Question

For each \(a \in \mathbb{R},\) let \(A_{a}=\left\\{\left(x, a\left(x^{2}-1\right)\right) \in \mathbb{R}^{2}: x \in \mathbb{R}\right\\} .\) Prove that \(\bigcap_{a \in \mathbb{R}} A_{a}=\\{(-1,0),(1,0)\\}\).

Short answer

The only common points for all $Af$ sets are $(-1,0)$ and $(1,0)$, which can be seen through the intersection operation. Therefore, it's proven that $\bigcap_{a \in R} Af = \{(-1,0), (1,0)\}$.

Step-by-step solution

Interpretation of the problem

The first step is understanding what set $Af$ represents: any real number \(a\) multiplied by the expression \((x^2-1)\, is the y-coordinate and \(x\) as the x-coordinate of points in $Af$. The interval of defined values is the set of real numbers.

Finding the common points in all sets $Af$

The intersection of all $Af$ sets will yield the common points that belong to all of them. If we solve the equation \(a(x^2-1) = 0\), we find that $x = -1$ and $x = 1$ are roots for all values of \(a\). This means that the following two points are common for all $Af$ sets: $(-1,0)$ and $(1,0)$.

Proving no other points can be in the intersection of all $Af$

To prove these are the only points that can be in the intersection, let's assume there's a point \((x', y')\) that belongs to every $Af$ and \(x' ≠ ±1\). According to the definition of $Af$, the y-coordinate \(y' = a(x'^2 -1)\) for every \(a\). Choosing two different real numbers, we get two distinct \(y\) values - contradiction! So there are no other common points.