equivalent definitions of transitive sets. The one we used here is stated in this book in the first exercise in this section.