# Is denseness a antisymmetric relation?

Upon discovering that denseness is transitive, I wondered if denseness is a partial ordering ($$\iff$$ reflexive, antisymmetric, transitive).

To be more precise: Let $$X$$ be a topological space. Then define $$R \subset \mathcal{P}(X) \times \mathcal{P}(X)$$ by $$(a,b) \in R \iff aRb \iff a \text{ is dense in } b,$$ where $$a,b \in \mathcal{P}(X)$$.

Since every set is dense in itself this relation is symmetric but is it also antisymmetric ($$aRb$$ and $$bRa \implies a = b$$)? Or do I have to define it a little bit more precise as $$(a,b) \in R \iff a \cap b \text{ dense in } b,$$ since $$a$$ doesn't have to be a subset of $$b$$?

• I think what you might be asking about is a binary relation $R$ on $P(X) ,$ the set of all subsets of $X.$ If so, please edit....BTW "$A$ is dense in $B$" usually means that $A$ is a dense subset of $B$, which is not the same thing as "$A\cap B$ is a dense subset of $B$". May 19, 2019 at 6:48
• @WilliamElliot I meant subset, I'll correct it. May 19, 2019 at 10:50
• @DanielWainfleet You are right, I'll edit. I don't quite understand the second part: If $A \subset B$, $B$ can't be dense in $A$, right? May 19, 2019 at 10:52

It depends on how you define relative denseness for subsets/subspaces of $$X.$$ For example, we might say:

Given two subsets $$A,B$$ of a topological space $$X,$$ we say that $$A$$ is dense in $$B$$ if $$A\subseteq B$$ and the closure of $$A$$ in $$B$$ (relative to the subspace topology on $$B$$) is equal to $$B.$$

In that case, your relation $$R$$ is trivially antisymmetric by double-inclusion. The above is the typical definition, as Daniel Wainfleet points out in the comments.

On the other hand, you're considering that we might instead say:

Given two subsets $$A,B$$ of a topological space $$X,$$ we say that $$A$$ is dense in $$B$$ if the closure of $$A\cap B$$ in $$B$$ (relative to the subspace topology on $$B$$) is equal to $$B.$$

In that case, we won't be able to prove antisymmetry. Consider for example $$X=\Bbb R$$ in the usual topology, $$A=X\setminus\{0\},$$ and $$B=X\setminus\{1\}.$$

Moreover, we won't be able to prove transitivity, either! Consider for example $$A=\Bbb Q,$$ $$B=\Bbb R,$$ and $$C=\Bbb R\setminus\Bbb Q$$ as subsets of $$\Bbb R$$ in the usual topology. Then $$A$$ is dense in $$B$$ and $$B$$ is dense in $$C,$$ but $$A$$ is not dense in $$C.$$

Consequently, it seems that the first definition is the one you want, which makes $$R$$ a partial order, and in fact, $$R$$ is a sub-relation of $$[\subseteq]_{\mathcal{P}(X)}.$$ You are correct that, given two sets $$A$$ and $$B$$ we don't necessarily have either of them as a subset of the other, which only means that $$R$$ won't be a total order (unless $$X$$ is empty).

• +1.... A rather trivial comment on your last paragraph: We also have $(A\subset B\lor B\subset A)$ for all $A,B\in P(X)$ if $X$ has exactly one member. May 19, 2019 at 16:25
• @DanielWainfleet: That's true, though $R$ still wouldn't be a total order in that case, since $\emptyset$ is not dense in $X$. May 19, 2019 at 16:31