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Optional Fun Problem: Infinity Minus Two
Let $[0, 1]$ denote the set $\Set{ x \in \mathbb{R} \ \vert \ 0 \le x \le 1 }$ and $(0, 1)$ denote the set $\Set{ x \in \mathbb{R} \ \vert \ 0 \lt x \lt 1 }$. That is, the set $[0, 1]$ is the set of all real numbers between $0$ and $1$, inclusive, and the set $(0, 1)$ is the set of all real numbers between $0$ and $1$, exclusive. These sets differ only in that the set $[0, 1]$ includes $0$ and $1$ and the set $(0, 1)$ excludes $0$ and $1$.
Give a definition of bijection $f : [0, 1] \to (0, 1)$ via an explicit rule (i.e. writing out $f(x) = \blank$ or defining $f$ as a piecewise function), then prove that your function is a bijection. This proves that $\abs{[0, 1]} = \abs{(0, 1)}$.