I am not sure if what I am looking for even makes sense (or) exists. Anyway I would be happy if someone can clear my confusion.

The set of real numbers $\mathbb{R}$ is obtained as completion of $\mathbb{Q}$. However, $\mathbb{Q}$ is not the only set which is dense in $\mathbb{R}$. $\mathbb{Q} \backslash \mathbb{Z}$ is also a dense subset of $\mathbb{R}$. I am wondering if it makes sense to talk of the "smallest" dense subset of $\mathbb{R}$. To phrase what I am looking for precisely and what I mean by smallest, I am looking for a dense set $A$ of $\mathbb{R}$ such that if $B$ is a proper subset of $A$ then $B$ is not dense in $\mathbb{R}$. I am able to "see" that the set $A$, I am looking for doesn't exist since any open interval consists of infinite rationals. But I am unable to precisely argue out to myself and convince why $A$ doesn't exist. Could some one throw more light on this?

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    $\begingroup$ There's no smallest dense subset of $\mathbb{R}$, but $\mathbb{Q}$ is the smallest subfield of $\mathbb{R}$, i.e. the smallest subset containing 0 and 1 that is closed under adding, subtracting, multiplying, and dividing. $\endgroup$ Aug 31, 2011 at 12:33
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    $\begingroup$ You are looking for minimal, not a smallest or least dense subset. $\endgroup$
    – Rasmus
    Aug 31, 2011 at 12:37
  • $\begingroup$ @Rasmus: Smallest corresponds to minimum. A priori it does not seem that there should not be a minimum. As Amitesh and Pete answered - there is no minimal dense subset, so in particular there is no minimum. $\endgroup$
    – Asaf Karagila
    Aug 31, 2011 at 13:30
  • $\begingroup$ small or big can not say uniquely as we know Q is countable but C cantor set not so Q is small and c is big in the sense of cardinality but topologically Q is dense but C is nowhere dense so Q is big and C is small But measure Q and C have measure Zero so both are small $\endgroup$
    – user83198
    Jun 20, 2013 at 6:49
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2 Answers 2


If $A$ is a dense subset of $\mathbb{R}$, then $A\neq \emptyset$. In particular, there exists an element $a\in A$. If $B=A\setminus \{a\}$, then is $B$ dense in $\mathbb{R}$? (Hint: if there is an open subset $U$ of $\mathbb{R}$ such that $U\cap A=\{a\}$, then $A$ is not dense in $\mathbb{R}$.)

Exercise 1: Prove or give a counterexample: a finite intersection of dense subsets of $\mathbb{R}$ is dense in $\mathbb{R}$.

Exercise 2: Prove or give a counterexample: if $\{A_i\}_{i\in I}$ ($I$ is an index set) is an infinite collection of dense subsets of $\mathbb{R}$ such that the intersection of any finite number of $A_i$'s is again dense in $\mathbb{R}$, then the intersection of all the $A_i$'s dense in $\mathbb{R}$. (If you wish to view a hint, hover your cursor over the grey region directly below:

(Hint: if $x\in \mathbb{Q}$, let $A_x=\mathbb{Q}\setminus \{x\}$; prove that the intersection of any finite number of $A_x$'s is dense in $\mathbb{R}$ and determine the intersection $\bigcap_{x\in\mathbb{Q}} A_x$.)

Exercise 3: If $A$ is dense in $B$ and if $B$ is dense in $\mathbb{R}$, $A\subseteq B\subseteq \mathbb{R}$, then is $A$ dense in $\mathbb{R}$?

Exercise 4 (if you are familiar with measure theory): Let $\epsilon>0$. Prove that there exists an open dense subset $U$ of $\mathbb{R}$ such that the Lebesgue measure of $U$ is at most $\epsilon$. (If you wish to view a hint, hover your cursor over the grey region directly below:

(Hint: enumerate $\mathbb{Q}$ as $q_1,q_2,\dots$. If $n\in\mathbb{N}$, let $I_n=(q_n-\frac{\epsilon}{2^{n+1}},q_n+\frac{\epsilon}{2^{n+1}})$; prove that the union $\bigcup_{n\in\mathbb{N}} I_n$ is an open dense subset of $\mathbb{R}$ and determine the Lebesgue measure of $\bigcup_{n\in\mathbb{N}} I_n$.)

Exercise 5 (if you are familiar with path-connected topological spaces): If $U$ is an open path-connected subspace of $\mathbb{R}$ and if $U$ is dense in $\mathbb{R}$, then prove that $U=\mathbb{R}$.

I hope this helps!

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    $\begingroup$ In exercise 2 you can even further the requirement of $A_i$ to form a chain in $\supseteq$ relation. $\endgroup$
    – Asaf Karagila
    Aug 31, 2011 at 9:59
  • $\begingroup$ @Theo Thanks! I have fixed this. $\endgroup$ Aug 31, 2011 at 10:03
  • $\begingroup$ @Asaf Thanks! Indeed, this is not hard to construct using the counterexample given in the hint. $\endgroup$ Aug 31, 2011 at 10:04
  • $\begingroup$ For the first one, a counter example is $\mathbb{Q}$ and $\mathbb{R} \backslash \mathbb{Q}$. For the second one again, it is a counter example. Third one is true. Fourth one yes I have done it before proving the Lebesgue measure of rationals is $0$. Fifth one I still need to try. So all these mean the set I asked for does not exist? $\endgroup$
    – Adhvaitha
    Aug 31, 2011 at 10:06
  • $\begingroup$ @Adhvaitha Yes, there is no smallest dense subset of $\mathbb{R}$. However, the fact that the Lebesgue measure of the rationals is $0$, although interesting, is not relevant to Exercise 4 ... Also, can you characterize all open path-connected (not necessarily dense) subsets of $\mathbb{R}$? $\endgroup$ Aug 31, 2011 at 10:08

Let $X$ be a topological space which is separated and without isolated points: that is, for all $x \in X$, the singleton set $\{x\}$ is closed and not open. Let $Y$ be a dense subset of $X$.

CLAIM: For every nonempty open subset $U$ of $X$, $U \cap Y$ is infinite.

Proof: First observe that $U$ is infinite: if not, removal of all but one of its points would make a singleton open set. Then similarly, if $U \cap Y$ were finite, removal of all of its points would leave a nonempty open subset of $X$ which is disjoint from $Y$, contradicting denseness of $Y$.

It follows immediately from the claim that removing any finite number of points from $Y$ leaves us with a subset which still meets every nonempty open subset of $X$ so is still dense. In particular, there is no minimal dense subset of $X$.

In particular the argument applies to any metric space without isolated points, like $\mathbb{R}$. It is a good exercise to check that both of the hypotheses imposed on $X$ are necessary. Especially, in a non-separated space one may well have singleton dense subsets: these are called generic points and are ubiquitous (and useful) in algebraic geometry.

  • $\begingroup$ Do you know if the generic points which you describe are in any way related to generic reals? I know, for example, that if we add a single Cohen real $r$, then in $V[r]$ the measure of $\mathbb R^V$ is zero. Quite intriguing, I'd think, and the reason is some sort of density of the singleton $\{r\}$ (not exactly density but it follows from the genericity pretty quick). $\endgroup$
    – Asaf Karagila
    Sep 13, 2011 at 22:09
  • $\begingroup$ @Asaf: (Sorry for the delayed response.) To be absolutely sure I would need to know something more about generic reals than I do....but, no, it seems quite unlikely that there is a direct connection here. The word "generic" gets a lot of different uses across mathematics. $\endgroup$ Feb 11, 2012 at 2:55
  • $\begingroup$ Incidentally just a few weeks ago I was sitting in a class about descriptive set theory, and we proved something via forcing. I asked Magidor (as he usually knows answers to historical questions like that) about the origin of the term generic, he said that it was coined by Solovay and the meaning was taken from AG. I don't have references for that yet, but there seem to have a vague connection between the different "generic" points in mathematics. $\endgroup$
    – Asaf Karagila
    Feb 11, 2012 at 11:07
  • $\begingroup$ Regarding @Pete's nice answer above, it seems that being $T_2$ (separated) is not a necessary condition; only $T_1$ is necessary -- the argument in his proof only used the property of singletons being closed. Did I miss anything? $\endgroup$
    – ALife
    Oct 11, 2019 at 14:18
  • $\begingroup$ @ALife You are right, the argument uses only $T_1$ (and that the space has no isolated points). $\endgroup$ Oct 11, 2019 at 22:18

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