Needing to prove that $|[0,1]| = |(4,7)|$.
I know by the Schroeder-Bernstein Theorem: if $f: A \rightarrow B$ is one to one then $|A| \le |B|$ and if $f: A \rightarrow B$ is onto then $|A| \ge |B|$ then we can conclude $|A| = |B|$.
I know $(4,7)$ is a open interval or can be view as a point, but what exactly is $[0,1]$ in this context? I know it's a intervals where it is closed to $0$ and $1$.
Don't how to show this is one-to-one or onto without have a formula. I was thinking I could do slope intercept form and get a equation that I could show one-to-one and onto, but I don't think I can do that $[0,1]$.
Any help would greatly be appreciated!
Best Answer
Another way: If $g(x) := \tfrac13 x - \tfrac43$ for $x \in (4,7)$, then $g$ is a one-to-one function from $(4,7)$ onto $(0,1)$. Now, if we define $$\begin{align} f : (0,1) & \to [0,1] \\ \tfrac12 & \mapsto 0 \\ \tfrac13 & \mapsto 1 \\ \tfrac14 & \mapsto \tfrac12 \\ \tfrac15 & \mapsto \tfrac13 \\ & \ \, \, \vdots \\ x & \mapsto x \quad \small \textrm{if $x$ cannot be written as $\tfrac1n$ for some integer $n \geq 2$,} \end{align}$$ then $f \circ g$ is a one-to-one function from $(4,7)$ onto $[0,1]$.