2
$\begingroup$

I am going through an introductory topology chapter and finding some difficulty in the following question. I had a chance to go through Daniel's and Brian's answers on this page. However, I am still a bit not convinced.

If $D_1$ is an open dense set in $\mathbb R^p$ and $D_2$ is any dense set in $\mathbb R^p$, then,Show that $D_1 \cap D_2$ are also dense in $\mathbb R^p$ .

Attempt: Given that $D_1$ is an open dense set in $\mathbb R^p$ which means if $x \in D_1^c$, then every neighborhood of $x$ contains at least one element of $D_1$.

But, this fact is contradictory because elements of $D_1$ are not supposed to be in every neighborhood of elements in $D_1^c$.

$(a)~$Hence, can we infer that $D_1^c$ must not have any non void open set else any open ball in $D_1^c$ will also contain elements of $D_1$ which is a contradiction?

Suppose that $D_1 \cap D_2$ is not dense in $\mathbb R^p$. Then, for some $y \in \mathbb R^p$, there is a neighborhood of $y = B(y,r)$ which does not contain any element of $D_1 \cap D_2$.

This is certainly possible because $B(y,r)$ can contain elements of $D_1/ (D_1 \cap D_2)$ and $D_2/ (D_1 \cap D_2)$.

How do I move forward?

Thank you for your help

$\endgroup$
1
$\begingroup$

To show $D_{1} \cap D_{2}$ is dense in $\mathbb{R}^{p}$, you can show it intersects every open subset of $\mathbb{R}^{p}$. That is one characterization of density.

So, let $O \subseteq \mathbb{R}^{p}$ be an arbitrary open set. Since $D_{1}$ is dense, $D_{1} \cap O \neq \emptyset$. Furthermore, $D_{1} \cap O$ is open, since both sets are open. So we know by the nonempty intersection $\exists x \in D_{1} \cap O$. Since the intersection is open, $\exists \epsilon_{x} > 0$ such that $B(x,\epsilon_{x}) \subseteq D_{1} \cap O$.

But $B(x, \epsilon_{x})$ is an open subset of $\mathbb{R}^{p}$, and thus since $D_{2}$ is dense, $D_{2} \cap B(x, \epsilon_{x}) \neq \emptyset$. But since this ball is contained in $O \cap D_{1}$, this implies $D_{2}$ intersects $O \cap D_{1}$. Written down, this says $D_{2} \cap (D_{1} \cap O) \neq \emptyset$.

Since set intersection is associative, this means $(D_{1} \cap D_{2}) \cap O \neq \emptyset$ for any arbitrary open set $O \subseteq \mathbb{R}^{p}$, which implies $D_{1}\cap D_{2}$ is dense by our characterization of density.

$\endgroup$
  • 1
    $\begingroup$ @VHP Since $D_{1}$ is open and dense, that means $D_{1}^{c}$ is nowhere dense (i.e., its closure has empty interior). To prove this, first note that $\overline{D_{1}^{c}} = D_{1}^{c}$ since $D_{1}^{c}$ is closed. Now, suppose that $\exists x \in (D_{1}^{c})^{o}$ (the interior of $D_{1}^{c}$). Since this interior is open, $\exists \epsilon > 0$ such that $B(x, \epsilon) \subseteq (D_{1}^{c})^{o}$. But since $B(x, \epsilon)$ is open, it intersects $D_{1}$, and so it cannot be contained in $(D_{1}^{c})^{o}$, which implies $(D_{1}^{c})^{o}$ is empty, and thus $D_{1}^{c}$ is nowhere dense. $\endgroup$ – layman Sep 7 '14 at 15:03
  • 1
    $\begingroup$ @VHP Is that what you were trying to say? If so, can you explain what the contradiction was you were trying to find? Were you assuming $D_{1} \cap D_{2}$ is not dense in the beginning in order to arrive at the contradiction you think you found? $\endgroup$ – layman Sep 7 '14 at 15:04
  • 1
    $\begingroup$ @VHP Yes, you are trying figure out whether or not $D_{1}^{c}$ is nowhere dense. If it is nowhere dense, then it has no non-void open set. If it is not nowhere dense, it has a non-void open set. But we know now that it is nowhere dense based on my proof above. And you are completely correct that the proof relies on the fact tht $D_{1}$ is dense. You are right. $D_{1}^{c}$ contains no non-void set (i.e., $D_{1}^{c}$ is nowhere dense), precisely because $D_{1}$ is dense. $\endgroup$ – layman Sep 7 '14 at 15:13
  • 1
    $\begingroup$ @VHP Now that you have established $D_{1}^{c}$ is nowhere dense, are you trying to use this fact to give an alternate proof to the one I posted of $D_{1} \cap D_{2}$ being dense? $\endgroup$ – layman Sep 7 '14 at 15:20
  • 1
    $\begingroup$ @VHP I'm not sure how to use the fact that $D_{1}^{c}$ is nowhere dense to prove $D_{1} \cap D_{2}$ is dense in $\mathbb{R}^{p}$, but in any case, we don't need to do that, because we have a pretty straightforward proof already. However, it's nice that you unlocked the feature that $D_{1}^{c}$ is nowhere dense. I didn't realize this immediately when proving the original statement. And it might be useful in an alternative proof to the statement. I'm just not sure. $\endgroup$ – layman Sep 7 '14 at 15:23

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.