The following books explicitly take the position that $\operatorname{diam}\varnothing =0$:
- C. Kuratowski, Topology, vol.I
- M. H. A. Newman, Elements of the topology of plane sets of points
The following books explicitly take the position that $\operatorname{diam}\varnothing =-\infty$:
- G. F. Simmons, Introduction to topology and modern analysis
- M. Ó. Searcóid, Metric spaces
(I never heard of either of these before Google Books search brought them up.)
The following books explicitly restrict the definition of diameter to nonempty sets:
- W. Rudin, Principles of mathematical analysis
- H. L. Royden, Real analysis.
- K. Falconer, Fractal geometry
It seems that W. Sierpiński, General topology, belongs to the second or third category, because the author says on page 110: "Thus the diameter of every non-empty set contained in a metric space is a uniquely defined real non-negative number, finite or infinite". But it's not very clear what Sierpiński's intention was when writing this.
Many books do nothing of the above: they define the diameter of a set as supremum of pairwise distances, and offer no further details.
If you allow the diameter of the empty set to be $−\infty$, does it lead to problems?
The definition of Hausdorff measure would become awkward. For example, the 1-dimensional measure involves the infimum of $\sum \operatorname{diam} U_i$ over certain families of sets. If $\operatorname{diam}\varnothing =-\infty$, we'd be able to make the infimum $-\infty$ by throwing in the empty set. (Note that the Wikipedia article explicitly says that $\operatorname{diam}\varnothing =0$). One can try to fix this by requiring $U_i$ to be nonempty, but then the measure of empty space becomes a special case (and the measure of $\varnothing$ definitely needs to be $0$).
Another issue is the inequality
$$
\operatorname{diam}(A\cup B)\le \operatorname{diam}A+\operatorname{diam}B+\operatorname{dist}(A,B)
$$
which should hold for all $A,B$. Suppose $B$ is empty but $A$ is not. The right-hand side becomes undefined due to presence of
$\operatorname{diam}\varnothing =-\infty$ and $\operatorname{dist}(A,\varnothing)=+\infty$. (And the latter definitely needs to be $+\infty$.)
Third issue: if one applies a metric transform, i.e., replaces metric $d$ with $\varphi(d)$ where $\varphi $ is an increasing concave function, the diameters of sets should transform accordingly. With $-\infty$ in the mix, one is led to awkward conventions ($\sqrt{-\infty}=-\infty$?).
That said, I can imagine some arguments in favor of $\operatorname{diam}\varnothing =-\infty$. One is that the following statement becomes true:
In a complete metric space, each decreasing sequence of closed sets $C_n$ with $\operatorname{diam}C_n\to 0$ has nonempty intersection.
(quoted from S. Willard, General topology). If $\operatorname{diam}\varnothing =0$, the above is false without additional requirement that $C_n$ are nonempty.
That said, it's probably best to put nonempty there. The absence of nonempty leads to wrong statements in a number of books, e.g., "if $N$ is compact, there exist $x,y\in N$ such that $\rho(x,y)=\operatorname{diam}N$". (G.T. Whyburn, Analytic topology).
Summary.
- It's safer to keep $\operatorname{diam}$ nonnegative, because it may appear in formulas that need nonnegative inputs.
- If the validity of what you write depends on the interpretation of $\operatorname{diam}\varnothing$, consider changing the statement.
we can prove (b) with a mathematical induction. when $n$ is 2, get $a_1$ and $a_2$ from $A_1$ and $A_2$. now when we are finding $sup\{d(x,y): x, y \in A\}$ if both of $x$ and $y$ be from one of the $A_1$ or $A_2$ then the calculated distance is'nt more than maximum of diameter of $A_1$ and $A_2$, and when $x$ and $y$ are'nt in a same set like when $x$ is from $A_1$ and $y$ is from $A_2$ then we have from triangle inequality:
$$d(x,y) \leq d(x,a_1)+d(a_1,a_2)+d(a_2,y)\leq diam (A_1)+d(a_1,a_2)+diam(A_2)$$
thus $d(x,y)$ is less than a fixed value and the calculated suprimum is less than this fix value.
for proving (c) it is enoght that we set a fix element in the intersection of all sets and transfer throght it with using triangle inequlity for bounding distants of two arbitrary point from the above.
Best Answer
The diameter of a nonempty subset $S\subseteq X$ is defined to be $$\sup_{x,y\in S}d(x,y),$$ which is always defined (though it may be infinite). Thus you can always determine the diameter of a nonempty subset.
A nonempty subset $S\subseteq X$ having finite diameter is equivalent to $S$ being contained in some ball $B(x,r)$ for some $x\in X$ and some $r>0$. To see this, first suppose that $S$ has finite diameter $\Delta<\infty$. Pick any $x\in S$. Then $S\subseteq B(x,2\Delta)$, since $d(x,y)\leq \Delta$ for every $y\in S$. On the other hand, suppose that $S$ is contained in some ball $B(x,r)$, where $x\in X$ and $r>0$. Then for any two points $y,z\in S$, one has $d(y,z)\leq d(y,x) + d(x,z)\leq 2r$, so the diameter of $S$ is $\leq 2r$. I hope that answered your question!