Does maximum or supremum of an infinite set exit? If I have a set: $S = \{ k_1, k_2,...,k_n,... : k_i < \infty \}$, which means that $S$ include infinite many elements, but each element is a finite real number. In this case does 
$\max(S)$ or $\sup(S)$ exit, and is finite? 
I can not find useful theorem to clarify this. Any help appreciate.
 A: Some examples:
$$ \mathbb{N} = \left\{1, 2, 3, 4, 5, \dots\right\} $$
$\mathbb{N}$ does not have a maximum, nor does it have a supremum.
$$ T = \left\{1-\frac{1}{n} : n \in \mathbb{N} \right\}$$
$T$ does not have a maximum but does have a supremum: $\sup(T) = 1$.
$$ U = \left\{\frac{1}{n} : n \in \mathbb{N} \right\}$$
$U$ has a maximum and a supremum, which are necessarily the same (a maximum of a set must be its least upper bound): $\max(U) = \sup(U) = 1$.
A: An infinite set which is bounded above always has a supremum by the completeness of $\mathbb{R}$; i.e. any set which has an upper bound has a least upper bound. A set $A \subseteq \mathbb{R}$ is said to be bounded above if there is $M \in \mathbb{R}$ such that $a \leq M$ for all $M$.
Examples: 


*

*The infinite set $\mathbb{N}$ is not bounded above so it does not have a supremum. 

*The set $[0, 1]\cap\mathbb{Q}$ is bounded above; for example, taking $M = 1$ we see that $a \leq M$ for every $a \in [0, 1]\cap\mathbb{Q}$. Therefore, $[0, 1]$ has a supremum.

*The set $\{-\frac{1}{n} \mid n \in \mathbb{N}\}$ is bounded above; for example, taking $M = 0$ we see that $a \leq 0$ for every $a \in \{-\frac{1}{n} \mid n \in \mathbb{N}\}$. Therefore, $\{-\frac{1}{n} \mid n \in \mathbb{N}\}$ has a supremum.


An element $x \in A \subseteq \mathbb{R}$ is said to be maximal (or just a maximum of $A$) if $a \leq x$ for all $a \in A$. The difference between this definition and the related definition above is that $x$ belongs to $A$ whereas $M$ need not belong to the set in question. A finite subset always has a maximum, but an infinite set may or may not, even if it is bounded above.
Examples:


*

*The set $[0, 1]\cap\mathbb{Q}$ has a maximum, namely $1$ because for any $a \in [0, 1]\cap\mathbb{Q}$, $a \leq 1$.

*The set $\{-\frac{1}{n} \mid n \in \mathbb{N}\}$ does not have a maximum, because for any $x \in \{-\frac{1}{n} \mid n \in \mathbb{N}\}$, there is $a \in \{-\frac{1}{n} \mid n \in \mathbb{N}\}$ such that $x < a$. More precisely, if $x = -\frac{1}{n}$, then $a = -\frac{1}{2n}$ satisfies $x < a$.

