0

I have a question when proving the empty set is open in a given metric space. Here is my proof: By definition of an open set, we have the statement "the empty set is open" is equivalent to " for every a in $\emptyset$, there is an $r>0$ s.t $B(a;r) \subset \emptyset$". It can be seen that there is no $a$ in $\emptyset$, so the latter statement is always true (this is what i understand as a vacuous truth) . As a result, the former is proven. But someone said that the vacuous truth was somewhat a pitfall. Besides, my lecturer did not accept my proof and he used contraposition to tackle this problem. Could anyone help me show my error? And am i misunderstanding something related? Thanks in advance.

Linda
  • 23
  • 6
  • 4
    There is absolutely nothing wrong with your proof (other than a typo where you wrote $a$ instead of $x$). I have no idea what was going on in lecturer’s mind when he said not correct, or what the person who complained about vacuous truth was on about (maybe some philosophers don’t like it for who knows what reason, but vacuous truth is as good as any other kind of truth in standard logic that working mathematicians use daily) – M W Sep 20 '23 at 14:19
  • 4
    The empty set is an open set, and I think your argument is good enough. As to whether it is an open ball, your $r>0$ suggests it could be convenient to say no. Related: https://math.stackexchange.com/questions/1638820/is-the-empty-set-an-open-ball-in-a-metric-space – Henry Sep 20 '23 at 14:22
  • 1
    Just read the title, I will note there is a distinction between “$A$ is an open ball” and “$A$ is an open set”, so if you were ambiguous about that issue somehow perhaps that was lecturers problem. Not having been there, I can’t really say. – M W Sep 20 '23 at 14:25
  • 1
    There are two reasonable tentative defintions of $X\subseteq (Y,d)$ being an open ball: 1) $[\exists y\in X,\exists r>0,\forall x\in Y, ( x\in X\Leftrightarrow d(x,y)<r)]$; 2) $[\exists y\in Y,\exists r\in\Bbb R,\forall x\in Y,( x\in X\Leftrightarrow d(x,y)<r)]$. Personally, I think the former is the "right one", in which case all open balls are non-empty. If it's the latter, then $\emptyset$ is an open ball of $Y$ if and only if $Y$ isn't the empty metric space. – Sassatelli Giulio Sep 20 '23 at 14:27
  • I would have given the same proof you did. Vacuously true statements are simple to argue and it's a valid method. The contrapositive still argues that there are no elements in $\emptyset$. It's a logically equivalent statement proved by appealing to exactly the same property. Their preference for the contrapositive form is arbitrary and unjustified. – CyclotomicField Sep 20 '23 at 14:36
  • I agree with @Sassatelli. An open ball has a center which precludes the empty set from being an open ball in a metric space. It is open because we cannot find a point in the empty set that isn't contained in an open ball contained in the empty set. – John Douma Sep 20 '23 at 16:41
  • Thanks for your help and i appreciate all of them. I had mistaken those concepts so i wrote the wrong title, but i can clearly understand then. The correct one is the empty set is an open set, not the open ball as written. – Linda Sep 21 '23 at 04:44
  • You should not edit your question (except correcting typos) after having received an answer. I was confused by the first sentence of the answer, but then I realized that you changed the original title. – Kritiker der Elche Sep 24 '23 at 14:39
  • @KritikerderElche Sorry for the confusion. I will rewrite the original one. – Linda Sep 24 '23 at 14:47

1 Answers1

1

Your proof is abolutely correct, but in the title of your question you write "Prove that the empty set is an open ball". This has nothing to do with your arguments in the text body.

Usually an open ball in a metric space $(X,d)$ is defined as a set of the form $$B(x;r) = \{x' \in X \mid d(x',x) < r \} \tag{1}$$ with $x \in X$ and $r > 0$. We say that $B(x;r)$ is the open ball with center $x$ and radius $r$. Since always $x \in B(x;r)$, open ball are never empty, hence the empty set is not an open ball.

However, if we allow $r = 0$ in definition $(1)$ - which would be an extremely unusual approach - we get $B(x;0) = \emptyset$ for all $x \in X$. With this interpretation the empty set is an open ball - unless $X =\emptyset$ ...

Paul Frost
  • 76,394
  • 12
  • 43
  • 125
  • And similarly, we get into definition details for the question of whether $X = \emptyset$ is a metric space at all. – aschepler Sep 20 '23 at 23:17
  • 1
    I don't like the idea of a zero radius open ball because one point sets are not usually open in metric spaces. In fact, every set would be open because every point would be contained in an open ball that lies completely within the set. – John Douma Sep 20 '23 at 23:44
  • 1
    @JohnDouma I agree that allowing zero radius open balls is not a very good idea - but all zero radius open balls are empty, thus you cannot conclude that one point sets are open. We can only say that singletons are zero radius closed balls. – Paul Frost Sep 21 '23 at 00:00
  • 1
    That's true. I was thinking the center would be in the ball but $d(x,x)=0\not\lt 0$. – John Douma Sep 21 '23 at 00:23
  • Thanks for pointing out the errow in the title, now i can discriminate between the open ball and the open set, i will edit it. – Linda Sep 21 '23 at 04:29