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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

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MathMan
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1 Answers1

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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.

layman
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  • 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 \$.

    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?

     – MathMan Sep 07 '14 at 14:53
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    @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. – layman Sep 07 '14 at 15:03
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    @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? – layman Sep 07 '14 at 15:04
  • Uhm .. no actually.. I was trying to find out if $D_1^c$ can have any non void open set? – MathMan Sep 07 '14 at 15:07
  • I think it can't because $D_1$ is dense, hence, if $x \in D_1^c$, then every nbd of $x$ contains an element of $D_1$. Is this possible for an element in $D_1^c?$ – MathMan Sep 07 '14 at 15:08
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    @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. – layman Sep 07 '14 at 15:13
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    @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? – layman Sep 07 '14 at 15:20
  • yeah, I was trying to move forward from there .. – MathMan Sep 07 '14 at 15:20
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    @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. – layman Sep 07 '14 at 15:23
  • Np :-) .. Thank you for the answer :-) – MathMan Sep 07 '14 at 15:26
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    @VHP By the way, I just wanted to make one clarification. In general, a set is nowhere dense if its closure has empty interior. In other words, if its closure has no non-void open sets. The reason we didn't talk about the closure of $D_{1}^{c}$ when saying it is nowhere dense is that the closure of $D_{1}^{c}$ is itself, because it is a closed set. But in general, to prove $A$ is nowhere dense, you have to show that the closure of $A$, $\overline{A}$, has no non-void open sets. If $A$ is already closed, then we can just show $A$ has no non-void open sets for $A$ to be nowhr dnse. – layman Sep 07 '14 at 15:28
  • Oh okay .. I get that – MathMan Sep 07 '14 at 15:48