Prove that $[0,1] \approx [0,a)$ where $a$ is a positive real number.
I know I need to fine a bijective function function between the two sets. The $a$ in the second set is throwing me off. Origninally I was thinking of doing $x=\frac{1}{a}$ that would get me $[0,x)$. But i am unsure if this is the correct route. Also how do I address the open interval on $a$ as oppose to a closed interval?
A typical approach is simply to use the Cantor-Bernstein theorem. That is, because we can find an injective function $[0,1] \to [0,a)$ and another injective function $[0,a) \to [0,1]$, we can deduce that the two sets must be equinumerous.
However, if we must find a bijection, one trick is to get use a bijection $[0,1) \to [0,a)$, and find a way to "sneak in" the element $1$. With that in mind: let $S$ denote the set $S = \{1,1/2,1/3,1/4, 1/5, \dots\} \subset [0,1] $. We can define a function $f:[0,1] \to [0,a)$ as follows: $$ f(x) = \begin{cases} ax & x \notin S\\ \frac 1{n+1} & x = \frac 1n \in S \end{cases} $$ You can verify that this is a bijection between the sets.
It can be shown that there is no continuous bijection between these sets by noting that any continuous bijection from a compact space to a Hausdorff space is necessarily a homeomorphism.
