Exercise 1.10

Question

Let ${\mathbb{R}}$ denote the set of real numbers.
For each of the following subsets of ${\mathbb{R}} 	imes {\mathbb{R}}$, determine whether it is equal to the cartesian product of two subsets of ${\mathbb{R}}$.
\begin{enumerate}[label=(\alph*)]
\item $\left\{(x,y) \mid 	ext{$x$ is an integer}ight\}$.
\item $\left\{(x,y) \mid 0 < y \leq 1ight\}$.
\item $\left\{(x,y) \mid y > xight\}$.
\item $\left\{(x,y) \mid 	ext{$x$ is not an integer and $y$ is an integer}ight\}$.
\item $\left\{(x,y) \mid x^2 + y^2 < 1ight\}$.
\end{enumerate}

Dan Whitman · Accepted Answer

(a) This is equal to the set ${\mathbb{Z}} imes {\mathbb{R}}$, which is trivial to prove. (b) It is easy to show that this is equal to ${\mathbb{R}} imes (0, 1]$, where of course $(a,b]$ denotes the half-open interval $\left\{x \in {\mathbb{R}} \mid a < x \leq b ight\}$. (c) We claim that this cannot be equal to the cartesian product of subsets of ${\mathbb{R}}$. \begin{proof} Let $A = \left\{(x,y) \mid y > x ight\}$ and suppose to the contrary that $A = B imes C$ where $B,C \subset {\mathbb{R}}$. Since $1 > 0$, we have that $(0,1) \in A$. Then also $0 \in B$ and $1 \in C$ since $A = B imes C$. We also have that $1 \in B$ and $2 \in C$ since $2 > 1$ so that $(1,2) \in A = B imes C$. Thus $1 \in B$ and $1 \in C$ so that $(1,1) \in B imes C = A$, but this cannot be since it is not true that $1 > 1$. Hence we have a contradiction so that $A$ cannot be expressed as $B imes C$. \end{proof} (d) It is trivial to show that this set is equal to $({\mathbb{R}} - {\mathbb{Z}}) imes {\mathbb{Z}}$. (e) We claim that this set cannot be expressed as the cartesian product of subsets of ${\mathbb{R}}$. \begin{proof} Let $A = \left\{(x,y) \mid x^2 + y^2 < 1 ight\}$ and suppose to the contrary that $A = B imes C$ where $B,C \subset {\mathbb{R}}$. We then have that $(9/10)^2 + 0^2 = 81/100 + 0 = 81/100 < 1$ so that $(9/10, 0) \in A = B imes C$, and hence $9/10 \in B$ and $0 \in C$. Also $0^2 + (9/10)^2 = (9/10)^2 + 0^2 = 81/100 < 1$ so that $(0, 9/10) \in A = B imes C$, and hence $0 \in B$ and $9/10 \in C$. Hence $(9/10, 9/10) \in B imes C = A$ since $9/10$ is in both $B$ and $C$. However, we have $(9/10)^2 + (9/10)^2 = 81/100 + 81/100 = 162/100 \geq 1$ so that $(9/10, 9/10)$ cannot be in $A$, so we have a contradiction. So it must be that $A$ cannot be equal to $B imes C$. \end{proof}

Exercise 1.10

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Exercise 1.10

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