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Here is a quote from Derbyshire's "Unknown Quantity"

The rational numbers are "dense." This means that between any two of them you can always find another one.

This being a pop. math book, this is not meant as a formal definition of dense sets but it got me wondering whether the dense subsets of $\mathbb{R}$ could be defined in this way. (It is clearly insufficient by itself since the empty set would be dense!)

So I wonder, is there a nice way to fill in the following statement:

$S \subset \mathbb{R}$ is dense iff $(\forall x,y \in S, x < y)(\exists z \in S)(x < z < y)$ and ...

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I should say: by 'nice' I mean that, for instance, adding the standard definition for density is not what I am looking for (though it would obviously work...) I think $S$ may need to be unbounded and contain its limits points –  nolion Jan 24 '11 at 8:02
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That isn't intended as a definition of "dense as a subset of $\mathbb{R}$". Derbyshire hasn't even mentioned $\mathbb{R}$ at the point he says this. He's just defining "dense" in the sense of dense orders. –  Chris Eagle Jan 24 '11 at 8:56
    
@Chris Thanks for clarifying that! Without looking at the book, I was slightly shocked by this "definition", even when regarded as a heuristic. –  Alex B. Jan 24 '11 at 9:25
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@nolion: note that an order is dense if and only if, as a topological space equipped with the order topology, it is dense in itself in the usual sense. –  Qiaochu Yuan Jan 24 '11 at 11:00
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@Qiaochu: I think it's just a matter of notation; you perhaps meant what is sometimes written as "dense-in-itself", i.e., no isolated points? –  Arturo Magidin Jan 24 '11 at 18:18

4 Answers 4

up vote 5 down vote accepted

It seems this line of definition leaves open the possibility of "holes" in $S$, even if one adds the condition that $S$ is unbounded in both directions. Consider for example any union of open intervals. Or the intersection of the set of rationals with such a union.

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There is another definition, which is more general. A subset $F \subset E$ is dense in $E$ if the set of all limits of sequences of elements of $F$ is $E$.

Where does this come from?

A way to answer this question is from the closure of $F$. The closure of $F$ can be thought of as what I called the set of all limits of sequences of elements of $F$. It's generally defined as the smallest closed set containing $F$. Concretely, it's what can be reached by the elements of $F$ without really going "outside" of $F$. What really matters here is that you'll be able to reach the boundaries (as limits of some sequences).

Just picture the real interval $]0,1[$ (often written $(0,1)$ in English litterature). Let's consider the sequence $(a_n)_{n \geq 2}$ defined by $a_n = 1 - \frac{1}{n}$. For any $n \geq 2$, $a_n \in ]0,1[$. However, its limit is $1$, which does not belong to $]0,1[$. Here, 1 is an example of what I meant by "can be reached from elements of $F$" earlier. You can guess that the closure of $]0,1[$ is $[0,1]$.

So here we can say for instance that $]0,1[$ is dense in $[0,1]$. This definition holds in much more general contexts than $\mathbb{R}$ but that's really a way to look at what that is all about.

Hope this helps.

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A better definition of denseness would be that "between any two points of the complement there is a point of the set". You will have no difficulty writing this down formally with quantifiers: $$ \forall x<y\in S^C\;\exists z\in S... $$

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As many have commented, the notion of $S$ being dense by itself (i.e., as an ordered set) is different from the notion of $S$ being a dense subset of $\mathbb{R}$.

  • For the former notion, the definition should refer only to the set itself: $(S, <)$ is dense if for any $x,y \in S$ such that $x<y$, there exists a $z \in S$ such that $x < z < y$.
  • For the latter notion, the definition needs to refer to the ambient set: $S \subset \mathbb{R}$ is dense if for any $x, y \in \mathbb{R}$ such that $x<y$, there exists a $z \in S$ such that $x < z < y$.

The latter notion is closely related to the former in the case of real numbers because the open intervals $(x, y)$ form a basis for the usual topology on $\mathbb{R}$.

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