# Is it true that subset D is dense in space X iff every nonempty open set in X contains one point of D

I have a proposition that I am not sure if it is right:

Let D be a subset of space X. D is dense in X if and only if every nonempty open set U of X contains at least one point of D.

When X is a metric space, then I could verify that this is true in both directions.

When X is a topological space, I could prove the ==> direction; however the <== direction is so difficult that I doubt the proposition is right. However, I cannot yet find a counter example.

This proposition is the generalization of a property of real numbers R:

Q is dense in R so every open interval of R contains a point of Q.

Thank you.

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what is your definition of density? – mookid Mar 5 '14 at 4:26
D is dense in X if closure of D is X, or equivalently, every point in X is the limit of a sequence in D. You think the proposition is exactly the definition of the density? I am taking "introduction to topology". – Thang Mar 5 '14 at 4:32
@Thang: It's worth noting, as Henno Brandsma mentions in the comments below, that sequential density (your second definition) is not equivalent in general to density (your first definition). Those definitions coincide for what are known as "sequential spaces," a class of topological spaces that includes metric spaces. – Cameron Buie Mar 5 '14 at 12:34
Many texts take the definition from the proposition as the definition of density, which is readily equivalent to the first one you gave. – Cameron Buie Mar 5 '14 at 12:37

Hint: Proving by contrapositive would probably be easier. If $D$ is not dense, then its closure is not all of $X$. What can you then say about the complement of its closure? On the other hand, if there is a non-empty open set disjoint from $D,$ then the complement of this open set is closed and contains $D,$ but isn't all of $X,$ so what can we say about the closure of $D$?

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$D$ is dense in $X$ if closure of $D$ is $X$, or equivalently, every point in $X$ is the limit of a sequence in $D$.

• Suppose $D$ is dense. Let $x\in X$, let $r>0$, $x_n\to x$ a $D$ valued sequence. As $$d(x_n,x) \to 0$$there is a $N$ such as $$n\ge N \Rightarrow d(x_n,x)< r$$ In particular, $x_N\in B(x,r)\cap D$.
• Suppose that every open ball of $X$ contains an element of $D$. Let $x\in X$, let $x_n \in D\cap B(x,2^{-n})$. This is a $D$ valued sequence, and $$d(x_n,x) < 2^{-n}$$ so $x_n\to x$.

Now in a general topologic context:

• Suppose $D$ is dense. Let $x\in X$ and $x_n\to x$ a $D$ valued sequence. Let $V$ an open set such as $x\in V$. As $x_n\to x$ there is a $N$ such as $$n\ge N \Rightarrow x_n\in V$$ In particular, $x_N\in V\cap D$.

• There is no hope of a better result. Take for example the discret topology. $O = X - D$ is (as every subset) an open set, containing $x$ when $x\notin D$, such as $O\cap D = \emptyset$.

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Thank you. However, this is the proof for metric space. I am asking if the proposition is still true when X is a general topological space. Then, there is no metric. There a lot of things that are true in a metric space, but it is very difficult to generalize it in a topological space. Anyway, thank you. – Thang Mar 5 '14 at 4:59
I edited for the general case (you were right). – mookid Mar 5 '14 at 5:15
Please do not use sequences for general spaces. They don't suffice for density in general spaces! – Henno Brandsma Mar 5 '14 at 6:51
I don't understand what you mean by "no hope of a better result." The (non-sequential) definitions are equivalent even in general topological spaces. This includes discrete spaces, each of which has precisely one dense subset (itself). – Cameron Buie Mar 5 '14 at 12:40
@HennoBrandsma: I don't understand "[sequences] don't suffice for density in general spaces". I think using sequences in general topological space is fine. The definition of limit of a sequence in a topological space is: sequence x_n approaches L if for every open set containing L, there exists a number N such that the open set contains all x_n for n > N. – Thang Mar 5 '14 at 19:34

As to your definition: $D$ is dense when the closure of $D$ (a.k.a. $\overline{D}$) in $X$, equals $X$. (this does not mean in general that every point of $X$ is a limit of a convergent sequence from $D$; it does in metric spaces, but not in general topological spaces). So as the definition I will use $\overline{D} = X$.

Of course, now the solution will depend on the definition of the closure of a set $A$. Let's use the definition, or characterization, that $x \in \overline{A}$ iff for every open set $O$ of $X$ that contains $x$, $O$ intersects $A$.

Now the statement about the density of $D$ is pretty clear: if $D$ is dense and $O \neq \emptyset$ is open, then pick $p \in O$, then $p \in \overline{D}$, as $\overline{D} = X$, and so every open neighbourhood of $p$ intersects $D$; in particular $O$ intersects $D$, i.e. $O$ contains a point of $D$.

On the other hand, if every non-empty open set $O$ of $X$ contains a point of $D$, let $x \in X$. Let $O$ be any open set that contains $x$. Then $O$ is non-empty (because of $x$) and so contains a point of $D$, i.e. $O \cap D \neq \emptyset$. But this just says that $x \in \overline{D}$ by the definition above. As $x \in X$ was arbitrary, $X \subset \overline{D}$. As $\overline{D} \subset X$ by definition, we have equality.

Note that nothing at all of a metric is used. This is just a general fact for topological spaces.

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Could you give me an example of a topological space X and a dense subset D of X such that there is a point x in X (and not in D) that is not the limit of any sequence in D. – Thang Mar 5 '14 at 21:21
A classical one: $D = \mathbb{N}$, $X = \beta\mathbb{N}$, the Cech-Stone compactification of $\mathbb{N}$. No point of $\beta\mathbb{N} \setminus \mathbb{N}$ is a limit of a sequence from $\mathbb{N}$, even though the latter dense set is dense in $\beta\mathbb{N}$. – Henno Brandsma Mar 5 '14 at 21:44