Determine $d(A)$, when $A = \{f_n:[0,1] \to \Bbb R \vert f_n(x) = x^n, n \in \Bbb N \}$. 
Let $E = ([0, 1], \mathbb{R})$ be the set of bounded functions equipped with the sup norm and the metric it creates. Determine $d(A)$, when $A = \{f_n:[0,1] \to \Bbb R \vert f_n(x) = x^n, n \in \Bbb N \}$.

Looking at the graphs of $x^n$ it looks as if they would be getting closer and closer to $1$. I'm not sure if this has anything to do with the problem. I'm new to topology and metric spaces and trying to learn about it before taking a class. I guess I should somehow look at the limiting function here and use the sup norm to find $d(A)$, but I haven't done these kind of problems before so I have little to no idea on how to start. How should I start or even look at the problem here?
 A: Another user just said that $d(A)$ isn't the derived set of $A$, but the diameter of $A$, defined by $d(A)=\sup\{\|g-h\|_\infty=\sup_{x\in[0,1]}|g(x)-h(x)|: g,h\in A\}$. I will keep my old answer at the end just in case.
We can actually calculate $\|g-h\|_\infty$ for two functions $g,h\in A$. Let's say $g(x)=f_m(x)=x^m$ and $h(x)=f_n(x)=x^n\quad\forall x\in[0,1]$. WLOG we can assume $m>n$ (we don't have to consider the case $m=n$, since then $\|g-h\|_\infty=0$ and we're looking for the supremum of these norms).
Since $x\in[0,1]$ and $m>n$, we have $|g(x)-h(x)|=|x^m-x^n|=x^n-x^m\quad\forall x\in[0,1]$. Since $g(0)-h(0)=g(1)-h(1)=0$, then $\sup_{x\in[0,1]}|g(x)-h(x)|=\max_{x\in(0,1)}|x^m-x^n|=\max_{x\in(0,1)}(x^n-x^m)$.
Let's consider the function $f(x)=x^n-x^m\quad\forall x\in[0,1]$. Then $f'(x)=nx^{n-1}-mx^{m-1}\quad\forall x\in(0,1)$, and setting $f'(x)=0$ we have $nx^{n-1}-mx^{m-1}=0\Rightarrow x^{n-1}(n-mx^{m-n})=0$, and since we're interested in $x\in(0,1)$ we get $n-mx^{m-n}=0\Rightarrow x^{m-n}=\dfrac{n}{m}\Rightarrow x=\left(\dfrac{n}{m}\right)^\tfrac{1}{m-n}$.
So for $m,n\in\Bbb N$ such that $m>n$ we know $\|f_m-f_n\|_\infty=\left(\dfrac{n}{m}\right)^\tfrac{1}{m-n}$. Since $m>n$ then $\dfrac{n}{m}<1$, so $\left(\dfrac{n}{m}\right)^\tfrac{1}{m-n}<1$. Therefore $\sup\{\|g-h\|_\infty: g,h\in A\}\le1$.
Now, consider $n=1$ and any $m>1$. We have $\|f_m-f_1\|_\infty=\left(\dfrac{1}{m}\right)^{\tfrac{1}{m-1}}$. We can make $m\to\infty$ to see what happen's when $m$ gets bigger and bigger: $\lim_{m\to\infty}\left(\dfrac{1}{m}\right)^{\tfrac{1}{m-1}}=1$, so $\sup\{\|g-h\|_\infty: g,h\in A\}=1$.
Therefore $d(A)=1$.

Now I'm assuming $d(A)$ is the derived set, but let's call it $A'$.
We're looking for the bounded functions $g$ on $[0,1]$ such that for every $\epsilon>0$ we have $B(g;\epsilon)\cap(A\setminus\{g\})\neq\emptyset$, where $B(g;\epsilon)=\{h\in E\mid \|g-h\|_\infty=\sup_{x\in[0,1]}|g(x)-h(x)|<\epsilon\}$. This means that for every $\epsilon>0$ there must be a function $h$ distinct from $g$ such that $h$ is in $A$ and $\|g-h\|_\infty<\epsilon$. In this case this is: for every $\epsilon>0$ there must be some $m\in\Bbb N$ such that $\|g-f_m\|_\infty<\epsilon$, where $f_m(x)=x^m\quad\forall x\in[0,1]$.
Imagine there is some function $g$ in $A'$. I claim that then $\|g-f_n\|_\infty\to0$ as $n\to\infty$, this is, the sequence of functions $(f_n)_{n\in\Bbb N}$ converges uniformly to $g$ in $[0,1]$.
It that weren't the case, then there would be some $\epsilon_0>0$ and some $n_0\in\Bbb N$ such that $\|g-f_n\|_\infty\ge\epsilon\quad\forall n\ge n_0$. Then, if $g\in A$ such that $g=f_k$ with $k<n_0$, take $\epsilon_1=\min(\epsilon_0,\|g-f_1\|_\infty,\|g-f_2\|_\infty,\ldots,\|g-f_{k-1}\|_\infty,\|g-f_{k+1}\|_\infty,\ldots,\|g-f_{n_0-1}\|_\infty)>0$; if that's not the case for $g$, take $\epsilon_1=\min(\epsilon_0,\|g-f_1\|_\infty,\|g-f_2\|_\infty,\ldots,\|g-f_{n_0-1}\|_\infty)>0$.
Now consider a generic $f_m\in A$ distinct from $g$ (in case $g$ is in $A$). If $m<n_0$ then whatever is $g$ we have $\|g-f_m\|_\infty\ge\epsilon_1$; if $m\ge n_0$ then $\|g-f_m\|_\infty\ge\epsilon_0\ge\epsilon_1$. So for $\epsilon_1$ there is no $m\in\Bbb N$ such that $\|g-f_m\|_\infty<\epsilon_1$, but this can't be right, since we had $g\in A'$.
Then, if $g\in A'$, the sequence $(f_n)_{n\in\Bbb N}$ must converge uniformly to $g$. So by now we can say already that there is at most one element in $A'$ (since there cannot be more than one limit for $(f_n)_{n\in\Bbb N}$).
But there are two well known facts which say that a uniform convergent sequence of continuous functions must be continuous, and that uniform convergence implies pointwise convergence. Thus, if $g\in A'$, it must be continuous. However, using the second well known fact for $x\in[0,1)$ we would have $g(x)=\lim_{n\to\infty}x^n=0$, and for $x=1$ we would have $g(1)=\lim_{n\to\infty}1^n=1$, so $g$ isn't continuous. Therefore the sequence $(f_n)_{n\in\Bbb N}$ can't converge unformly to $g$, and thus $g\notin A'$.
Finally we can say $A'=\emptyset$.
