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?

Thanks in advance.


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 $\emptyset$.

  • $\begingroup$ 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?? $\endgroup$ – Celine Harumi Jul 5 at 21:50
  • $\begingroup$ @Danmat Yes, indeed. We want it to mean $\overline{A}=X$ and the correct formulation does that. $\endgroup$ – Henno Brandsma Jul 5 at 21:51
  • $\begingroup$ $P \iff Q$ 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? $\endgroup$ – Celine Harumi Jul 5 at 21:55
  • $\begingroup$ @Danmat Fine, but what's your point with that argument? what is $P$ and $Q$ here? $\endgroup$ – Henno Brandsma Jul 5 at 21:56
  • 2
    $\begingroup$ @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. $\endgroup$ – 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.


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