I was reading through the appendix of Lee's Introduction to Topological Manifolds and came across the following exercise in the section on metric spaces:
Exercise B.11 Let $M$ be a metric space and $A \subseteq M$ be any subset. Prove that the following are equivalent:
A is bounded.
A is contai...
He simply forgot to assume that $M$ is nonempty, that's all. If you care enough, you can email him about this issue so he can include it in errata for the book.
I don't see a problem with a ball in $A$ being empty. Topologically the empty set is open and closed so $A$ would be contained in every open and closed ball. The ball is defined as a subset of the elements in the metric space and the empty set is the only set that meets that criteria. $A$ would then be bounded since it's contained in every ball.
I suppose such a definition might make sense. But the definition of an open/closed ball that I am familiar with (and which is used in the book) requires a point $x$ and a radius $r$. (See also Wikipedia.) There does not seem to be a notion of an empty ball.
@CyclotomicField no, an (open) ball is defined as a special subset of the form $B(x,r)=\{y\in M\ |\ d(x,y)<r\}$ for $x\in M$ and $r>0$. This definition is not valid if you can't choose $x$ and $r$ (of course you always can choose $r$). The empty metric space does not have any balls. Also typically a ball is nonempty because $x\in B(x,r)$, because $r>0$, sometimes $r\geq 0$. But even if you allow negative radius, the empty set still doesn't have balls. Having an empty ball and not having balls are different things.
@freakish Metric spaces come with a metric and that's also defined using the nonexistent elements of $M$. If we can define the distance then we can define the ball. I feel like if you allow one you're allowing the other.
@CyclotomicField no, there does exist precisely one function $\emptyset\times\emptyset\to\mathbb{R}$ that satisfies all metric axioms (formally: the empty function, which technically is just the empty set itself), because none of them require existence. In particular $\emptyset$ is a well defined, valid metric space. This is a subtle maths, but very important formally. We want to avoid inconsistencies and other paradoxes.
Also, definition of boundness doesn't require existence of elements in $M$. But definition of ball does.
@CyclotomicField i meant a general approach where you define things intuitively, e.g. "you need points to define distance", no, you don't. You can define a ball as either $B(x,r)$ or empty set. But then you need to need extra care everywhere. Maybe not inconsistent or paradoxical, but annoying, confusing and nonstandard approach.
@CyclotomicField I think one would have to be very careful if defining a notion of empty ball, for the reasons pointed out in this answer, though it isn't directly relevant in my case (where the metric space $M$ is empty). That is, an empty ball around a point, say $B_0(x)$, is problematic to allow, because it would make every set open under the standard definition of an open subset of a metric space. Overall it seems like it would bring too much trouble to be useful.
@ummg the empty set is open in every metric space, but it's only a ball in the empty metric space. I'm not convinced there are any problems by allowing this.
@freakish I find arbitrarily excluding it confusing, not the other way around. If there is no compelling reason to do so why single it out for exclusion?
@CyclotomicField arbitrarily excluding it? You do realize that the same can be said about anything? Like, why is an open ball arbitrarily excluded to mean, I don't know, an elephant? No, ball is ball. It has short, compact, useful definition, that doesn't allow empty set. There is no need of extending it. Otherwise you will start writing "consider a nonempty ball", "consider a ball that is not an elephant". Its a convention, but a useful one. The real question is: why would I allow an empty set to be a ball? What does it give me? Not much. Maybe one less problematic exercise...
@freakish balls can only be empty if the metric is empty. You don't have to state the ball is non-empty in a metric space that has elements. All the complaints you have about the empty ball apply to the empty metric in exactly the same way. We gain nothing useful by extending metrics to the empty set yet we do. Including the empty metric and excluding the empty ball is internally inconsistent. Do both or do neither but this mix-and-match approach is arbitrary and confusing.
@CyclotomicField including empty space as a metric space is not a choice: it is a fact that follows from the definition. Excluding empty space as ball is not a choice: it is a fact that follows from the definition. The only way to change this is to change definitions. But you gain nothing from that, those definitions are good as they are. You are incorrect claiming that the empty space is not useful as a metric space, it very much is. Unlike empty ball. I think I'm starting to repeat myself, so I will end this conversation now.
@freakish your position just doesn't hold up to scrutiny. You keep claiming there is utility to it but aren't able to provide a single example of that utility. Until you can your reasoning is unconvincing.
@CyclotomicField first of all: I am not obligated to provide any examples. Definition is definition, one is satisfied by empty set, the other is not. End of story. You don't accept that? I don't care. Secondly: me and OP already provided multiple examples for you. You just refuse to listen. Good luck, I'm done.
@freakish here is a problem given where they ask if the empty set is a ball but they explicitly state that the metric space is non-empty. I would suggest they did because they have to. The author of this problem didn't exclude the empty metric or the empty ball. I would suggest because they didn't have to. Allowing the empty metric to have an empty ball doesn't cause any complications that aren't inherited from the empty metric. math.stackexchange.com/questions/1638820/…
@CyclotomicField I think you have some deep misunderstandings considering empty functions and vacuous truths. Everything freakish said is perfectly valid.
@Trebor the "trust me bro" line of reasoning isn't sufficient. If you have a mathematical reason to exclude it then provide it. Until then I'm unconvinced.
@CyclotomicField Since a correct answer is accepted, there is not much risk that future visitors will get confused. So I don't feel particularly compelled to write that down because it would probably only specifically help one person (comments are not easily searchable so if someone had the same misunderstandings as you they can't find this post). You can ask a separate question on MSE if you want.
Discussion on question by ummg: Techn…
Discussion on question by ummg: Techn…