# 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 $2.3.16 \ \text {(b)}$

## 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
 $\ds \qquad \ \$ $\ds S$ $=$ $\ds g^{-1} \set {\tuple {0, 2}, \tuple {0, 1} }$ $\ds$ $=$ $\ds \set {\tuple {x, y} \in \R^2: 0 < x^2 + y^2 < 2 \text { and } 0 < x y < 1}$
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.

## 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:

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 $x^2 + y^2 = 3$.