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March 16th, 2019, 09:30 AM   #21
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Quote:
 Originally Posted by Maschke Meanwhile, did you find the link on the hereditarily finite sets of interest? Is your idea different in some way? Also can you please say in English what #5 means? What is the domain of f?
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
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Quote:
 Originally Posted by Maschke Meanwhile, did you find the link on the hereditarily finite sets of interest? Is your idea different in some way? Also can you please say in English what #5 means? What is the domain of f?
Yes, I was interested in the hereditarily finite sets that you linked to in your initial post, however, the first time I read your initial post, I don't recall seeing that link. You may have edited the post after I read it as I am just now seeing that for the first time.

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 non-computable 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:

Quote:
 Originally Posted by Maschke Oh I see. You're just saying they're finite. Ok.
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
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Quote:
 Originally Posted by AplanisTophet Yes, I was interested in the hereditarily finite sets that you linked to in your initial post, however, the first time I read your initial post, I don't recall seeing that link. You may have edited the post after I read it as I am just now seeing that for the first time.
Could be. I generally edit each post a few times.

Quote:
 Originally Posted by AplanisTophet ... Constructible Universe ...

Ah. Tricky subject. All I know about it is that at each stage, you add in those sets that are "first-order 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:
 Originally Posted by Wiki The hereditarily finite sets are a subclass of the Von Neumann universe. They are a model of the axioms consisting of the axioms of set theory with the axiom of infinity replaced by its negation, thus proving that the axiom of infinity is not a consequence of the other axioms of set theory.
https://en.wikipedia.org/wiki/Heredi...mal_definition

Ok!

Last edited by Maschke; March 22nd, 2019 at 08:26 PM. Tags comprehension, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shaharhada Algebra 1 July 16th, 2018 10:20 AM shaharhada Geometry 1 June 30th, 2018 09:44 AM sdparsons Algebra 3 October 1st, 2017 03:12 AM sCoRPion Differential Equations 0 March 16th, 2015 01:05 PM pudro Algebra 0 November 10th, 2008 08:12 PM

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