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Is the following argument correct?

Let $D_1$ and $D_2$ be any open discs in $\mathbf{R}^2$ with $D_1\cap D_2\neq\varnothing$. If $(a,b)$ is any point in $D_1\cap D_2$, show that ther exists an open disc $D_{(a,b)}$ with center $(a,b)$ such that $D_{(a,b)}\subset D_1\cap D_2$.

Proof. Let $D_1,D_2$ denote two arbitrary discs in $\mathbf{R}^2$ having $\alpha = (a_1,b_1)$ and $\beta = (a_2,b_2)$ as there centers and radii $r_1$ and $r_2$ respectively, that is \begin{align*} D_1 = \{(x,y)\in\mathbf{R}^2:(x-a_1)^2+(y-b_1)^2<r_1^2\}\\ D_2 = \{(x,y)\in\mathbf{R}^2:(x-a_2)^2+(y-b_2)^2<r_2^2\} \end{align*} Let $\gamma = (a,b)\in D_1\cap D_2$. We define the disc $D_{(a,b)}$ similar to $D_1,D_2$ but having center $(a,b)$ and radius $r = \min\{\frac{r_1-d_1}{8},\frac{r_2-d_2}{8}\}$, where $d_1$ and $d_2$ are defined as follows \begin{align*} d(\gamma,(\alpha,\gamma)) = d_1 = \sqrt{(a-a_1)^2+(b-b_1)^2}\\ d(\gamma,(\beta,\gamma)) = d_2 = \sqrt{(a-a_2)^2+(b-b_2)^2} \end{align*} Now let $(x,y)$ be an arbitrary point inside the disc $D_{(a,b)}$, appealing to the triangle inequality then yields \begin{align*} \sqrt{(x-a_1)^2+(y-b_1)^2} &=d(\alpha,(x,y))\leq d(\alpha,\gamma)+d(\gamma,(x,y))\\ &=d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{8}<r_1 \end{align*} \begin{align*} \sqrt{(x-a_2)^2+(y-b_2)^2} &= d(\beta,(x,y))\leq d(\beta,\gamma)+d(\gamma,(x,y))\\ &=d_2+r<d_2+\frac{r_2-d_2}{8}=\frac{r_2}{8}<r_2 \end{align*} Since our choice of $(x,y)$ was arbitrary, we have $D_{(a,b)}\subset D_1\cap D_2$.

$\blacksquare$

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    $\begingroup$ The task is much easier if you prove it topologically. An intersection of two open sets is open. Since open balls form a base in $\mathbb{R}^2$, there will be an open ball inside that intersection. Although, you have to know the fact that open balls form a base in advance. $\endgroup$
    – Zeekless
    Oct 15 '18 at 18:21
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    $\begingroup$ It is correct for me $\endgroup$ Oct 15 '18 at 18:22
  • $\begingroup$ @Zeekless I am just starting topology, so would you care to elaborate what you mean by prove "topologically". $\endgroup$ Oct 15 '18 at 18:22
  • $\begingroup$ @Zeekless Thanks. $\endgroup$ Oct 15 '18 at 18:28
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    $\begingroup$ @Zeekless I think if you want to prove that the open balls form a base then you need to prove the statement from Atif Farooq (of course without using the fact that open balls form a base). So I dont think there is a way around doing the calculations that he did. $\endgroup$
    – supinf
    Oct 15 '18 at 18:36
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The proof is correct and mostly good, there are only minor mistakes and typos.

You write $$ d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{8}<r_1 $$ and here the equality is false. But this can be easily fixed, for example (using $d_1< r_1$) $$ d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{4}<r_1 . $$ The same issue exist with $d_2,r_2$ below.

In my opinion it would be nicer if you choose $$ r= \min\{r_1-d_1,r_2-d_2\}, $$ then your proof would still work, but your choice is not wrong!

Small typos: You probably mean to write $d(\gamma,\alpha)$ or $d(\gamma,(a_1,b_1))$ instead of $d(\gamma,(\alpha,\gamma))$ and $d(\gamma,\beta)$ or $d(\gamma,(a_2,b_2))$ instead of $d(\gamma,(\beta,\gamma))$.

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  • $\begingroup$ @Thank you so much for response, and i agree your choice of $r$ is much better than mine, aesthetics matter i guess. $\endgroup$ Oct 15 '18 at 18:38
  • $\begingroup$ Is it possible to prove it if the intersection is quadrilateral ambrsoft.com/TrigoCalc/Circles3/Intersection.htm (see case 12). Then for the other triangle,since Atif proved it for case 11 you would have to define d($a_3,b_3$)and d($a_4,a_4$) $\endgroup$
    – Eudoxus
    Dec 24 '20 at 13:06

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