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...

-
Thank you for your input! I think I don't quite understand that... the $max d(x_j,x_k)$ would be $a(n-2) + (a/2) + (a/2) = a(n-1)$ but $2a + a(n-1) = a(n+1)$ which doesn't intuitively make sense to me. Am I thinking about this incorrectly? If you take a closed, bounded interval of R, and line up the balls to cover that set you get that the distance between two points could be at most $a\times n$. I can't picture a bigger distance than that right now. – lillian Oct 9 '11 at 17:25
Suppose $X$ has $n$ points, with any two points being a fixed distance $d$ apart. Then $X$ has diameter $d$, but can be covered by $n$ open balls with radius as small as we like. So there's no function of $n$ and $a$ which will work as an upper bound here. – Chris Eagle Oct 9 '11 at 17:38
I've been assuming that "a" is a fixed radius but I think it is too ambiguously written to know either way. Thanks for the input, I think I need further direction to understand how to think about this problem. – lillian Oct 9 '11 at 17:41
The radius $a$ is fixed. The question is not at all ambiguous: it’s asking for the smallest number (as a function of $a$ and $n$ that is guaranteed to be at least as large as the diameter of $X$. Then, once you have that, you’re to find a specific example of $n$ balls of radius $a$ covering a space whose diameter really is that big; this will show that no smaller number would have met the guarantee. – Brian M. Scott Oct 9 '11 at 17:50
If $X$ is not connected, then there is no upper bound from the information given. – robjohn Oct 9 '11 at 17:52

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.