Does there exist an uncountable Well-ordered set $(X,<)$ and a function $f: X \times X \to \mathbb R$ such that $f(x,y)<f(x,z),\forall x<y<z$ ?
Note that if we want $X$ to be countable, then such a set up exists, for instance take $f: \mathbb N\times\mathbb N \to \mathbb R$ to be $f(a,b)=|a-b|,\forall a,b \in \mathbb N$.
Unfortunately, I am unable to find such an uncountable $X$.
Please help
Suppose $(X,<)$ is an uncountable well-ordered set.
Claim:
There does not exist a function $f: X{\,\times\,}X \to \mathbb{R}$ such that for all $x,y,z\in X$, if $x < y < z$ then $f(x,y) < f(x,z)$.
Proof:
Suppose such a function $f$ exists.
For $x\in X$, let $x^{*}$ denote the successor of $x$.
Let $x_0$ denote the least element of $X$, and let $S=X{\,\setminus}\{x_0\}$.
Then $f(x_0,s) < f(x_0,s^{*})$, for all $s\in S$.
For each $s\in S$, let $g(s)$ be a rational number such that $f(x_0,s) < g(s) < f(x_0,s^{*})$.
Then $g$ is a function from $S$ to $\mathbb{Q}$.
Since $X$ is uncountable, so is $S$, and since $\mathbb{Q}$ is countable, so is $g(S)$.
Hence $g$ is not injective.
Let $a,b\in S$ with $a < b$.
\begin{align*} \text{Then}\;\;&x_0 < a < b\\[4pt] \implies\;&x_0 < a < a^{*}\le b < b^{*}\\[4pt] \implies\;&f(x_0,a) < f(x_0,a^{*})\le f(x_0,b) < f(x_0,b^{*})\\[4pt] \implies\;&g(a) < g(b)\\[4pt] \end{align*} so $g$ is strictly increasing, hence injective, contradiction.