@ZuhairAl-Johar isomorphic as models "bijection between the domains of those models that preserves the relational sets"

I am not sure I understand your definition: $M$ is size preserving model if for each $a≠b∈M$ and $f:a\to b$ bijection in the universe, then $f∈M$? In this case take $M_0,M_1$ be 2 equinumerous non-isomorphic models of $T$ such that for each $a,b\in M_i$ we have $|a|≠|b|$, then both models are trivially equinumerous size preserving models that are not isomorphic

What I wrote still stands. Your new definition of ">" doesn't change the definition of "<". Also your "definition" is a "preorder" of the cardinals, not "partial order", so I am not even sure what do you mean by lub with your new definition...

@M.Solomon no, "A does not inject to B" means "$|A|\not ≤ |B|$", while $|B|<|A|$ means "B inject into A (but not surject)", it is not the same without the axiom of choice

@M.Solomon how does it change the example of Emil? If there exists an infinite D-finite set then there is no least upper bound for $\{V_n\mid n\in\omega\}$

oh right, I forgot how tetretion works for x<1. a transformation between the m_k to the M_k looks possible, but I'm not sure what it would be

> which can be interpreted as the area gained over (0,1) is greater than that lost over (1,∞) $$$$ Well, if we say $M_k=\int_1^\infty{dx\over f_k(x)}$ and $m_k=\int_0^1{dx\over f_k(x)}$, we have $0<M_k \to 0$ and $m_k\to\infty$, moreover $m_k$ grows rate is above linear, so after a certain point $M_{k}-M_{k+1}<m_{k+1}-m_k$, which is exactly "$\int_0^\infty{dx\over f_k(x)}$ is eventually monotonic increasing". It just looks like $k=1$ is that "certain point"

@ZuhairAl-Johar Where did I use replacement on $J$? $j_0$ is a set because it is the empty set, if $α∈ω_1$ then $J(α)∈ω$ so it is also a set, hence $(α,J(α))$ (from normal ZF arguments) is a set, so $j_{α+1}$ is a set. And if $β∈ω_1$ is limit, we can define the functional relation $φ$ such that $φ(x,y,β)⇔(x∈β\text{ and }y=j_{x})\text{ or }(x∉β\text{ and }y=0)$, this is a formula only on sets, then there exists a set $b$ such that $a∈b$ if and only if there exists $c∈β$ such that $φ(a,c,β)$, that is, $a=j_{α}$ for some $α∈β$.

Can't you take x,y to be aleph numbers, e.g. aleph1, aleph0, and then by induction on aleph1 show that $J$ is a set, which will lead to inconsistency? e.g. If $J:\omega_1↣\omega$, let $j_0=\emptyset$ and for $\alpha\in\omega_1$ let $j_{\alpha+1}=j_\alpha\cup\{(\alpha, J(\alpha))\}$, and at limit points take the union. Every stage is a set, as well as the union of all stages. That union is injective iff $J$ is injective, which is impossible

@ZuhairAl-Johar thanks for the link, very interesting that axiom of choice can be treated as a weak form of foundation here

@ZuhairAl-Johar What is Coret's principle? I couldn't find anything on this online

@EmilJeřábek do you know if Quine atoms consistent with global choice? I can't see why they would fail, but still

@EmilJeřábek oops, that is embarrassing lol