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

- $2.3$: Open sets in metric spaces:

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

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.16 \ \text {(b)}$