$f$ is a monotone increasing, but not necessarily continuous, on $\mathbb{R}^n$, $A$ is compact. Is $f$ always has a maximum on $A$?

Question

Call a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ nondecreasing if $x,y \in \mathbb{R}^n$ with $x \geq y$ implies $f(x) \geq f(y)$. Suppose $f$ is a nondecreasing, but not necessarily continuous, function on $\mathbb{R}^n$, and $D \subset \mathbb{R}^n$ is compact. Show that if $n = 1$, $f$ always has a maximum on $D$. Show also that if $n > 1$, this need no longer be the case.

I'm stucked at second afirmative. How to get a nondecreasing function ($f: \mathbb{R}^n \rightarrow \mathbb{R}$), with compact domain and show that there is not maximum to it?

Any thoghts would be appreciated. Tks

Answer 1

So if you define $f(x) = -\infty, x \leq b$ and $f(x) = 1/(x-b),b < x < a$ and $f(x) = + \infty, x \geq a$ then $f$ is increasing on $(b,a)$ but does not attain a finite maximum.

Answer 2

Hint: Consider the set $$ A = \{(t,-t) \in \Bbb R^2: t \in [0,1]\} $$ note that any function on $A$ is, by definition, non-decreasing.