Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola $ xy\equal{}1$ and both branches of the hyperbola $ xy\equal{}\minus{}1.$ (A set $ S$ in the plane is called [i]convex[/i] if for any two points in $ S$ the line segment connecting them is contained in $ S.$)
1. **Identify the problem and the constraints:**
   We need to find the least possible area of a convex set in the plane that intersects both branches of the hyperbolas \(xy = 1\) and \(xy = -1\). A convex set is defined such that for any two points in the set, the line segment connecting them is also contained in the set.

2. **Determine the points of intersection:**
   The convex set must contain at least one point on each of the four branches of the hyperbolas. Let these points be:
   \[
   (x_1, y_1), (-x_2, y_2), (-x_3, -y_3), (x_4, -y_4)
   \]
   where \(x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4\) are positive numbers and satisfy the hyperbola equations:
   \[
   x_1 y_1 = 1, \quad x_2 y_2 = 1, \quad x_3 y_3 = 1, \quad x_4 y_4 = 1
   \]

3. **Calculate the area of the convex hull:**
   The smallest convex set containing these points is the convex hull, which in this case is a quadrilateral. The area \(A\) of a quadrilateral with vertices \((x_1, y_1), (-x_2, y_2), (-x_3, -y_3), (x_4, -y_4)\) can be calculated using the shoelace formula:
   \[
   A = \frac{1}{2} \left| x_1 y_2 + (-x_2) y_3 + (-x_3) (-y_4) + x_4 y_1 - (y_1 (-x_2) + y_2 (-x_3) + y_3 x_4 + y_4 x_1) \right|
   \]

4. **Simplify the area expression:**
   Substitute \(y_i = \frac{1}{x_i}\) for \(i = 1, 2, 3, 4\):
   \[
   A = \frac{1}{2} \left| x_1 \frac{1}{x_2} + (-x_2) \frac{1}{x_3} + (-x_3) \frac{1}{x_4} + x_4 \frac{1}{x_1} - \left( \frac{1}{x_1} (-x_2) + \frac{1}{x_2} (-x_3) + \frac{1}{x_3} x_4 + \frac{1}{x_4} x_1 \right) \right|
   \]
   \[
   = \frac{1}{2} \left| \frac{x_1}{x_2} - \frac{x_2}{x_3} - \frac{x_3}{x_4} + \frac{x_4}{x_1} - \left( -\frac{x_2}{x_1} - \frac{x_3}{x_2} + \frac{x_4}{x_3} + \frac{x_1}{x_4} \right) \right|
   \]
   \[
   = \frac{1}{2} \left| \frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{x_2}{x_3} + \frac{x_3}{x_2} + \frac{x_3}{x_4} + \frac{x_4}{x_3} + \frac{x_4}{x_1} + \frac{x_1}{x_4} \right|
   \]

5. **Apply the AM-GM inequality:**
   For any positive \(u\), we have \(u + \frac{1}{u} \geq 2\). Applying this to each term:
   \[
   \frac{x_1}{x_2} + \frac{x_2}{x_1} \geq 2, \quad \frac{x_2}{x_3} + \frac{x_3}{x_2} \geq 2, \quad \frac{x_3}{x_4} + \frac{x_4}{x_3} \geq 2, \quad \frac{x_4}{x_1} + \frac{x_1}{x_4} \geq 2
   \]
   Summing these inequalities:
   \[
   \frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{x_2}{x_3} + \frac{x_3}{x_2} + \frac{x_3}{x_4} + \frac{x_4}{x_3} + \frac{x_4}{x_1} + \frac{x_1}{x_4} \geq 8
   \]

6. **Determine the minimum area:**
   Therefore, the minimum area is:
   \[
   A \geq \frac{1}{2} \times 8 = 4
   \]
   Equality holds when \(x_1 = x_2 = x_3 = x_4\), meaning the quadrilateral is a rectangle with sides parallel to the axes.

The final answer is \(\boxed{4}\).