Prove the sets have the same cardinality by finding a mapping For each pair of sets below, prove they have the same cardinality $$A = [0, 1]$$ and $$B = [0, 1/100]$$
So they're both infinite sets with some bijection but I can't figure out that bijection.
 Can someone help?
 A: Try with the linear function $$f(x)=ax+b$$ where $f(0)=0$ and $f(1)=1/100$.
A: As @Chinnapparaj says, one can always find a bijection between two $\phi\neq[a,b],[c,d]\subset \mathbb{R}$ subsets of $\mathbb{R}$ by defining $f:[a,b]\to[c,d]$ as $f(x)=c+(\frac{d-c}{b-a})(x-a)$. Let's prove that this is actually a bijection:
Let $x_1,x_2$ such that $f(x_1)=f(x_2)$, let's see that $x_1=x_2$ and so $f$ will be injective.
If $f(x_1)=f(x_2)$ then $c+(\frac{d-c}{b-a})(x_1-a)=c+(\frac{d-c}{b-a})(x_2-a)$ so $(\frac{d-c}{b-a})(x_1-a)=(\frac{d-c}{b-a})(x_2-a)$ so $x_1-a=x_2-a$ so $x_1=x_2$.
Let $p\in [c,d]$, then $x=\frac{(p-c)(b-a)}{(d-c)}+a$ is a preimage for $p$.
A: It takes a bit of practice and a ... mathematician's.... mindset to get used to these things.
But what do $[0,1]$ and $[0,\frac 1{100}]$ have in common and what do they have that are different?  And is the difference important?
Well, they are closed intervals.  And one is bigger than the other.
So, does size matter?  Can you convert one size to the other?
Well, the other is $\frac 1{100}$ as big, so I guess we can shrink one to the other.
How would you do that?
Well,  I guess I'd divide each number by $100$....
Can you write that out?
Um... $f(x) = \frac x{100}$....?
Is $f(x)= \frac x{100}$ a bijection between $[0,1]\to [0,\frac 1{100}]$?
Um... I guess so?
The answer is yes.  
You could map $0\mapsto 0$ and $1\mapsto \frac 1{100}$ and all points in between to points in between.  And $f(x) = \frac 1{100} x$ is the easiest and most evident way to do that.  
