1. ## Equinumerous

Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the two sets in each pair.

a. S= [0,1] and T = [1,3]

b. S= [0,1] and T = [0, infinity]

2. Originally Posted by GoldendoodleMom
Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the two sets in each pair.

a. S= [0,1] and T = [1,3]
Think geometrically. Length of $[0,1]$ is $1$ and length of $[1,3]$ is $2$. Thus, strech $[0,1]$ to $[0,2]$. Now shift this interval by one to the right this gives $[1,3]$.

Now the streching can be desribed by function $f :[0,1] \to [0,2]$ by $f(x) = 2x$. And the shifting can be described by $g:[0,2]\to [1,3]$ by $g(x) = x+1$. Thus, $g\circ f: [0,1] \to [1,3]$ is what you are looking for. Finally $g(f(x)) = 2x+1$ is the mapping you want.

3. Originally Posted by GoldendoodleMom
Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the two sets in each pair.

b. S= [0,1] and T = [0, infinity]
Consider $f:[0,1) \to [0, \infty)$ defined by $f(x)=\frac{x}{1-x}$ for $x \in [0,1)$

RonL

4. Originally Posted by CaptainBlack
Consider $f:[0,1) \to [0, \infty)$ defined by $f(x)=\frac{x}{1-x}$ for $x \in [0,1)$

RonL
Also an alternative is to use $\tan$ and/or $\tan^{-1}$ (inverse of $\tan$), these really help too much.
$f:[0,1)\to[0,\infty)$
......... $t\to f(t)=\tan\bigg(\frac{\pi}{2}t\bigg)$