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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 $b$?

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  • $\begingroup$ 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 $B$". $\endgroup$ May 19, 2019 at 6:48
  • $\begingroup$ @WilliamElliot I meant subset, I'll correct it. $\endgroup$
    – Ramanujan
    May 19, 2019 at 10:50
  • $\begingroup$ @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? $\endgroup$
    – Ramanujan
    May 19, 2019 at 10:52

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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).

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  • $\begingroup$ +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. $\endgroup$ May 19, 2019 at 16:25
  • $\begingroup$ @DanielWainfleet: That's true, though $R$ still wouldn't be a total order in that case, since $\emptyset$ is not dense in $X$. $\endgroup$ May 19, 2019 at 16:31

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