I'm wondering if $\sup_{x \in M} f(x) = \max_{x \in M} f(x)$ holds when $f$ is some arbitrary function and $M = \{0,1,\dots,n\}$ for some $n \in \mathbb N$.
My idea is that $M$ is closed and bounded as a (finite) union of closed and bounded sets containing one element each, so its compact due to Heine-Borel and if $f$ is continous then this equality will hold. But what if $f$ is not continous? Also what happens in the limit $n \to \infty,$ so that $M = \mathbb N_0$?
Thanks!
I will assume that $f$ is a map from $\mathbb N$ into $\mathbb R$.
If $M=\{1,2,\ldots,n\}$ then $\{f(1),\ldots,f(n)\}$ is a finite set of real numbers and for every finite non-empty subcet $F$ of $\mathbb R$, there is always some element greater or equal than all the otheres. That element will be both $\max F$ and $\sup F$.
On the other hand, if the domain of $f$ is $\mathbb N$, then, even if $\sup\{f(n)\,|\,n\in\mathbb N\}$ exists, it doesn't have to be a maximum. Take $f(n)=-\frac1n$, for instance. Then $\sup\{f(n)\,|\,n\in\mathbb N\}=0$, but $0\notin\{f(n)\,|\,n\in\mathbb N\}$.
