0
$\begingroup$

I am trying to figure out that the set of rationals are open or not. My problem is that I do not see how would the density of irrationals help me.

so the density of irrationals mean that an interval in the real numbers contain an irrational number, does this mean that an interval in the rationals will contain an irrational number?

I appreciate any help!

$\endgroup$
  • 2
    $\begingroup$ This has no sense, or well, it's trivially false: "an interval in the rationals" (or, what I think you're meaning) is a set of the form (for example) $\{x\in \mathbb{Q}: a\leq x \leq b\}$ with $a,b$ rationals. This is a subset of $\mathbb{Q}$ so it does not contain any irrational number. $\endgroup$ – Daniel Mar 21 '15 at 19:55
1
$\begingroup$

For $\mathbb{Q}$ to be open, every point would need to be an interior point. Since any open interval $(a,b)$ with $a,b\in \mathbb{Q}$ contains an irrational, no point is interior. So $\mathbb{Q}$ is not open.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ when you say that " any open interval (a,b) with a,b∈Q contains an irrational" would this mean that the irrationals are dense in the rationals? Or am I far off? $\endgroup$ – Hanna7777 Mar 21 '15 at 20:16
  • $\begingroup$ Rationals are not a subset of irrationals, therefore $\mathbb Q$ couldn't be dense in irrationals. $\endgroup$ – iiivooo Mar 21 '15 at 20:20
  • 1
    $\begingroup$ @Hanna7777 You are not far off. Recall that a set $M$ is dense in a set $X$ if the closure of $M$ is $X$ (i.e. $cl(M)=X$, $\bar{M}=X$, or whatever notation you use). The argument here is that the irrationals are dense in $\mathbb{R}$. It should be proven that if a set $M$ is dense in $\mathbb{R}$ then the intersection of any interval with $\mathbb{R}$ contains a point of $M$. $\endgroup$ – Eoin Mar 21 '15 at 20:21
  • 1
    $\begingroup$ The definition of an open set is that every point is interior, (or equivalently, a set $U$ is open if $\forall x\in U$ there is an open neighborhood of $x$, $N_x$, contained in $U$). In $\mathbb{R}$ open sets are equivalent to the union of open disks with some radius. So suppose $x\in \mathbb{Q}$ and consider an open neighborhood (an open set of $\mathbb{R}$) of $x$. Every open neighborhood of $x$ is either a disk or the union of disks, so it is sufficient to consider the former. $\endgroup$ – Eoin Mar 21 '15 at 20:25
  • 1
    $\begingroup$ But an open disk of $\mathbb{R}$ is just an interval which, by the first comment, must contain an irrational point. Thus, any open neighborhood of a rational point contains an irrational point. This shows that no open neighborhood of $x\in \mathbb{Q}$ can be contained in $\mathbb{Q}$ or, equivalently, no point of $\mathbb{Q}$ is interior. Therefore, $\Bbb{Q}$ is not open. $\endgroup$ – Eoin Mar 21 '15 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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