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In Avigad's lecture notes, we are given a set $U$, a subset $B \subseteq U$, and some functions $f_{1}, \dots, f_{k}$. Furthermore, following Enderton (p. 22) say a set is inductive if it contains $B$ and is closed under the $f_{j}$'s and let $C^{*}$ be the intersection of all the inductive subsets of $U$. I want to fill out the missing part of showing that $C^{*}$ is also inductive.

Here is what I have to offer:

Let $$C^{*}:=\bigcap\{S \mid S \text{ is inductive}\}.$$ If we choose an arbitrary sequence that belongs to $C^{*}$ then this will surely be contained within a subset, say $B$ because of the fact that $C^{*}$ is the intersection of all inductive sets. Moreover, the set is closed for any function $f_{j}$. This is because each inductive set satisfy, by definition, the property of being closed under any arbitrary function and so when we take its intersection we have that any function $f_{j}$ maps within $C^{*}$. Therefore, $C^{*}$ is inductive.

I realize that this looks more of a sketch rather than an actual proof. However, I'm unable make any progress as to write it clearer, and so my question is: How would you formalize this?

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  • $\begingroup$ The right hand side of your formula for $C^*$ doesn't make sense. There is no set $\{S \mid S$ is inductive$\}$ in ZFC. $\endgroup$ – kahen Sep 18 '13 at 19:59
  • $\begingroup$ That is one of the criticism you could raise against this problem - it is not acceptable to constructivists. $\endgroup$ – user94284 Sep 18 '13 at 20:03
  • $\begingroup$ I'd have written $\{S \subseteq U \mid S$ is inductive$\}$. Just to be completely clear that I'm not talking about something like $\mathbb{ON}$, the class of ordinals. $\endgroup$ – kahen Sep 18 '13 at 20:05
  • $\begingroup$ Fair enough, thanks! $\endgroup$ – user94284 Sep 20 '13 at 8:48
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You have all of the ingredients and a clear informal statement of the proof. If you want to make it a bit more formal, you could do it like this:

Let $f_j$ be any of the functions. If $f_j$ is an $n$-ary function, let $\langle c_1,\ldots,c_n\rangle\in (C^*)^n$ be arbitrary. Let $S$ be any inductive subset of $U$. Clearly $\langle c_1,\ldots,c_n\rangle\in S^n$, so $f_j(c_1,\ldots,c_n)\in S$; since $S$ was arbitrary, $f_j(c_1,\ldots,c_n)\in S$ for each inductive $S\subseteq U$. Thus, $f_j(c_1,\dots,c_n)\in C^*$ for $j=1,\ldots,k$, and $C^*$ is inductive.

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