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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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    Possible duplicate of [if A is a countable dense subset of $\mathbb{R}$ then there exists $B\subset A$ dense and countable in $\mathbb{R}$](https://math.stackexchange.com/questions/832810/if-a-is-a-countable-dense-subset-of-mathbbr-then-there-exists-b-subset-a)2018-09-14

2 Answers 2

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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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    Open path-connected subsets of $\mathbb{R}$ are the open intervals.2011-08-31
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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.

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    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.2012-02-11