Find a countable sequence of proper subclasses of $Ord$ such that $Ord\supset C_0\supset C_1\supset...$ and $\bigcap_{n\in\omega} C_n=\emptyset$

Question

As in the question I am thinking about this excercise. My idea is to take a decreasing sequence of cofinal subclasses of $Ord$. The fact is it must be a countable sequence.

Hence I would try with something like:
$C_0=Ord\setminus(Ord\cap V_{f(0)})$
$C_n=Ord\setminus(Ord\cap V_{f(n)})$ where $f$ should be "cofinal" $f:\omega\rightarrow Ord$

But does something like this even exist? Could we maybe exploit the $\aleph$ function and put $f(n)=\aleph_{\omega+\omega+...+\omega}$ where $\omega$ is repeated $n$ times?

Answer 1

Hint. Take $$C_n=\{\alpha+m\mid m\ge n, \text{$\alpha$ is a limit ordinal}\}.$$