Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm using the definition (without much topology) that a set $S$ of real numbers is dense in $\mathbb{R}$ if $S \cap (a,b) \neq \varnothing$ for all $a,b \in \mathbb{R}$ and $a<b$.

My questions:

a.) If a set $S$ is dense in $\mathbb{R}$, what can you conclude about the set $A$ that contains $S$ as a subset?

I was thinking that I need to show that as long as $cl(S) \subset A$ that this would be fine. However, topology isn't a requirement for this question and I am unsure how to do it.

b.) If 2 sets $B1$ and $B2$ are both dense in $\mathbb{R}$, what can be said about the set $B1 \cap B2$?

Here I have been thinking about comparing the rationals and irrationals. I understand that they are both dense in $\mathbb{R}$ but one is countable and the other isn't. Here they have no intersection too. But, if I considered $(-\infty, 10) \cap (0,+\infty)$ there would be a non-empty intersection.

So I am not sure what can be said about such a general question....

Thanks for any guidance!

share|improve this question
    
I am not sure what you are saying in a.), can you rephrase it? What do you mean "this would be fine"? As for b.) you have the right idea to look at the irrationals and the rationals. They are both dense, but what about their intersection? Countable and uncoutable actually don't matter much in this context. –  Chris Janjigian Feb 15 '12 at 15:53
    
(a) If $S \subseteq A$, then notice that $\phi \not= S \cap (a,b) \subseteq A \cap (a,b)$. So $A$ is dense as well. As for (b), you're right. Not much can be said. $B_1 \cap B_2$ could be dense or possibily not. Your counter-example (rational and irrationals) does a great job of showing how far from dense an intersection can be. –  Bill Cook Feb 15 '12 at 15:53
1  
Also, (referring to (a)), the condition about the closure is unnecessary. In fact, a dense set $S$ is defined (in any topological space) as a subset such that $\mathrm{cl}(S)$ is the whole space. Thus if you demanded that $\mathrm{cl}(S) \subseteq A$, then $A$ would have to be $\mathbb{R}$ itself! –  Bill Cook Feb 15 '12 at 15:55
1  
For (a), certainly if $S\subseteq A$ and $S$ is dense, then $A$ is, one step from the definition. –  André Nicolas Feb 15 '12 at 15:56
    
Thanks all - I thought there might be more to the question - especially (ii) - but apparently not :) –  nate Feb 15 '12 at 15:58
show 1 more comment

1 Answer

up vote 3 down vote accepted

a) $A$ must be dense, since any open interval intersects $S$, and thus intersects $A$.

b) The intersection of two dense sets could be empty. Consider, as you did, the set of rationals and the set of irrationals. The intersection could possibly be dense by taking a dense set $S$ and setting $S=B_1=B_2$. Nothing can be said in general about the intersection of two dense sets. The intersection could possibly be dense, empty, or even non-empty and non-dense (for instance the rationals and the union of the irrationals with a non-dense set $A$ of rationals).

share|improve this answer
    
That last intersection is quite imaginative! Thanks for the food for thought! –  nate Feb 15 '12 at 16:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.