2
$\begingroup$

The question at hand is to prove the following inclusion: $\delta (A\cap B)\cap (\bar A \cup (\overline{\delta B})) \subseteq \delta A$ for $A,B \subset \mathbb R^n$

NOTE: For my notation, $\bar X$ is the complement of $X$ and $\delta X$ is the boundary of $X$.

I can draw a picture of this and make sense of it, but I have issues showing this is true mathematically. One of the theorems we know is that $\delta (A \cap B) \subseteq \delta A \cup \delta B$ if that helps.

Attempt at a solution:

$$\delta (A\cap B)\cap (\bar A \cup (\overline{\delta B})) \subseteq \delta A$$

$$= (\delta (A \cap B) \, \cap \, \bar A) \ \cup \ ( \ \delta(A \cap B) \cap (\overline{\delta B}))$$

$$\subseteq (\delta (A \cap B) \, \cap \, \bar A) \ \cup \ ( \ (\delta A \cup \delta B) \cap (\overline{\delta B}))$$

$$= (\delta (A \cap B) \, \cap \, \bar A) \ \cup \ ( \ (\delta A \cap \overline{\delta B}) \cup (\delta B \cap \overline{\delta B}))$$

$$= (\delta (A \cap B) \, \cap \, \bar A) \ \cup \ (\delta A \cap \overline{\delta B}) $$

But at this point, I feel lost. Perhaps I didn't approach this the right way from the beginning, or I just don't know how to continue. Any help would be greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ An additional condition is that A and B are subsets of $\mathbb R^n$ $\endgroup$ – BSplitter Feb 8 '17 at 6:37
  • 1
    $\begingroup$ Hm. Let $A=\mathbb R$, $B=\mathbb Q$. Then $\delta(A\cap B)\cap(\overline A \cup\overline{\delta B})=\mathbb R$, but $\delta A=\varnothing$. $\endgroup$ – Sergei Golovan Feb 8 '17 at 6:38
  • 1
    $\begingroup$ @SergeiGolovan: The bars here denote complement, not closure. I made the same mistake. :) $\endgroup$ – Eric Wofsey Feb 8 '17 at 6:39
  • $\begingroup$ Yeah I apologize for that. My professor has been using this notation, so I just followed suit. I'm getting the impression that this isn't the norm. $\endgroup$ – BSplitter Feb 8 '17 at 6:41
  • $\begingroup$ Okay, it seems to be correct then. $\endgroup$ – Sergei Golovan Feb 8 '17 at 6:43
2
$\begingroup$

You're actually almost done. You've shown that $$\delta (A\cap B)\cap (\bar A \cup \overline{\delta B})\subseteq (\delta (A \cap B) \cap \bar A) \cup (\delta A \cap \overline{\delta B})$$ (though your notation is a bit screwy; you seem to be using $\iff$ in place of equals signs). To finish, you just have to prove the right-hand side is contained in $\delta A$. That is, you need to prove that $\delta (A \cap B) \cap \bar A$ and $\delta A \cap \overline{\delta B}$ are each contained in $\delta A$. These are both not too hard. If you're stuck on the first one, you may find it useful to remember that $\delta(A)=\operatorname{cl}(A)\cap\operatorname{cl}(\bar{A})$, so you just have to prove that if $x\in\delta (A \cap B) \cap \bar A$ then both $x\in\operatorname{cl}(A)$ and $x\in \operatorname{cl}(\bar{A})$.

$\endgroup$
  • $\begingroup$ Thank you! I have fixed the equals sign notation! $\endgroup$ – BSplitter Feb 8 '17 at 7:42
3
$\begingroup$

I'd take some $x\notin\delta A$. It belongs either to the interior or to the exterior of $A$.

  1. Let $x\in\mathop{\mathrm{ext}} A$. Then $x\notin\delta(A\cap B)$ (there's a neighborhood of $x$ without any point from $A$, hence without any point from $A\cap B$.

  2. Let $x\in\mathop{\mathrm{int}}A$. If $x\notin\delta(A\cap B)$ then it's okay. If $x\in\delta(A\cap B)$ then $x\in\delta B$. Therefore, $x\notin\overline A$ and $x\notin\overline{\delta B}$, which finishes the proof.

$\endgroup$

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.