2

In Jech's Set Theory book it has the following problem where $N$ is the intersection over all inductive sets.

Prove that every non-empty $X\subset N$ has a $\in$-minimal element. Hint: Let $n\in N$ and consider $X\cap n$.

I've seen almost this asked elsewhere, but other questions like here and here either don't actually ask for the solution, or get the solution by a means other than the hint--and I'm really wondering how you do it with the hint.

Here's what I have so far. I've already proved that $N$ is transitive so $n\subset N$. I also proved that if $X$ is inductive then so is $$\{x \in X: x \text{ is transitive and every nonempty } z \subset x \text{ has an $\in$-minimal element}\}$$ which I believe implies $n$ has a $\in$-minimal element (since $N$ is inductive hence we apply the above to $N$ but then since $N$ is the intersection of all inductive sets then so is that one and hence every non-empty element is transitive and has a $\in$-minimal element).

From here I'm tempted to show that $X$ has a minimal element directly, but even if I could, it seems like I'm missing something that could complete the proof along the lines of the hint in a few short steps.

Community
  • 1
Addem
  • 6,058
  • What about $\emptyset \subset \mathbb N$? – Moritz Oct 06 '16 at 23:25
  • $X$ is restricted to nonempty subsets – C7X Mar 20 '22 at 18:13
  • I'm not sure how rigorous the following is, but maybe one line of reasoning could be that assuming $X\cap n$ is nonempty, $(X\cap n,\in)$ is an initial segment of $(X,\in)$, and I suppose that implies the former contains the latter's minimal element. Since $(X\cap n,\in)$ is finite and nonempty, it must have a minimal element under the relation in question ($\in$), which is equal to $(X,\in)$'s minimal element. – C7X Mar 21 '22 at 01:37

0 Answers0