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 want to show that this result is absurd. Firstly, I'm not a logician at all, but I want to go in this way. So suppose the above result is valid.

My attempt to translate this: Given $A,\mathcal{O}\subseteq E$

$B_1 :=[Q_1(A)] \iff B_2 :=[\forall \mathcal{O}(P_2(\mathcal{O})\rightarrow Q_2(\mathcal{O},A))]$


$Q_1(A):=$"$A$ is dense in $E$"

$P_2(\mathcal{O}):=$"$\mathcal{O}$ is open"

$Q_2(\mathcal{O},A):=$"$\mathcal{O} \cap A\neq \emptyset$"

Since $\emptyset$ is an open set in $E$ which does not contain any element of $A$, we have that $B_2$ will always be false. Then $\neg B_1 \iff \neg B_2$ must always hold (which is an equivalent to what was stated). But this implies that $A$ is not dense in $E$. As $A$ is an arbitrary subset of $E$, it means that no subset of $E$ is dense. However, clearly $E$ is dense in itself ($closure(E)=E$ in the space $(E,d)$). Contradiction.

My questions:

1) Is my translation into logical language wrong?

2) The negation $\neg B_1 \iff \neg B_2$ is equivalent to $\neg Q_1(A) \iff \exists \mathcal{O}(P_2(\mathcal{O})\land \neg Q_2(\mathcal{O},A))$?

3) In case of everything above were wrong, how would you proceed to prove the result is wrong using formal logic?

Thanks in advance!

  • $\begingroup$ The asker asked a previous question about this particular problem, so that everyone is up to speed. @Danmat, when asking multiple questions about a single problem, it's usually a good idea to include a link to the previous question in your current question. $\endgroup$ – Theo Bendit Jul 6 at 1:57
  • $\begingroup$ @TheoBendit Ok, thanks! $\endgroup$ – Celine Harumi Jul 6 at 2:01

1) Your logic is fine, 2) you're correct, and 3) I wouldn't use formal logic for something like this.

I think you're over-dressing your argument a little. There's no need to delve into symbolic logic here. A good mathematical proof is no longer than it needs to be, and uses symbols only to make it more readable. A trap for beginners is to think symbols make your argument more formal or more correct, but words are frequently preferable. Unless you're planning to enter your proof into proof verification software, you should always preference human readability.

To prove that this exercise is wrong, you have the right idea: the full set $E$ (considered as a subset of $E$) satisfies the premises of the theorem, but fails to satisfy the conclusion of the proposition. Essentially, we have that $E$ is dense in $E$, that $\emptyset$ is open, and that $E \cap \emptyset = \emptyset$. I think these three statements can be stated without justification.

If you feel like expanding any of them, I would recommend proving $E$ is dense in $E$ (using whatever definition of density you have). At least that feels relevant to a course on topology. On the other hand, proving $E \cap \emptyset = \emptyset$ is a basic exercise in set theory, and has little to do with metric spaces, so I definitely wouldn't bother writing up a full proof of it.


The statement is wrong as written. You need not (and should not) translate the plain English into formal logic to see that: the empty set is open and does not contain an element of $A$ - whether or not $A$ is dense.

Now try to prove what I think the problem should ask: $A$ is dense if and only if every nonempty open set contains an element of $A$.

  • $\begingroup$ This means the necessary condition for denseness is always false, but $A$ can be dense in $E$ (like the example I gave above). Is it a valid proof? $\endgroup$ – Celine Harumi Jul 6 at 1:51

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