Metric Spaces and $\sup$ of Diameter Question: Let $(X, d)$ be a metric space such that there is a positive $a$ and $n$ open balls $B(x_1, a),\ldots, B(x_n, a)$ such that together these balls cover $X$.
Find an upper bound $M$ for the diameter of $X$. Find "the smallest" upper bound $M$ for the diameter of $X$ in the following sense: the formula for $M$ is valid in all cases, and there is a metric space $X$ such that $\operatorname{diam} (X) = M$.
Thoughts: I find the question a little hard to parse. I think it's asking what is the biggest diameter that this can have, so my answer then would be $\sup M = an$.
Thinking about the real number line, if there are $n$ # of balls, the two furthest points from each other would be $a\cdot n$ length from each other. Still trying to figure out what the rest of the questions asks...

Let $x,y\in X$. Then we can find $j,k\in\{1,\ldots,n\}$ such that $d(x,x_k)<a$ and $d(y,x_j)<a$. We get 
$$d(x,y)\leq d(x,x_k)+d(x_k,x_j)+d(x_j,y)< a+d(x_k,x_j)+a\leq 2a+\max_{1\leq j,k\leq n}d(x_k,x_j),$$
therefore $\displaystyle \operatorname{Diam}(X)\leq 2a+\max_{1\leq j,k\leq n}d(x_k,x_j)$. 
We can't hope a better bound. Indeed, taking $X:=\left]0,(n+1)a\right[$ with the usual metric, and $x_j=ja$, we get that $\displaystyle X=\bigcup_{j=1}^nB(x_j,a)$, $\operatorname{Diam}(X)=(n+1)a$ and $\displaystyle \max_{1\leq j,k\leq n}d(x_k,x_j)=\max_{1\leq j,k\leq n}|ka-ja|=(n-1)a$, and the above inequality is an equality.
