ƒ is a monotonic increasing function under two total orders, R and S, of a set A. How are R and S related?

Question

This is problem #9 from section 7.4 of Axiomatic Set Theory by Patrick Suppes. Intuitively, it seems to me that either R=S, or S is R's converse. However, I don't see how one would approach proving this. I would appreciate it if someone could clarify whether this is correct and if so, where to begin with proving it.

Answer 1

If by increasing you mean "strictly increasing", i.e. $x R y \rightarrow f(x) S f(y)$, where $R, S$ are strict order relations, then :

$R$ embeds into $S$, i.e. the map $f$ defines a copy of the order $R$ on $A$, using the set $Im(f)$ and the order induced by $S$ on this set.

Proof : Take two points $a, b$ in $Im(f)$. Then there exist $c,d$ in $A$ such that $f(c) = a$ and $f(d) = b$. Then, $a S b$ if and only if $c R d$. Moreover, $f$ is injective since the orders are total.

However, you do not necessarily have equality of the orders. For instance : $R$ is the usual order on $\mathbb{N}$, $S$ is the usual one on $\mathbb{N} \setminus \lbrace 0 \rbrace$, and $0$ is bigger than everyone else. Then, if $f(x) = x+1$, the map $f$ is increasing from $(\mathbb{N}, R)$ to $(\mathbb{N}, S)$. But the orders are not equal.