Let us assume $<$ is some class relation without minimal elements, meaning $\forall a, \exists b, b< a$. This means that for every $n\in\omega$, one can build a decreasing function $f$ with domain $n+1$, meaning that $f(k+1)<f(k)$ for every $k\in n$. This is a very simple inductive argument.


My question is, how can we show the existence of a decreasing function whose domain is all $\omega$?


If I had access to a global choice function $\tau$, I could do the following. For every $n\in\omega$, inductively show the existence of functions with domain $n+1$ such that $f(0)=\varnothing$ and $$f(k+1)=\tau\{x\in V\_\alpha|x<f(k)\}$$ for $k\in n$, where $\alpha$ is the least ordinal such that $V\_\alpha$ contains some set less than $f(k)$, as in Scott's trick. Then I could simply use replacement and union over all of these to get the desired set.


Since ZFC with global choice is a conservative extension of ZFC, I understand that there should be some way to prove this result without invoking global choice, at least for a fixed class relation $<$. But how?



Infinite decreasing sequence for class relation without minimal elements
Why do you need a global choice function, though? Yes, it's neater, but unnecessary.


Define, for each $x\in V\_\alpha$ an ordinal $\alpha\_x$ such that it is the minimal for which $V\_{\alpha\_x}$ contains witnesses that $x$ is not minimal in $<$. Now let $\alpha\_0$ be defined for witnesses that $\varnothing$ is not minimal; then $\alpha\_{n+1}=\sup\{\alpha\_x\mid x\in V\_{\alpha\_n}\}$. Finally, $\alpha=\sup\alpha\_n$.


Now all you need is to have a choice function for $V\_\alpha$. This can be modified, of course, to start with any set, not just $\varnothing$.


Alternatively, use Reflection to find some $V\_\alpha$ such that $(V\_\alpha,<\restriction V\_\alpha)$ reflects the "no minimal elements", which is essentially what we did before, and then work with a choice function for that specific $V\_\alpha$.


