# The non-empty intersection of two open discs contains an open disc.

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 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 \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 Since our choice of $$(x,y)$$ was arbitrary, we have $$D_{(a,b)}\subset D_1\cap D_2$$.

$$\blacksquare$$

• 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. Oct 15 '18 at 18:21
• It is correct for me Oct 15 '18 at 18:22
• @Zeekless I am just starting topology, so would you care to elaborate what you mean by prove "topologically". Oct 15 '18 at 18:22
• @Zeekless Thanks. Oct 15 '18 at 18:28
• @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. Oct 15 '18 at 18:36

You write $$d_1+r and here the equality is false. But this can be easily fixed, for example (using $$d_1< r_1$$) $$d_1+r 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 $$d(\gamma,(\beta,\gamma))$$.
• @Thank you so much for response, and i agree your choice of $r$ is much better than mine, aesthetics matter i guess. Oct 15 '18 at 18:38
• 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($a_4,a_4$) Dec 24 '20 at 13:06