Preimage of Subset under Mapping/Examples/Preimages of f(x, y) = (x^2 + y^2, x y)/Mistake

Source Work

 * $2$: Continuity generalized: metric spaces:
 * $2.3$: Open sets in metric spaces:
 * Example $2.3.16 \ \text {(b)}$
 * Example $2.3.16 \ \text {(b)}$

Mistake

 * Let $g: \R^2 \to \R^2$ be given by $\map g {x, y} = \tuple {x^2 + y^2, x y}$. Then for example


 * We know from (a) that $\set {\tuple {x, y} \in \R^2: 0 < x y < 1}$ is the shaded region in Fig. 2.2. Also, $\set {\tuple {x, y} \in \R^2: 0 < x^2 + y^2 < 2}$ is the interior of a disc of radius $2$ with its centre removed. Hence $S$, the set of all $\tuple {x, y}$ satisfying both conditions, is the intersection of the shaded region with the outlined disc, not including any of the boundary curves or lines.


 * Example-x2+y2-and-xy-mistake-1.png

Correction
The graph as given does not match the definition of the problem.

The graph shows distinct intersections between $x^2 + y^2 = 2$ and $x y = 1$.

But in fact the curves are tangent to each other:


 * Example-x2+y2-and-xy-mistake-2.png

In order to make the diagram correct, and to illustrate the point which is being made, it may be suggested that the first curve be $x^2 + y^2 = 3$.