Supremum of a sequence of functions I have a problem understanding the function $\sup f_n(x)$ on a domain of convergence. Given a sequence of functions $f_n(x)$ I guess that $\sup f_n(x)$ is the function such that for each x we find the lub of the values of all $f_n$ on $x$. 
To make thinks more simple, say that the functions $f_n$ are all bounded,then $\sup f_n$ means that for each $x$ of the domain of definition and convergence one finds the max of the values of $f_n$ on the particular $x$. Am I right ?
Now I present an example:
$$\sup\{|x^k|: -1 < x < 1, k \in \mathbb{N} \} = 1$$ which I don't understand. If I follow the thinking scheme of above, I should find $|x|$ and not $1$.
So where is the difficulty? Is my understanding of the definition of supremum of a sequence of functions wrong?
Thanks for any comments. 
 A: In general, we take suprema of sets of real numbers, which is written $\sup S$ for a subset $S\subseteq \mathbb R$. Notations $\sup_{n\in\mathbb N}a_n$ can be rewritten as $\sup\{\,a_n: n\in\mathbb N\,\}$. Likewise $\sup_{n\in\mathbb N}f_n(x)=\sup\{\,f_n(x): n\in\mathbb N\,\}$ where the set on the right depends on the parameter $x$, so that the whole supremum-expression becomes a function of $x$. Your final example
$$\sup\{\,|x|^k: -1\le x\le 1, k\in\mathbb N\,\} $$
is already given with a subset of $\mathbb R$; the $x$ occuring in it does not make us consider functions for $x$ (like $k$) is not a parameter (or "free varaiable") here. As cleraly $|x|^k\le 1$ for all $x,k$ with $-1\le x\le 1$ and $k\in\mathbb N$, we see that $1$ is an upper bound for the set; and as $1$ actually is in the set, we have thus found th eleast upper bound ...
A: I think you may have a slight misunderstanding. 
The supremum is taken on numbers; you take the supremum of a set of numbers.
When you find something like 
$$\sup_{x \in I} \ f_n(x)$$ it means that, for each fixed $n$, you take the supremum of the values $f_n(x)$ for each $x \in I$.
Note that it is the exact same way it is used to find the supremum of a normal function; here the function depends on $n$, so the supremum too will depend on $n$, but it is not conceptually different. 
Example:
$f_n = \frac 1n$
Each $f_n$ is a constant function, so $\sup f_n = \frac 1n$. 

For your example, again you are taking the supremum of a set of numbers; all the numbers in the form $x^k$ for $x \in (-1, 1)$ and $k \in \mathbb N$.
The easiest way to find out what the supremum is, is noticing that $x^k \le 1$ and you can get arbitrarily close to $1$. Hence the supremum is $1$. 
A: Here it says you take the supremum not just over all $k$ but all $x$ as well.
In this case, with, say, $k = 1$ and $x$ close to $1$ you can get $|x|^k$ to be arbitrarily close to $1$.  You obviously can't do better than that, so the sup is $1$. If it said $\sup \{x^k\; : \; k \in \mathbb N\}$, that would be what you were talking about before: for each $x$ you take the supremum of the values $x^k$ over all $n \in \mathbb N$. 
