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This question is related to Why are metric spaces non-empty? . I think that a metric space should allowed to be empty, and many authorities, including Rudin, agree with me. That way, any subset of a metric space is a metric space, you don't have to make an exception for $\varnothing$, and you can ascribe certain properties to $\varnothing$, such as, $\varnothing$ is compact and connected. By definition of diameter of a metric space, the diameter of $\varnothing$ should be $-\infty$, since the $\sup$ of the empty set is $-\infty$. This "feels wrong", since the diameter is a measure of the "size" of a metric space or a subset thereof, and its seems like the diameter of $\varnothing$ ought to be zero.

I guess I really have two questions:

If you allow the diameter of the empty set to be $-\infty$, does it lead to problems? For example, $-\infty$ plus any real number is $-\infty$, and I could imagine how that might lead to a problem, but I haven't seen an actual situtation where that happens.

In practice, what do expert analysts (such as Rudin, Folland, Royden, etc.) use for the diameter of $\varnothing$?

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    $\begingroup$ I would say the space relevant for the supremum is $[0,\infty]$, so $\sup \varnothing = 0$. $\endgroup$ – Daniel Fischer Nov 24 '13 at 16:26
  • $\begingroup$ @DanielFischer : that seems to make the most sense. I'd like a reference to back that up. $\endgroup$ – Stefan Smith Nov 24 '13 at 16:34
  • $\begingroup$ Boto v. Querenburg, Mengentheoretische Topologie, definition 13.8. That was the first book I looked in. $\endgroup$ – Daniel Fischer Nov 24 '13 at 16:38
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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.

  1. It's safer to keep $\operatorname{diam}$ nonnegative, because it may appear in formulas that need nonnegative inputs.
  2. If the validity of what you write depends on the interpretation of $\operatorname{diam}\varnothing$, consider changing the statement.
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    $\begingroup$ user111213 : Wow! thanks for taking the question seriously and providing such a complete answer. I think that if possible, the diameter of the empty set should be defined somehow. Based on your answer, it seems like it's best to define the diameter of $\varnothing$ to be zero, since allowing it to be $-\infty$ may create problems, and the one argument you gave in favor of using $-\infty$ is not terribly compelling when one considers the problems may arise if you use $-\infty$. $\endgroup$ – Stefan Smith Nov 25 '13 at 13:57
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    $\begingroup$ I'm in favour of diam $\phi=0$ for reasons that Qiaochu Yuan mentions in the other thread, but I disagree with the reason involving that inequality of unions. Why should the latter dist$(A,\phi)$ need to be $+\infty$? This can well be the matter for another question and so on, but the point I wish to make is this: arguments for diam $\phi=-\infty$ would most naturally and easily be arguments for dist$(A,\phi)=-\infty$ as well. In my opinion, this choice between 0 and $-\infty$ is more or less like the choice between $-1$ and $-\infty$ for dimension of null set or degree of the zero polynomial. $\endgroup$ – N Unnikrishnan Jan 7 '17 at 21:08
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I've been reading through Kolmogorov & Fomin, Introductory Real Analysis, which has the following two problems at the end of section 7 (note: $\rho(\cdot,\cdot)$ is the distance in the metric space):

Problem 4. By the diameter of a subset $A$ of a metric space $R$ is meant the number $$ d(A) = \sup_{x,y\in{A}} \rho(x,y). $$ Suppose R is complete, and let ${A_n}$ be a sequence of closed subsets of $R$ netsted in the sense that $$ A_1 \supset A_2 \supset \ldots \supset A_n. $$ Suppose further that $$ \lim_{n\rightarrow{}\infty}d(A_n) = 0. $$ Prove that the intersection $\bigcap\limits_{n=1}^{\infty}A_n$ is nonempty.


Problem 6. Give an example of a complete metric space $R$ and a nested sequence ${A_n}$ of closed subsets of $R$ such that $$ \bigcap\limits_{n=1}^{\infty}A_n = \emptyset. $$ Reconcile this example with Problem 4.

To answer Problem 6:

If $R=\emptyset$, then $R$ is a complete metric space. And if we define $A_n=\{\emptyset,\emptyset,\ldots\}$, then $A_n$ is a nested sequence of closed subsets of $R$. And clearly $\cap_{n=1}^{\infty}A_n = \emptyset$. In order to reconcile this with Problem 4 we may well ask, what is the diameter of the empty set? As suggested by @user111213, this cannot be reconciled if the empty set has diameter zero. If the diameter is $-\infty$, or if it is not defined, then the limit in problem 4 cannot be zero, thus reconciling our example.

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