March 16th, 2019, 09:30 AM  #21 
Senior Member Joined: Jun 2014 From: USA Posts: 506 Thanks: 40  Yes, I will answer your above (and/or former) questions. Just really busy working tax season hours and wrapping up a class I’m teaching. Typing this now on a cell phone...

March 22nd, 2019, 07:48 PM  #22  
Senior Member Joined: Jun 2014 From: USA Posts: 506 Thanks: 40  Quote:
No, my idea is not different. I was more interested in knowing what the difference was between the $A$ I was describing and $L_{\omega}$ (or $V_{\omega}$, as $L_{\omega} = V_{\omega}$). But, as clarified in the wiki article you cited, $A = V_{\omega} = \bigcup_{k=0}^{\infty} V_k$. I understand now why $L_{\omega}$ is the set of all hereditarily finite sets, but my first reading of the definition of the Constructible Universe had me thinking that $L_{\omega}$ could contain infinite sets so long as every element of each infinite set in $L_{\omega}$ was an element of some $L_{k}$ where $k \in \mathbb{N}$. Further, I understand why $L_{\alpha + 1}$ is a subset of $V_{\alpha + 1}$ for all $\alpha \geq \omega$ (because there are more elements of $\mathcal{P}(L_{\omega})$ than there are in $\text{DEF}(L_{\omega})$ on account of there being noncomputable elements of $\mathcal{P}(L_{\omega})$ while every element of $\text{DEF}(L_{\omega})$ must be the result of some first order formula using only variables contained in $L_{\omega}$. #5 said nothing different than #'s 1 through 4. I was just continuing the pattern in an initial effort to define in 'MyMathForum' style the hereditarily finite sets. With a little more thought I could have given a better definition, but you understood it anyways as you wrote: Where $f$ was only defined for the inputs '$\{$' and '$\}$', that was its domain. I used $f$ within an alternative definition of $A$ in which I applied $f$ to every $x$ such that $x$ was an element of some model of ZF set theory so as to say, if $f[x]$ was a finite binary string, then $x$ was in $A$. This all is trivial anyways. Using $f^{1}$ across the finite binary strings is simply an alternative way of defining $A$.  
March 22nd, 2019, 08:22 PM  #23  
Senior Member Joined: Aug 2012 Posts: 2,265 Thanks: 690  Quote:
Ah. Tricky subject. All I know about it is that at each stage, you add in those sets that are "firstorder definable with parameters." I gather that's some inside baseball in the field of logic. I don't know if it's wise to try to extrapolate too much from this subject unless you know the logic. I don't, maybe you do. Of course I have an intuition about what it means, but not the technical details. So regarding your original post(s), you had some condition #5 that talked about a and b and c and d and I think you might have had e, and your exposition really got lost, at least as far as I could follow it. But is there a difference between what you're trying to express and the hereditarily finite sets? I am not sure about this, but I've always been under the impression that the hereditarily finite sets are all the sets you get if you reject the axiom of infinity. But maybe that's not right. ps  Aha. It is right. Wiki to the rescue. Quote:
Ok! Last edited by Maschke; March 22nd, 2019 at 08:26 PM.  

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