# Simple question about “vacuous truth”.

In my homework there is an exercise that asks to show the following result:

Let $$(E,d)$$ be a metric space. Show that a subset $$A$$ is dense in $$E$$ iff every open set in $$(E,d)$$ contains an element of $$A$$.

I was thinking in the case of the empty set. My question:

"$$\emptyset$$ contains an element of $$A$$" is false or is vacuously true?

If it is false, then the necessary condition for the denseness of $$A$$ will always be false, because there will always be the (open) empty set in $$E$$ which does not contain any element of $$A$$. In this case, logically, $$A$$ would never be a dense subset of $$E$$. Is my argument right or am I going crazy?

The formulation you quoted is slightly wrong, it should have been:

Let $$(E,d)$$ be a metric space. Show that a subset $$A$$ is dense in $$E$$ iff every non-empty open set in $$(E,d)$$ contains an element of $$A$$.

So you're not going crazy. In the formulation you gave no set will ever be dense and we've defined a "vacuous property". And the corrected formulation (by vacuous truth, as there are no non-empty open subsets to check) indeed allows us even to say that $$\emptyset$$ is dense in the empty space $$\emptyset$$.

• It is what I was thinking. If I dont exclude the empty set, it would imply that no subset in $E$ would be dense. Do you agree?? – Celine Harumi Jul 5 at 21:50
• @Danmat Yes, indeed. We want it to mean $\overline{A}=X$ and the correct formulation does that. – Henno Brandsma Jul 5 at 21:51
• $P \iff Q$ is equivalent to $\neg P \iff \neg Q$. So if $Q$ is always false, it means that $P$ is always false too. What do you think of this argument? – Celine Harumi Jul 5 at 21:55
• @Danmat Fine, but what's your point with that argument? what is $P$ and $Q$ here? – Henno Brandsma Jul 5 at 21:56
• @Danmat Sure. But you still have to show that $E$ always has a dense subset to finish the argument. (e.g. by noting $E$ is always dense in $E$). But the exercise is false, let's just stop flogging a dead horse, shall we? Just do the corrected exercise instead. – Henno Brandsma Jul 5 at 22:15

Let $$p \in X \subset E$$ open. So, there exists $$r > 0$$ such that $$B(p, r) \subset X$$.

As $$\bar{A} = E$$, $$p$$ is adherent to $$A$$. Therefore $$B(p,r) \cap A \neq \emptyset$$. Thus, there is $$q \in A$$ and $$q \in B(p,r) \subset X$$, this is, $$A \cap X \neq \emptyset$$

reciprocally, suppose it absurd that $$\bar{A} \neq E$$. Therefore, there is an element $$x \in E$$ such that $$x \not \in \bar{A}$$. Thus, $$x \in E - \bar{A} \Rightarrow x \in \mbox{int}(E - A) \subset E - A$$. So, $$B(x, \varepsilon) \cap A = \emptyset$$, absurd ! Therefore, $$\bar{A} = E$$.

The empty set contains no elements, from $$A$$ or any other set. This is not an instance of vacuous truth, this is just false.

The vacuous truth that I think you're thinking of is of the form $$\forall x \in \emptyset, P(x)$$, where $$P$$ is some predicate. It doesn't matter what the predicate is, or how laughably false it might be (e.g. "$$x$$ is a square prime"), the preceding statement is considered true simply by virtue of there being nothing preventing it from being false. Its negation, $$\exists x \in \emptyset : \neg P(x)$$, is always false, simply because it asserts the existence of an element $$x$$ of the empty set.

But this is not the case here. You do have a "for all" statement; you are considering all open subsets of $$E$$, which indeed includes the empty set, and hence is a non-empty set! You therefore do not get vacuous truth. Instead, you now have a counterexample: the empty set does not intersect with any $$A$$, so according to this (false) result, no set is dense.