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:

  1. A is bounded.
  2. A is contained in some closed ball.
  3. A is contained in some open ball.

Now, this is straightforward if $M$ is nonempty, even in the case $A = \varnothing$, but what happens if $M = A = \varnothing$? We still have that $A$ is bounded (vacuously), but can we still say that $A$ is contained in a closed/open ball? The only subset of $M$ is the empty set, which is not a ball (at least from how I read the definition of an open/closed ball). It seems to me that the result must be false in this degenerate case. Am I correct in this assertion?


1 Answer 1


You are correct. The empty set $\emptyset$ does not have any balls. Not under standard definitions anyway.

An open ball of radius $r>0$ centered around $x\in M$ is defined as

$$B(x,r)=\{y\in M\ |\ d(x,y)<r\}$$

Then an open ball (without specifying $x$ and $r$) is typically defined as $B(x,r)$ for some $x\in M$ and $r>0$. So it forces existence of $x\in M$ (and $r>0$ which is not relevant here). In particular the collection of all open balls is given by

$$\mathcal{B}(X)=\{B(x,r)\ |\ x\in X, r>0\}$$

What I said so far is that $\mathcal{B}(\emptyset)=\emptyset$, while $\mathcal{B}(X)\neq\emptyset$ when $X\neq\emptyset$.

Also note that typically balls are nonempty because $r>0$ and so $x\in B(x,r)$. If you allow $r=0$ or even $r<0$ (for closed balls) then it might be empty. This kind of definition is problematic though, is $B(x,-1)=\emptyset$ really an open ball around $x$? But still the empty set won't contain balls, even if we allow them to be empty. In other words it may happen that $\emptyset\in\mathcal{B}(X)$, depending on whether we allow negative radius or not, but $\emptyset\not\in\mathcal{B}(\emptyset)$, even if we allow negative radius. Because the existence of some $x\in M$ is needed. The only way to allow this is to modify the definition of ball to mean: $B(x,r)$ or $\emptyset$ (note: which is different than allowing negative radius, we decouple $\emptyset$ from $x$). Which can be done, but is nonstandard and likely to lead to lots of boilerplate in proofs, e.g. "let $B$ be a nonempty ball", why "nonempty" word needed? Currently it is redundant.

Note: it is a big difference between "being empty set" (i.e. $\mathcal{B}(X)=\emptyset$) and "having empty set" (i.e. $\mathcal{B}(X)=\{\emptyset\}$). So as you can see, we need to be extra careful when dealing with the empty set. As small difference in wording may have big consequences.

On the other hand the empty set does have a bounded subset: itself. That's because the standard definition of being bounded (i.e. $A\subseteq M$ is bounded if there is $K$ such that $d(x,y)<K$ for all $x,y\in A$) does not require existence of elements in $M$, unlike the definition of ball.

So in the exercise there should be the additional assumption that $M\neq\emptyset$. It's a minor, technical mistake.

  • $\begingroup$ Yes, this is what I was thinking. I consider this settled now. $\endgroup$
    – ummg
    Feb 27 at 14:44

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