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Proof

(related to Corollary: Minimal Inductive Set Is Subset Of All Inductive Sets)

By the axiom of foundation, we have to show that every element of the minimal inductive set $\omega$ is also an element of any given inductive set $X$. From this, it will follow immediately that $\omega\subseteq X.$

• Let $X$ be an inductive set.
• By the definition of the minimal inductive set $\omega$, the empty set $\emptyset$ is an element of $\omega$ and thus, we have $\emptyset\in \omega\Rightarrow \emptyset\in X.$
• The same argument holds for all sets $z\in \omega$; thus $z\in \omega\Rightarrow z\in X.$
• But if $z$ is element of $\omega$, so is the singleton $\{z\}$; thus $\{z\}\in \omega\Rightarrow \{z\}\in X.$
• Finally, since $z\in\omega$ and $\{z\}\in \omega,$ then $z\cup \{z\}\in \omega$, and $z\cup \{z\}\in X$, by the same argument.
• Altogether, we have shown $\omega\subseteq X$ for any given inductive set $X.$
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References

Bibliography

1. Hoffmann, Dirk W.: "Grenzen der Mathematik - Eine Reise durch die Kerngebiete der mathematischen Logik", Spektrum Akademischer Verlag, 2011
2. Ebbinghaus, H.-D.: "Einführung in die Mengenlehre", BI Wisschenschaftsverlag, 1994, 3th Edition