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

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Source Work

1975: W.A. Sutherland: Introduction to Metric and Topological Spaces:

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


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.



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