This might sound like a very trivial question. Given an infinite chain of subsets indexed by ordinals $\lambda \leq \mu$:
$$X_\mu\subseteq ...\subseteq X_\lambda\subseteq...X_1\subseteq X_0$$ where $\mu$ is any ordinal.
Is it possible to get a refinement where all the inclusions are strict (where only the repeated subsets are gone)?
This is, is it possible to get a second chain $$X_{\mu'}\subsetneq ...\subsetneq X_{\lambda'}\subsetneq...X_1\subsetneq X_0$$ where for any $X_i$ in the first chain there exists a $X_{i'}$ in the second chain where $X_i=X_{i'}$?
I know that $\mu'$ will be smaller than $\mu$, and that even the cardinal of $\mu$ and $\mu'$ might be different.
My way to do this is the following. I start with $X_0$ if $X_0=X_1$ then I remove $X_1$. If $X_0=X_2$ then I remove $X_2$ and so on. If $X_1\not= X_\lambda$ for some $\lambda$ then I move to $\lambda$ and start again the process. I believe that this works because all the indexes are ordinals and thus have a successor. But I am not very confident working with infinite chains and I might be doing something wrong.
If you don’t care about the subscripts, just let $\mathscr{X}=\{X_\xi:\xi\le\mu\}$; then $\left\langle\mathscr{X},\supsetneqq\right\rangle$ is a well-ordered chain that contains exactly one representative of each set in the original chain.
If you want to extract a specific subsequence of $\left\langle X_\xi:\xi\le\mu\right\}$ that contains exactly one representative of each set, let
$$\varphi:\mu+1\to\mu+1:\xi\mapsto\min\left\{\eta\le\xi:X_\eta=X_\xi\right\}\,;$$
then $\left\langle X_\xi:\xi\in\operatorname{ran}\varphi\right\rangle$ is a subsequence of $\left\langle X_\xi:\xi\le\mu\right\}$ that contains exactly one representative of each set. If you want nicer indexing, you can let $\nu$ be the order type of $\operatorname{ran}\varphi$ and $h:\nu\to\operatorname{ran}\varphi$ the associated order-isomorphism and write
$$\left\langle X_\xi:\xi\in\operatorname{ran}\varphi\right\rangle=\left\langle X_{h(\xi)}:\xi<\nu\right\rangle\,.$$