March 9th, 2019, 11:20 PM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  Comprehension Question
Assuming it contained only sets that could exist within a model of ZF, a set of finite sets that contain only finite sets must be countable, correct? My reasoning, assuming I understand correctly, is that it follows from $L_{\omega} = V_{\omega}$ where $L$ and $V$ are as generally described here: https://en.m.wikipedia.org/wiki/Constructible_universe Would $\{x \in L_{\omega} : x < \mathbb{N} \}$ in fact equal the set of finite sets that contain only finite sets? Last edited by AplanisTophet; March 9th, 2019 at 11:23 PM. 
March 10th, 2019, 09:17 AM  #2 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43 
I should be more specific with what I mean by a set of finite sets containing only finite sets. If $A$ is the set of finite sets containing only finite sets, I mean to imply that: 1) $a \in A \implies a \in \mathbb{N}$ 2) $a \in A$ and $b \in a \implies b \in \mathbb{N}$ 3) $a \in A$, $b \in a$, and $c \in b \implies c \in \mathbb{N}$ 4) “ ...$c \in b$, and $d \in c \implies d \in \mathbb{N}$ 5) “ ...$d \in c$, and $e \in d \implies e \in \mathbb{N}$ . . . There are no infinite sets involved except of course $A$ itself. I trust that makes sense. Can anyone confirm the countability of $A$ (seems every element of $A$ could be computed with a finite formula of which there are only countably many) and my above question involving a set containing the finite elements of $L_{\omega}$ (again with the clarification that the only infinite set I mean to involve is a subset of $L_{\omega}$ equal to $A$)? 
March 10th, 2019, 11:43 AM  #3 
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740 
See https://en.wikipedia.org/wiki/Hereditarily_finite_set. I don't believe this has anything to do with constructibility. Last edited by Maschke; March 10th, 2019 at 11:52 AM. 
March 10th, 2019, 02:29 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics 
This isn't true. For example if $A = \{\{r\} : r \in \mathbb{R} \}$ then every element of $A$ has cardinality 1 and $A$ has cardinality of $\mathbb{R}$.

March 10th, 2019, 03:15 PM  #5 
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740  I can't imagine what that notation means, with the leading dots and unbalanced quotes. But as SDK notes, you have to limit yourself to finite unions. If you allow arbitrary unions you leap out of the class of hereditarily finite sets.

March 10th, 2019, 06:28 PM  #6 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  Given that I restricted the elements of $A$ to those finite sets that may appear within a model of ZF set theory, it follows that each element of $A$ contains no urelements. Rather, each element of $A$ is a finite set containing only finite sets by definition. How then do you intend to express each $r \in \mathbb{R}$ as a finite set so as to make your above assertion true?

March 10th, 2019, 06:37 PM  #7  
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  Quote:
If $A$ is the set of finite sets containing only finite sets, I mean that to imply: 1) $a \in A \implies a \in \mathbb{N}$ 2) $a \in A$ and $b \in a \implies b \in \mathbb{N}$ 3) $a \in A$, $b \in a$, and $c \in b \implies c \in \mathbb{N}$ 4) $a \in A$, $b \in a$, $c \in b$, and $d \in c \implies d \in \mathbb{N}$ 5) $a \in A$, $b \in a$, $c \in b$, $d \in c$, and $e \in d \implies e \in \mathbb{N}$ . . . ...Does this make sense? No infinite sets allowed (except for $A$ itself)!!!  
March 10th, 2019, 08:32 PM  #8  
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740  Quote:
Meanwhile, did you find the link on the hereditarily finite sets of interest? Is your idea different in some way? The Wiki article does say that the hereditarily finite sets are countable. Also can you please say in English what #5 means? One thing I'm getting is that you are only putting natural numbers in your class, not the sets themselves. I'm getting something along the lines of, if a set has cardinality 5, you put the natural number 5 into your class, not the set itself. Am I understanding that? Oh I see. You're just saying they're finite. Ok. I'll leave the previous paragraph up just so you can see an example of where a readers is struggling with your notation. And I can't understand what that crazy chain of set membership in #5 is supposed to be about. Last edited by Maschke; March 10th, 2019 at 08:39 PM.  
March 11th, 2019, 05:07 PM  #9 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43 
I thought about it and just figured out a way to count $A$. Each element of $A$ is a finite set constructed of left brackets, $\{$, and right brackets, $\}$, so in replacing each left bracket with a $0$ and each right bracket with a $1$, we can convert each element of $A$ to a unique finite binary string: Let $f(\{) = 0$ and let $f(\}) = 1$. Arbitrarily, let $a = \{ \{ \{ \} \}, \{ \{ \}, \{ \{ \} \} \} \}$. Then $f[a] = f(\{) f(\{) f(\{) f(\}) f(\}), f(\{) f(\{) f(\}), f(\{) f(\{) f(\}) f(\}) f(\}) f(\}) = 00011001001111$. Define $A = \{ x : \text{ there exists some model of ZF set theory such that } x \text{ is an element of the model and } f[x] = \text{ a finite binary string} \}$. Now that my definition is clear, I am curious, what is the difference between $L_w$ and $A$? 
March 11th, 2019, 05:34 PM  #10 
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740  I'll wait for comment by someone to whom that remark applies. From where I sit this most recent post did not add clarity to your presentation. But maybe that's just me. I'm experiencing restricted comprehension. What is the domain of $f$? Surely you can't be serious about that notation. 

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