Exercise 5.1

Question

Show that there is a bijective correspondence of $A 	imes B$ with $B 	imes A$.

Dan Whitman · Accepted Answer

\begin{proof}
  We define a function $f: A 	imes B 	o B 	imes A$.
  For any element $(a,b) \in A 	imes B$ we set $f(a,b) = (b,a)$, noting that of course $a \in A$ and $b \in B$.
  It should be obvious then that $f(a,b) = (b,a) \in B 	imes A$ so that $B 	imes A$ can be the range of $f$.

$\def\opx{(a_1, b_1)}
  \def\opy{(a_2, b_2)}
  \def\pox{(b_1, a_1)}
  \def\poy{(b_2, a_2)}$
  First we show that $f$ is injective.
  To this end consider $\opx$ and $\opy$ in $A 	imes B$ where $\opx 
eq \opy$.
  Of course we have that $f\opx = \pox$ and $f\opy = \poy$.
  Since $\opx 
eq \opy$ clearly either $a_1 
eq a_2$ or $b_1 
eq b_2$.
  In either case it should be clear that $f\opx = \pox 
eq \poy = f\opy$, which shows that $f$ is injective since $\opx$ and $\opy$ were arbitrary.

It is very easy to that $f$ is also surjective since, for any $(b,a) \in B 	imes A$, clearly $(a,b) \in A 	imes B$ and $f(a,b) = (b,a)$.
  Hence $f$ is a bijection as desired.
  Note that if $A 	imes B = \varnothing$ then $f = \varnothing$ as well, which is vacuously a bijective function since it must be that $B 	imes A = \varnothing$ as well (because either $A = \varnothing$ or $B = \varnothing$).
\end{proof}

Exercise 5.1

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Exercise 5.1

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