If $g(.)$ is a monotonically increasing function and $a <b$, is $a<g(a)<g(b)<b$?

Question

My question is in relation to a problem I am trying to solve here. If $g(.)$ is a monotonically increasing function and $a <b$, is it always true that $a<g(a)<g(b)<b$? Why or why not?

Answer 1

In fact, it is impossible to have any function $g : \mathbb{R} \to \mathbb{R}$ which satisfies that inequality for all pairs $a<b$. To see this, suppose we have a triple $a < b < c$. Then from $a < b$ we conclude $$a < g(a) < g(b) < b$$ whereas from $b < c$ we conclude $$ b < g(b) < g(c) < c. $$ Now, combining $b < g(b)$ from the second and $g(b) < b$ from the first, this implies $b < b$, which is a contradiction.

(In fact, essentially the same argument works if the domain of $g$ is any linear order with at least three elements.)

Answer 2

You can't say $a < g(a)$ and also $g(b)$. But you can say $g(a) < g(b)$ as it is monotinocally increasing.

As a counterexample, you can suppose $g(x) = \frac{x}{2}$.

Answer 3

This is not true. For instance, the exponential function satisfy the inequalities $$2<3<e^2<e^3$$

More easily, the function $g(x)=2x$. Then you have

$$2<3<g(2)=4<g(3)=6$$

Answer 4

Why would it be true? Let $g(x) = 10^{5634789} + x$. The $g'(x) = 1 > 0$ so it is monotonically increasing. Let $0 < 1$. Does it follow that $0 < 10^{5634789} < 10^{5634789} + 1 < 1$?

... In fact, it's hard for me to imagine a function were this is always true.

You are essentially saying that for any $a < b$ then $a < g(a)$ but $b < g(b)$. Well then what about $a < b < c$? If it's true that $a < g(a) < g(b) < b$ then it can not be true that $b < g(b) < g(c) < c$ because $g(b) < b$.

Not only does this not need to be true, it can't be true.