Is $\max$ a valid order for $\cup_{n\in\mathbb{N}}\Gamma_n$, each $\Gamma_n$ is finite. The question is:

Consider the partitions $(\Gamma_n)_{n\in\mathbb{N}}$, where  $\Gamma_n=\{z,\cdots,z+\frac{1}{n} \}$ is a partition of $[z,z+\frac{1}{n}]$, for each $n\in\mathbb{N}$ (exclude $0$). Can we order decreasingly using $\max$ the set  $$ \bigcup_{n\in\mathbb{N}}\Gamma_n\ \ \ ?$$

Is it valid for example define: $$y_1=\max \left\lbrace\bigcup_{n\in\mathbb{N}}\Gamma_n\right\rbrace =z+1$$ and for $n\geq 2$ define $$y_n=\max \left\lbrace \left\lbrace\bigcup_{n\in\mathbb{N}}\Gamma_n\right\rbrace \backslash \left\lbrace \bigcup_{k\leq n-1 }\{y_k\} \right\rbrace \right\rbrace$$ so we should have $y_n<y_{n+1}$ for all $n$. You can think of this problem also as (not partitions but sets of finite elements), at infinite we have $\lim_{n\to\infty}y_n=z$.
 A: Yes, this will work. The crucial observation is that for each $\alpha\in(0,1)$ there are only finitely many elements of $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ that are greater than $\alpha$. To see this, let $m\in\Bbb Z^+$ be large enough so that $\frac1m\le\alpha$; then $\Gamma_n\subseteq[0,\alpha]$ for all $n\ge m$, so
$$(\alpha,1]\cap\bigcup_{n\in\Bbb Z^+}\Gamma_n\subseteq\bigcup_{1\le n<m}\Gamma_n\,.$$
This means that the set $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ is well-ordered by $\ge$: every non-empty subset of it has a largest element. Your procedure simply picks off the largest remaining element at each stage.
Note, though, that it is possible for some member of $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ to belong to more than one of the sets $\Gamma_n$ (though only, as we’ve just seen, to a finite number). As long as you care only about the points themselves and not the specific sets $\Gamma_n$ to which they belong, you’ll be in good shape. If you need to distinguish a point $z$ in some $\Gamma_n$ from the same point in some different $\Gamma_m$, however, you’ll need to work with ordered pairs $\langle z,n\rangle$ and replace $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ with $\bigcup_{n\in\Bbb Z^+}(\Gamma_n\times\{n\})$. You can then order the pairs by setting $\langle z_1,n_n\rangle\preceq\langle z_2,n_2\rangle$ if and only if $z_1<z_2$, or $z_1=z_2$ and $n_1\ge n_2$. You can still pick off the maximum element at each step, but it should be the maximum element with respect to the order $\preceq$.
