cartesian product

if $\text{[math]}$ and $\text{[math]}$ are sets, then we define the cartesian product $\text{[math]}$ to be the collection of ordered pairs, whose first component lies in $\text{[math]}$ and second component lies in $\text{[math]}$ , thus
$\text{[math]}$ or equivalently
$\text{[math]}$ [cite:;taken from @tao_analysis_1 definition 3.5.4]

some stuff from college

cartesian product of 2 sets is the set of all the possible ordered pairs that can be obtained by taking an element from $\text{[math]}$ as the first in the pair and an element from $\text{[math]}$ as the second
$\text{[math]}$

assume $\text{[math]}$
$\text{[math]}$

$\text{[math]}$

the cartesian product of $\text{[math]}$ where $\text{[math]}$ is:
$\text{[math]}$

assume $\text{[math]}$
$\text{[math]}$

power:
$\text{[math]}$

$\text{[math]}$ or $\text{[math]}$ or $\text{[math]}$

$\text{[math]}$ assume $\text{[math]}$ or $\text{[math]}$ or $\text{[math]}$
need to prove: $\text{[math]}$
we split into cases:
case 1: assume $\text{[math]}$
$\text{[math]}$ case 2: assume $\text{[math]}$ , same as $\text{[math]}$
case 3: assume $\text{[math]}$
$\text{[math]}$ $\text{[math]}$ assume $\text{[math]}$
need to prove: $\text{[math]}$ or $\text{[math]}$ or $\text{[math]}$
we assume in contradiction that $\text{[math]}$ and $\text{[math]}$ and $\text{[math]}$
$\text{[math]}$ exists an $\text{[math]}$ so that $\text{[math]}$ and $\text{[math]}$ or there exists an $\text{[math]}$ so that $\text{[math]}$ and $\text{[math]}$
we split into cases:
case 1: there exists an $\text{[math]}$ so that $\text{[math]}$ and $\text{[math]}$
$\text{[math]}$ exists $\text{[math]}$ so that $\text{[math]}$
$\text{[math]}$ and there exists an $\text{[math]}$ so that $\text{[math]}$ and $\text{[math]}$
$\text{[math]}$ there exists $\text{[math]}$ so that $\text{[math]}$ and $\text{[math]}$
$\text{[math]}$
$\text{[math]}$
$\text{[math]}$
case 2: there exists $\text{[math]}$ so that $\text{[math]}$ and $\text{[math]}$ , a similar path to the previous case
we arrived at a contradiction so the theorem is true