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This is a problem from the book Set theory by You-Feng Lin.

Principle of Transfinite Induction

Let $(A,\le)$ be a well-ordered set. For each $x \in A$, let $p(x)$ be a statement about $x$. If for each $x \in A$, the hypothesis "$p(y)$ is true for every $y \lt x$" implies that "$p(x)$ is true," then $p(x)$ is true for every $x \in A$.

I'm trying to prove this theorem directly using this lemma.

Let $(A, \le)$ be a well-ordered set, and let $\mathscr T$ be a family of segments of $A$ such that

(1) any union of members of $\mathscr T$ belongs to $\mathscr T$.

(2) if $A_x \in \mathscr T$, then $A_x \cup \{x\} \in \mathscr T$.

Then $\mathscr T$ contains all segments of $A$.

How may I use this lemma to prove the principle of transfinite induction?

I'm not sure how to form a family of segments to satisfy those conditions, and how that would guarantee the theorem. Any help?

Martin Sleziak
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nomadicmathematician
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1 Answers1

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I think it is easier to prove the claim directly without auxiliary lemma.

Denote $B:=\{x\in A; p(x)\}$.

If we assume that $A\setminus B$ is non-empty, then there exists the smallest element $m:=\min(A\setminus B)$.

What can you say about $m$? Can you get a contradiction from this?

If you (or the authors of the book) insist on using the lemma stated in your post, then maybe you could choose $\mathscr T$ to be the system of all lower segments $S$ such that each element of $S$ fulfills $p(x)$.

$S\in\mathscr T \Leftrightarrow S$ is an initial segment of $A$ and $(\forall x) (x\in S \Rightarrow p(x))$

If you can show that $\mathscr T$ fulfills the assumptions of the lemma, then you get that $A\in\mathscr T$. (Since all initial segments belong to $\mathscr T$ and one of them is $A$.) This means that $p(x)$ is true for each $x\in A$.

Martin Sleziak
  • 56,060