Two definitions of one set being dense in the other

Question

Assume $\langle X,\mathscr{O}\rangle$ is a topological space. Let $A,S\subseteq X$. Let $\langle S,\mathscr{O}\rangle$ be a subspace of $X$ with $\mathrm{Cl}_S$ being its closure operation.

There are two definitions of $A$ being dense in $S$:

$A$ is dense in $S$ iff $A$ is dense in $\langle S,\mathscr{O}\rangle$ iff $\mathrm{Cl}_S\,(A\cap S)=S$
$A$ is dense is $S$ iff $S\subseteq\mathrm{Cl}\,A$.

By elementary porperties of subspaces we have: $$ \mathrm{Cl}_S\,(A\cap S)=S\longleftrightarrow\mathrm{Cl}\,(A\cap S)\cap S=S\longleftrightarrow S\subseteq \mathrm{Cl}\,(A\cap S) $$ In answers to both these questions: the first and the second, the following equivalence is referred to: $$S\subseteq \mathrm{Cl}\,(A\cap S)\longleftrightarrow S\subseteq\mathrm{Cl}\,A$$ in order to establish equivalence between 1 and 2. Yet simple counterexample shows that the implication from right to left is false: take the set of reals with the standard order topology and put $S=\{1\}$ and $A=(0,1)$.

EDIT: The sets may even have a non-empty intersection: put $S=\{\frac{1}{2},1\}$ and the rest as in the example above.

My question is: what am I missing from the picture to see that 1 and 2 are indeed equivalent?

OK, but neither of the definitions requires it. That is the point. — Rafał Gruszczyński, Aug 17 '15 at 14:47
Right, it's just how the idea might arise. People apparently tend to think of subsets of $S$ when the words "dense in $S$" occur. — Daniel Fischer, Aug 17 '15 at 14:53
Possibly nicer counterexample, $\Bbb{R\setminus Q}\subseteq\operatorname{Cl}(\Bbb Q)$ when considering the real numbers with the standard topology. — Asaf Karagila, Aug 18 '15 at 07:56

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

I think your counterexample is correct. Note that the answer to your second link has a comment that gives a similar counterexample. I think the equivalence is wrong.

There is a bit of ambiguity in the definition of denseness. In most cases (see the first link as well as wikipedia), $A$ is assumed to be a subset of $S$ and I suspect that the source of your second definition assumes this condition (in which case both definitions are clearly equivalent).

I haven't encountered cases where this terminology is used when $A$ is not necessarily a subset of $S$, but if you do, just use the definition given in whatever context you're in. Your first definition is a reasonable generalization: if $A \not\subset S$, then $A$ is defined to be dense in $S$ if $A \cap S$ is dense in $S$ (in the sense of the second definition).

score 1 · Answer 2 · answered Aug 17 '15 at 14:48

1

1 and 2 are not equivalent. Your counterexample S={1} and A =(0,1) shows that: it satisfies 2 but not 1.

answered Aug 17 '15 at 14:48

Nate Eldredge

101,664

Well, indeed :) But "missing from the picture" I thought about possible additional assumptions (except from the obvious one that $A$ is a subset of $S$) in which the equivalence holds. – Rafał Gruszczyński Aug 17 '15 at 14:59

Two definitions of one set being dense in the other

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