Is the empty set compact in $\mathbb R$?

Question

Here they say that the emptyset is compact. Nevertheless, in $\mathbb R$, I know that compact sets are closed and bounded. So, indeed $\varnothing $ is closed, but we have that $$\inf_{\mathbb R}\varnothing =+\infty \quad \text{and}\quad \sup_{\mathbb R}\varnothing =-\infty ,$$ in partuclar, it doesn't seem bounded. So, who is correct ?

@kp9r4d That's circular logic, though. We are trying to prove that $\varnothing$ is bounded in the first place. — Arthur, Sep 14 '20 at 10:40
There are several related Q&As already, e.g. Why is the empty set bounded? and Why is it false that empty set is not bounded? — Martin R, Sep 14 '20 at 10:58

score 12 · Answer 1 · answered Sep 14 '20 at 10:33

12

Boundedness is not defined in terms of $\inf$ and $\sup$ (it's actually the other way around; $\inf$ and $\sup$ are defined in terms of bounds).

A subset of $\Bbb R$ is bounded if there is a positive real number $M$ such that each element of the subset is contained in the interval $(-M, M)$. The empty set fulfills this vacuously.

answered Sep 14 '20 at 10:33

Arthur

204,511

With same vacuous logic empty set is unbounded also. – zkutch Sep 14 '20 at 11:17
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@zkutch Only if your definition of "unbounded" is something other than "not bounded". And it really shouldn't be. What is your definition, by the way? – Arthur Sep 14 '20 at 12:01
Let's speak about unboundedness only from right. May be I am wrong, but I think about following, let call it, "strong" unboundedness: $\forall M \in \mathbb{R}, \forall x \in \emptyset, x>M$. As you see I change $\exists$ with $\forall$ for $x \in \emptyset$. Now if this sentence we can consider as correct, then we can obtain by it usual unboundedness, because it use $\forall \Rightarrow \exists$. By the way, unique set, which I can imagine with "strong" unboundedness is ${ +\infty }$. – zkutch Sep 14 '20 at 20:14
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@zkutch I think $\forall x \in A,P(x) \implies \exists x \in A, P(x)$ only if $A$ is nonempty. – somitra Aug 07 '21 at 06:35

score 7 · Answer 2 · answered Sep 14 '20 at 11:17

The empty set is a subset of any set.

In particular it is included in open balls $\varnothing\subset B(0,r)$ so it is bounded.

But you do not really need a metric, since it is included in any open set, for a given open covering you can just take any set in it and it becomes a finite covering, making the empty set compact.

Is the empty set compact in $\mathbb R$?

2 Answers2