The purpose of doing this is to understand whether 0, 1, ε should be included in the language or not. I have come up with the following conversion:
$$ w \in L \iff \forall \text{positions} (\text{is even (position)} \implies \text{is zero (position)}) $$
According to this, $0, 1, ε$ make the statement vacuously true, so they must be included in $L.$
My questions are:
- Is this approach correct to determine $0, 1, ε$ belong to $L$?
- I'm confused if a there exists should be included in the predicate or not, like so: $$ w \in L \iff \exists w[\forall \text{positions (is even (position)} \implies \text{is zero (position)})] $$ why/why not should I include it?
Answer:
I'm going to assume you're talking about the set of natural numbers. Let $\omega \in \Omega$ such that$^{0}$: $$\text{even}(\omega, \omega_0) \; \; :\iff \; \omega = \omega_0 \cdot 2$$ With $\omega_0 \in \Omega$. If we define any element in $\omega$ to be the natural numbers then we write (for $\alpha, \beta \in \omega$): $$\psi(\alpha) \; \; :\iff \; \forall \alpha \; (\text{even}(\alpha, \beta) \; \to \; \text{min}_{\in}(\alpha,\omega))$$ And if I may remind you that (also, don't worry about $\psi$ there, it will get used in just a moment): $$\text{min}_{\in}(z,x) \; \; :\iff \; z \in x \; \land \; \forall y \; \big(y \in z \; \to \; \lnot(y \in z)\big)$$ Also, remember that an ordinal by nature represents the "position" of an element with respect to its ordering. In the natural numbers, they're well-ordered which means that they have a least element, and also that they don't break many rules (okay, the last one is a bit satirical). The arithmetic for ordinals represents the arithmetic for the naturals exactly, so we can just "copy-paste" the rules (keeping in mind that commutativity doesn't always hold). If you want it more algebraically adherent , then you sort of don't have a choice other than listing its ordering in a tedious way (a sequence that satisfies an ordering up to some position for its least element would be the "position") to not break commutativity but you can assume any system that you want really unless it prevents something important in your construction. It should be noted that this is not predicate calculus as the membership relation is not part of predicate calculus' language, and needless to say, any set notation. If you want to write this in set notation then simply say: $$X = \{\alpha \in A ~|~ \psi(\alpha)\}$$ Adhering to the axiom of restricted comprehension under Zeremlo-Frankael set theory (ZF [otherwise called the standard foundations]).
Appendix:
I'm going to explain some of the stuff that I assumed, like, your knowledge about what ordinals really are, and your understanding of the $\in$-minimal element, and lastly, what is restricted comprehension? An ordinal informally represents the numbering of the order of elements within a set, and this ordering comes from whatever rules they satisfy for their own ordering, except for two rules which make up what an ordinal is formally, we call a set $\omega$ an ordinal if it is both transitive (every element of the set is a subset of itself) and it it has a $\in$-minimal element; which is a great segway to what we want to discuss... an $\in$-minimal element. You saw the formula above and it translates roughly into English by saying an element $z$ belongs to $x$ and all element of $z$ belonging to $z$ implies that they don't, means that there is no element in $z$, and that it is the first set, the empty set$^{1}$. Restricted comprehension is a way to make a notation for building sets, and also a way to avoid paradoxes like the set of all sets not belonging to themselves belonging to itself if and only if it belongs to itself{^2}, and instead of saying previously directly that {z ~ | ~ \phi(z)}, we now say $\{z \in x ~ | ~ \phi(z)\}$, which translates into: "all element in $x$ are if and only if all set-theoretic formulae $\phi(z)$" roughly$^{3}$.
Footnotes:
0: $\omega \in \Omega$ is just notation for $\omega$ satisfying transitivity and $\text{min}(z,x)$.
1: This is called Zermelo's ordinal construction. Note that there are other ordinal constructions (e.g. Von Neumann's ordinal construction).
2: This is called Russell's paradox.
3: By saying this, it avoids said paradox because now it belongs to some set $x$ if all elements don't belong to themselves (with some extra details... check the axiom out for yourself for a better picture).
Lastly, I highly encourage you to look up things that I mentioned if you didn't know them already, otherwise... hope this answered helped!