I need to use MK to prove that: If $T$ is a set of ordinals then there cannot be an injection $f:Ord$ \ $T \to U$ for any set $U$.
I have just seen a proof use ZF:
Else, $V= \{u\in U:\exists a\in Ord$ \ $T\; (u=f(a))\}$ is a set, so $\{f^{-1}(v):v\in V\} \cup T=Ord$ is a set. (But then $Ord +1=Ord \cup > \{Ord\}$ is an ordinal that's larger than any ordinal.)
but now I am asking if it is possible to modify the argument above to get the desired result by MK.
I think the key point is to conclude that $V$ and the preimage are both sets. In the ZF version of the proof, they are sets because:
Since $f$ is injective we have $∀v∈V∃!a(f(a)=v)$.
By Replacement $∃W∀a(ψ(a)⟹a∈W)$, where $ψ(a)$ is:$ ∃v∈V(f(a)=v).$
By Comprehension there exists $X={a∈W(ψ(a))}$... Of course $X=f^{−1}V$
So could someone give a proof using MK to conclude that $V$ and the preimage are sets,or otherwise, prove the statement? Thanks a lot!