It is inconsistent. Call an ordinal b an independent critical point if for every for every Universe X,
if c∈X then there is a function f with domain X and an ordinal α, such that "f is an elementary
embedding from X to Vα" holds, for all x∈X f(x)∈X, and b is the least ordinal with f(b)≠b.
Note that the critical point of j is an independent critical point because the restriction of j to any
Universe X has the above property of f. Let c be the least independent critical point. By the
axiom schema of Reflection, there is a universe K such that "c is the least independent critical
point" holds relativized to K. By the definition of independent critical point, there is a function
g with domain K and ordinal γ such tha g is an elementary embedding from K to Vγ, and for all x∈K g(x)∈K
and c is the least ordinal with g(c)≠c. Let F(x) be a formula expressing
"x is the least independent critical point". Define a sequence s by
s0=c and s(n+1) is the least ordinal α such that α is greater than g(s(n)) and for K, Vα reflects all
subformulas of F. Let t=U{sn|n∈𝜔}. Then g(t)=t, and F(c) holds in V(t). By elementarity,F(g(c)) holds
in V(g(t)). That is F(g(c)) holds in V(t). But this is impossible.
It proves AC. For this, recall it's enough to see that for every ordinal $\alpha$, $\mathcal{P}(\alpha)$ is wellorderable, and for that it's enough to see that $\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$ has a choice function. Supposing this fails for some $\alpha$, let $\alpha$ be the least such; then $\alpha$ is an infinite cardinal. Let $\pi:3\times\alpha\to\alpha$ be a bijection. Let $\pi^+:\mathcal{P}(\alpha)^3\to\mathcal{P}(\alpha)$ be the induced bijection, i.e. $$\pi^+(x,y,z)=\pi[x]\cup\pi[\alpha+y]\cup\pi[2\times\alpha+z],$$ where $\alpha+y=\{\alpha+\beta\bigm|\beta\in y\}$ etc. Now for $x\in\mathcal{P}(\alpha)$ and $X\in\mathcal{P}(\mathcal{P}(\alpha))$ and $z\in\mathcal{P}(\alpha)$ define $(x,X,z)^*\in\mathcal{P}(\mathcal{P}(\alpha))$ by $$ (x,X,z)^*=\{\pi^+(x,y,z)\bigm|y\in X\}.$$
Clearly $(x,X,z)\mapsto(x,X,z)^*$ is injective in all 3 arguments $x,X,z$. For $x\in\mathcal{P}(\alpha)$ and $X\in\mathcal{P}(\mathcal{P}(\alpha))$ let
$$(x,X)^{**}=\{(x,X,z)^*\bigm|z\in X\}.$$
Note that if $X,Y\in\mathcal{P}(\mathcal{P}(\alpha))$ and $x,y\in\mathcal{P}(\alpha)$ with $(x,X)\neq(y,Y)$ then $(x,X)^{**}\cap(y,Y)^{**}=\emptyset$.
Also if $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$ then $(x,X)^{**}\neq\emptyset$.
Now let $$P=\Big\{(x,X)^{**}\Bigm|x\in\mathcal{P}(\alpha)\wedge X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}\Big\}.$$
So by the remarks above, we can apply Deep Choice to the family $P$. Let $f:P\to V$ be an injection witnessing this.
I claim that for some $x\in\mathcal{P}(\alpha)$, we have $f((x,X)^{**})\notin\alpha$ for all $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$. Otherwise, let $\beta_x$ be the least $\beta$ of form $f((x,X)^{**})$ for some such $X$; then $x\mapsto\beta_x$ is injective with domain $\mathcal{P}(\alpha)$, so $\mathcal{P}(\alpha)$ is wellorderable, contradiction.
Fix $x$ witnessing the claim. Then note that for each $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$, either
- (i) $f((x,X)^{**})=(x,X,z)^*$ for some $z\in X$, or
- (ii) $f((x,X)^{**})=\pi^+(x,y,z)$ for some $y,z\in X$
(this is because the only other options are further down in the transitive closure of $(x,X)^{**}$, but such objects are in the transitive closure of some $\pi^+(x,y,z)$, but these are only ordinals ${<\alpha}$, which have been ruled out already).
But now we can define a choice function $c$ for $\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$: given $X\in\mathcal{P}(\mathcal{P}(\alpha))\backslash\{\emptyset\}$, let $c(X)=$ the unique $z\in X$ witnessing (i), if (i) holds; or if (i) is false, then let $c(X)=$ the unique $z\in X$ witnessing (ii). The existence of $c$ contradicts our hypothesis.
Best Answer
Sure.
Start with countably many inaccessible cardinals, $\kappa_n$, and now take the full support product adding $\kappa_n^+$ subsets to each $\kappa_n$. Then the $n$th model is the symmetric extension given by adding all the generics for the first $n-1$ coordinates, and violating choice in the $n$th one (e.g. a Cohen style model).