Q: $\bigcap_{a\in\Bbb{R}} A_a\subseteq\{(-1,0),(1,0)\}$. Not sure how to formally state the obvious? For each $a\in\Bbb{R}$, let $A_a=\{(x,a(x^2-1))\in\Bbb{R}^2:x\in\Bbb{R}\}$. Prove that $\bigcap_{a\in\Bbb{R}} A_a=\{(-1,0),(1,0)\}$.
I'm having trouble getting started with this one. I can see that this is clearly true for the roots of $x^2-1$ and I know that I'm going to begin by showing that $\bigcap_{a\in\Bbb{R}} A_a\subseteq\{(-1,0),(1,0)\}$. However I'm not sure how to formally state this. Any advice is appreciated, thanks!

A: To show $\{(-1,0),(1,0)\} \in \cap A_a$ we just not that $a((-1)^2 - 1)=0$ for all $a$ so $(\pm 1, 0) \in A_a$ for all $a$.
To show $\cap A_a \in\{(-1,0), (1,0)\}$ I'd show that if $(x,u) \ne (\pm 1, 0)$ then there is an $a$ where $a(x^2 -1)\ne u$ so $(x,y) \not \in A_a$.
Case 1:  $x \ne \pm 1$ then $x^2 -1 \ne 0$.  Let $a$ be any real number not equal to $\frac u{x^2 -1}$ then $u \ne a(x^2-1)$ and $(x,u) \ne A_a$
Case 2:  $x =\pm 1$ but $u \ne 0$.  Then $a(x^2 -1) =0 \ne u$ and $(x,y) \not \in A_a$ for any $a$.
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Or maybe more to the point.
If $(x,y) \in \cap A_a$ then $y = a(x^2 -1)$ for all $a$.  This can only occur if and only if $x^2 -1=0$.
