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1: Axiom of extensionality: $\forall x \forall y [[\forall z [z \in x \iff z \in y]] \implies x=y]$ 2: Axiom of unordered pairs: $\forall x \forall y \exists z \forall w [z \in w \equiv [z=x \lor z=y]]$ 3: Axiom of union: $\forall x \exists y \forall z [z \in y \equiv \exists t [z \in t \land t \in x]]$ 4: Axiom of infinity: $\exists x [\varnothing \in x \land \forall y [y \in x \implies y^+ \in x]]$ 5: Axiom schema of replacement: $[\forall x \exists! y \varphi(x,y)] \implies \forall u \exists v \forall r [r \in v \iff \exists s [s \in u \land \varphi(s,r)]]$
A few minutes ago I read Von Neumann–Bernays–Gödel set theory which suggests that if Axiom of limitation of size is present in the set theory in some form, eventually it will imply the axiom of global choice, which is something much stronger than AC which is not what we want because we want to throw away AC to gain access to the more weird infinite sets
so that means some other way is needed to define a proper class
All except 5 looks fine, I am still trying to digest 5 since it is the most complicated one to write in 1st order logic
If I understood correctly, ZF allow us to do the following: 1. Check whether two sets are equal 2. Existence of unordered pairs 3. Performing unions 4. Constructing infinite sets 5. Produce sets by using maps or relations 6. Produce power sets
according to wiki, Axiom of Limitation of Size (ALS) implies A5, A3, that V the class of all sets is well-ordered, and the axiom of global choice. ALS is also equivalent to (A3 + A5 + AGC)
@Secret arbitrary intersection is obtained by using 5 on one of the sets (since the intersection is a subset of any original set)
Yes, we want to discard axiom of global choice, else we can only have non measurable and vitali sets as the weird sets
so that means we need to establish proper class some other way...
I am currently figuring how to construct the ordinals using the above 6 axioms. My initial guess is start with the empty set, then use 5 with the relation being $\in$
that will ensure well ordering because the nesting is done in a specific way in each step
and then we can have other types of sets not necessary well ordered
1: Axiom of extensionality: two sets are equal if for all z, "z is in the first set" is equivalent to "z is in the second set" 2: Axiom of unordered pairs: if x and y are sets, then there is a set z such that for all w, "w is in z" is equivalent to "w=x or w=y", i.e. z is the set containing only x and y. 3: Axiom of union: if x is a set, then there is a set y that is the union of the elements of x. 4: Axiom of infinity: $\Bbb N$ exists
Use the above recipe to construct the required nested well ordered sets. Call these the natural numbers $n$. Then 4 guarenteed the existence of $\omega$ where $n \in \omega$ for all the ns will have constructed so far.
Use pairing again to construct the sets $\omega +n$ for each $n$. Then use 4 to construct $\omega 2$
So in our "box of sets constructed" we have $\varnothing$, $n$ and $\omega$. We have used pairing and unions to construct $\omega + n$ so the last step requires gluing $\omega +n$ and all the $ns$ together in the correct ordering somehow...
We might be able to use pairings to make sets $\{\omega,\omega+1\}$ , then union them, and then pair the resulting set with the sucessors $\omega + (n+1)$. Repeat this indefinitely to get $\omega 2$?
The issue with sup here is that while it can be defined as an arbitrary union of elements in an order, we need to first construct such set in a finite number of steps before we can apply 3 to produce the union
because then we can easily construct a large number of countable ordinals by using the maps $\omega\to f(n,\omega)$, which seemed... too easy to be true?
in that case, we can use 5 to pick the map $n \to \alpha_n$ where $n\in \omega$ and $\alpha_n$ recursive and countable to produce the collection that will later be applied 3 to form $\omega_1^{CK}$
Ok so the next step is $\omega_2^{CK}$. Now we seemed to have a problem as this ordinal contains non recursive ordinals, those we cannot define in a finite string, thus we cannot use 5. What should we do to produce the whole collection of it so we can apply 3 to produce $\omega_2^{CK}$?
The FAQ says:
Each room has owners, who can change various room settings and transfer ownership to other users.
And:
There are not a lot of responsibilites connected to room ownership, just a few extra permissions: The owner can add and remove RSS feeds being posted into the room, chang...
Mathworks (Not the main chat!)
Mathworks (Not the main chat!)