Exercise 21.10

Question

Using the closed set formulation of continuity (Theorem~18.1), show that the following are closed subsets of ${\mathbb{R}}^2$:
  \begin{align*}
    A &= \left\{x 	imes y \mid xy = 1ight\} \,, \
    S^1 &= \left\{x 	imes y \mid x^2 + y^2 = 1ight\} \,, \
    B^2 &= \left\{x 	imes y \mid x^2 + y^2 \leq 1ight\} \,.
  \end{align*}
  The set $B^2$ is called the (closed) 	extbf{	extit{unit ball}} in ${\mathbb{R}}^2$.

Dan Whitman · Accepted Answer

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\begin{proof}
    Regarding the set $A$, the function $f: {\mathbb{R}} 	imes {\mathbb{R}} 	o {\mathbb{R}}$ defined by $f(x 	imes y) = xy$ is simply the multiplication function, which is continuous by Lemma~21.4.
    Clearly the set $A = {f}^{-1}(\left\{1ight\})$ by definition.
    We also have that $\left\{1ight\}$ is closed in ${\mathbb{R}}$ by Theorem~17.8 since ${\mathbb{R}}$ is Hausdorff.
    Thus $A = {f}^{-1}(\left\{1ight\})$ is closed in ${\mathbb{R}} 	imes {\mathbb{R}} = {\mathbb{R}}^2$ by Theorem~18.1 since $f$ is continuous.

For $S^1$, let $d$ denote the usual euclidean metric on ${\mathbb{R}} 	imes {\mathbb{R}}$, which we know induces the same topology as the product topology.
    We also know from Exercise~20.3 that $d: {\mathbb{R}} 	imes {\mathbb{R}} 	o {\mathbb{R}}$ is continuous.
    The function $f: {\mathbb{R}} 	imes {\mathbb{R}} 	o {\mathbb{R}}$ by
    \begin{gather*}
      f(x 	imes y) = d(x,y) \cdot d(x,y) = [d(x,y)]^2 = \left[\sqrt{x^2 + y^2}ight]^2 = x^2 + y^2 \,.
    \end{gather*}
    is also then continuous by Theorem~21.5 being the product of two continuous functions.
    As previously mentioned, the finite set $\left\{1ight\}$ is closed in ${\mathbb{R}}$.
    We also have that $S^1 = {f}^{-1}(\left\{1ight\})$ by definition, which then also must be closed in ${\mathbb{R}} 	imes {\mathbb{R}} = {\mathbb{R}}^2$ again by Theorem~18.1 since $f$ continuous.

Regarding the closed unit ball $B^2$, define $f = d \cdot d$ as above for $S^1$, which is of course still continuous.
    Define the subset $C = \left\{x \in {\mathbb{R}} \mid 0 \leq x \leq 1ight\}$ of ${\mathbb{R}}$.
    We claim that $B^2 = {f}^{-1}(C)$.
    This is pretty obvious since clearly $0 \leq f(x 	imes y) = x^2 + y^2 \leq 1$ for any $x 	imes y \in B^2$ by definition so that $f(x 	imes y) \in C$ and hence $x 	imes y \in {f}^{-1}(C)$.
    Conversely, for any $x 	imes y \in {f}^{-1}(C)$, we have that $f(x 	imes y) \in C$ so that $0 \leq f(x 	imes y) = x^2 + y^2 \leq 1$ so that $x 	imes y \in B^2$ by definition.
    This shows the desired equality of $B^2$ and ${f}^{-1}(C)$.
    As it is trivial to show that $C$ is closed in ${\mathbb{R}}$, and $f$ is continuous, it follows that $B^2 = {f}^{-1}(C)$ is also closed in ${\mathbb{R}} 	imes {\mathbb{R}} = {\mathbb{R}}^2$, again by Theorem~18.1.
  \end{proof}

Exercise 21.10

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Exercise 21.10

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