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I have difficulties with this problem and I have no idea how to solve it:

Suppose $A$ is a subset of $\Bbb R^2$ with this property: if $(x, y)$ belongs to $A$ then $(\|x\|-1,\|y\|^2)$ belongs to $A$. How can I demonstrate that the closure of $A$ has the same property? And the interior part?

fuglede
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Domenico Vuono
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1 Answers1

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The claim is not true: $A = \mathbb{R}^2$ has the given property, and the interior of $\mathbb{R}^2$ also has.

Regarding the closures, define $f : \mathbb{R}^2 \to \mathbb{R}^2$ by $f(x, y) = (\lVert x \rVert -1, \lVert y \rVert^2)$. Then $f$ is continuous, and by assumption, $f(A) \subseteq A$. Now, one characterization of continuity is that for every subset of $\mathbb{R}^2$, the image of the closure is contained in the closure of the image, so $f(\overline{A}) \subseteq \overline{f(A)} \subset \overline{A}$, which shows that $\overline{A}$ has the desired property.

fuglede
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  • Sorry I edited my question – Domenico Vuono Nov 21 '16 at 18:33
  • The claim is still not true after the edit: if $A = [-1, \infty) \times \mathbb{R}$, then $A$ has the desired property, but the interior of $A$ does not (since $(0, 0)$ belongs to the interior, but $f(0, 0) = (-1, 0)$ does not). – fuglede Nov 21 '16 at 18:44