Are open discs centered at some point $(x,0)\,\,,x\in\mathbb{R}$ form a basis for topology on $\mathbb{R}^{2}$? 
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*I actually have a doubt regarding this question which I found in one of my assignments. My thinking is that if the center $(a,0)$ is fixed then the collection of open discs should form a basis for topology as we can express the entire $\mathbb{R}^{2}$ as the union of all sets from the basis.
As well as the if we have a finite intersection of sets then it would be just the disc with the smallest radius which is itself a member of the basis. So by my logic this does form a basis if I am not completely wrong right?


*But if the question means that the center is not fixed . Then we might have a problem because there could exist a point in the intersection of two circles which cannot be contained inside a circle which is centered on the x axis in  $\mathbb{R}^{2}$ .
The answer given was as "no" .
Then was any of my reasoning correct?. Is the collection of discs with fixed center not a basis for topology?.
 A: Fix $a \in \mathbb{R}$. Let $\mathscr{B} = \{D((a,0), r) : r > 0\}$. Then $\mathscr{B}$ is a base, but it is not the base for the usual topology on $\mathbb{R}^2$.
To see that it's a base, we check that it covers $\mathbb{R}^2$ and that for any $B_1, B_2 \in \mathscr{B}$, there is a $B_3 \in \mathscr{B}$ with $B_3 \subset B_1 \cap B_2$.
For any $(x,y) \in \mathbb{R}^2$, let $r = d((x,y), (a,0))$. Then $D((a,0), r+1)$ is a basis element that contains $(x,y)$. So it covers $\mathbb{R}^2$.
Let $B_1 = D((a,0), r_1)$ and $B_2 = D((a,0), r_2)$ be two basis elements. Then $B_1 \cap B_2 = D((a,0), r')$ where $r' = \min(r_1, r_2)$. For any $(x,y) \in B_1 \cap B_2$, it will have distance less than $r'$ from $(a,0)$. Denote it $s$. Let $r_3 = \frac{r' + s}{2}$. Then $B_3 = D((a,0), r_3)$ will be a basis element containing $(x,y)$ and will be contained in the intersection.
This base does not generate the usual topology on $\mathbb{R}^2$. For example, consider the open disc of radius $1$ centered at $(0,2)$. You cannot write that open set as a union of basic open sets from your base.
To your edit: The answer is no, that this does not form a basis. Consider the discs centered at $(0,0)$ and $(1,0)$ with radii $\frac{2}{3}$. Then there will be a point in the intersection that has distance further than $\frac{1}{3}$ from the $x$-axis. Therefore, you cannot have any disc inside their intersection that contains that point.
