So a definition of a strictly increasing function is $~x_1 < x_2 \implies f(x_1) < f(x_2)$.

Can this be extended to be a two-way implication, namely, $~x_1 < x_2 \iff f(x_1) < f(x_2)$?



Yes, it is a two-way implication. Suppose that $f(x_1)<f(x_2)$. Could we have $x_1=x_2$? No, because then $f(x_1)=f(x_2)$. Could we have $x_1>x_2$? No, because$$x_1>x_2\iff x_2<x_1\implies f(x_2)<f(x_1).$$So, we must have $x_1<x_2$.


Say $f$ is strictly increasing.

To show $f(x_1)<f(x_2)\implies x_1<x_2$, assume to the contrary $f(x_1)<f(x_2)$ and $x_2\ge x_1$.

But $x_2>x_1$ implies $f(x_2)>f(x_1)$, a contradiction,

and $x_2=x_1$ implies $f(x_2)=f(x_1)$, also a contradiction.


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