I am wondering can I please get some hints on how to prove the following.
total pre-order on $X$ is isomorphic to a total order of sets of $X$
Note that:
For example given $\{r_1 \lesssim r_2 \lesssim r_3 \}$:
- we can not have duplicate things: WRONG => $\{ \{ r_1 \lesssim r_2 \} < \{ r_3 \lesssim r_1 \} \}$
- we don't lose anything: WRONG => $\{ \{ r_1 \} < \{ r_3 \} \}$
- we don't have empty inner sets WRONG => $\{ \{ r_1 \lesssim r_2 \} < \{ r_3 \} < \{ \} \}$
Notation: I used a single line to represent $a \lesssim b \lesssim c$ but that is generally not correct because in total pre-order we can have cycles. I meant a set of pair of relationships $a \lesssim b$.
My attempt:
To prove this we need a deterministic way of going from: $$T= \{ r_0 \lesssim \dots \lesssim r_i \lesssim \dots \lesssim r_{i+j} \lesssim \dots \lesssim r_n \}$$ to (for example): $$T'= \Big \{ \{ r_0 \lesssim \dots \lesssim r_i \} < \dots < \{ r_{i+j} \lesssim \dots \lesssim r_n \} \Big \}$$ and the opposite direction.
For the forward direction, for each $r_i \in T$ we can map it to some set $s_k \in T'$ in a way that the result of this mapping is compatible with $T$. Because for all $(r_i \lesssim r_j) \in T$ then its one of two cases:
- $r_i$ and $r_j$ belong to two different sets $s_k, s_l \in T'$, $r_i \in s_k$ and $r_j \in s_l$ such that $s_k < s_l$. This happens when $r_i < r_j$ and ($<$) notation is equivalent of $r_i \lesssim r_j \wedge r_j \not \lesssim r_i$
- Otherwise they belong to the same set $r_i, r_j \in s_k$.
Conversely, we can take reconstruct $T$ from $T'$ in two steps:
for all $s_k \in T'$, for each $(r_i \lesssim r_j) \in s_k$ yield a rule $r_i \lesssim r_j$
for all $s_k, s_l \in T'$, for each $r_i \in s_k$ and $r_j \in s_l$ yield a rule $r_i \lesssim r_j$
And then summarize all the rules to create a total pre-ordered set such that $\bigcup T' = T$. The end result is compatible with $T'$ because for any $(r_i \lesssim r_j) \in T'$ its one of two cases:
they belonged to two different sets: $r_i \in s_k$ and $r_j \in s_l$ and $s_k < s_l$. In this case $r_i < r_j$ (see above definition)
they belonged to the same sets.
Later thoughts:
I think it's a weak isomorphism because we are enforcing a constraint for the isomorphism to be true.