if A is a countable dense subset of $\mathbb{R}$ then there exists $B\subset A$ dense and countable in $\mathbb{R}$

Question

I found this result in so many books and I tried to demonstrate it but I'm not convinced with my proof,

if A is a countable dense subset of $\mathbb{R}$ then there exists $B\subset A$ dense and countable in $\mathbb{R}$.

score 6 · Answer 1 · answered Jun 13 '14 at 12:10

6

HINT: What happens if you remove a single point from $A$? Or any finite number of points from $A$?

answered Jun 13 '14 at 12:10

Asaf Karagila

I guess not a big change since A is dense – Houda Jun 13 '14 at 12:12
But can you prove it? – Asaf Karagila Jun 13 '14 at 12:15
I don't have a rigorous demonstration but I thought to take $B=A\cap \mathbb{Q}$ !! – Houda Jun 13 '14 at 12:18
And what if $A\subseteq\Bbb Q$? Or what if $A\cap\Bbb Q=\varnothing$? – Asaf Karagila Jun 13 '14 at 12:19
oh! you're right, but then we can choose $B=A\cap \mathbb{R}\backslash \mathbb{Q}$ can we ? – Houda Jun 13 '14 at 12:24
But what if $A=\Bbb Q$? Then $B=\varnothing$. In general, there are countable dense sets which are far strangers than you can imagine. It's best to try and come up with a general method, and not ad hoc intersections. – Asaf Karagila Jun 13 '14 at 12:25
okay, that's what it seems to me. – Houda Jun 13 '14 at 12:27
For that matter, take $B=A$. Usually "$\subset$" does not mean "$\subsetneq$", but some have different conventions in that regard. – MPW Jun 13 '14 at 12:45
@MPW: If you allow equality then this entire question doesn't make any sense. – Asaf Karagila Jun 13 '14 at 12:50
Of course it does. In fact, it's included in your answer, isn't it? Wouldn't you say that zero points is a finite number of points? – MPW Jun 13 '14 at 12:53
@MPW: Wow, have you ever heard the term "context" before? – Asaf Karagila Jun 13 '14 at 13:00

score 4 · Answer 2 · answered Jun 13 '14 at 12:17

4

With some care you can find an infinite and co-infinite subset of $A$ which is dense in $\mathbb{R}$.

Enumerate the open intervals with rational endpoints as $\{ I_n \}_{n \in \mathbb{N}}$.
Inductively pick for each $n$ distinct $a_n, b_n \in (A \cap I_n) \setminus \{ a_i , b_i : i < n \}$. (For this note that $I_n \cap A$ is infinite for all $n$.)
Consider $B = \{ b_n : n \in \mathbb{N} \}$.

answered Jun 13 '14 at 12:17

user642796

thanks for this procedure, but I was wonder what if we found that for a $n$ we found that $A \cap I_n$ – Houda Jun 13 '14 at 12:36
@houda, your above comment seems to be incomplete. If you meant to ask "...we found that $A \cap I_n \neq \varnothing$, note that this is impossible, since a dense set has nonempty intersection with every nonempty open set. – user642796 Jun 13 '14 at 14:40
oh! I didn't pay attention to that, coming back to your method I honestly can't figure out how to prove that B is dense. Can you please give me an indication ? – Houda Jun 13 '14 at 14:58
@houda: Can you show that $B$ has nonempty intersection with every nonempty open subset of $\mathbb{R}$? – user642796 Jun 13 '14 at 17:39

2 Answers2