Prove that $A = (0,1) \times (0,1)$ and $B = (2,4) \times (3, \infty)$ are equinumerous by explicitly define a bijection $h: A \rightarrow B$ Show that h is a bijection.
What I have thus far.
Let X = (0,1), Y = (0,1), Z = (2,4) and W = $(3, \infty)$.
There's a bijection between X and Z, such that $f(x) = 2 \cdot x +2$. therefore, we can list the sets as follows
X = {$x_1$, $x_2$,...,$x_n$}
Z = {$z_1$, $z_2$,...,$z_n$}
There's also a bijection between Y and W, such that $g(y) = tan(\frac{\pi y}{2}) + 3$ therefore, we can list the sets as follows
Y = {$y_1$, $y_2$,...,$y_m$}
W = {$w_1$, $w_2$,...,$w_m$}
Then |X x Z| = $m \cdot n$
Also |Y x W| = $m \cdot n$
Therefore we can see that |X x Z| = |Y x W|, which defines a bijection between A and B, as asked.
The question I have about this is weather or not my argument works for |Y x W| as W is an infinite set. I halfway assume it does, since there's a bijection between Y and W, but I've just started covering this material, and I'm unsure of myself at this point. Any help is greatly appreciated.