Find an explicit bijection (i.e. give a formula that explains what f(x) is given any x ∈ [0, 1]) between [0, 1] and [0, 1).

1. Break Down the Problem

We need to find a function $f: [0, 1] \to [0, 1)$ that is a bijection. This means:

• The function must cover every point in the target interval.
• Each point in the domain must map to a unique point in the codomain.

2. Relevant Concepts

A common approach is to use a function that approaches the upper limit of $[0, 1)$ but never actually reaches it for inputs within $[0, 1]$. One suitable function is: $f(x) = x \quad \text{for } x \in [0, 1)$ and we must define $f(1)$ to be any point in $[0, 1)$. A typical choice is: $f(1) = 1 - \epsilon$ where $\epsilon$ is an arbitrarily small positive number.

3. Analysis and Detail

Let’s define the function explicitly: $f(x) = \begin{cases} x & \text{if } 0 \leq x < 1 \\ 1 - \epsilon & \text{if } x = 1 \end{cases}$

• Injective: For any two $x_1, x_2 \in [0, 1]$:

  • If both $x_1, x_2 < 1$, then $f(x_1) = x_1$ and $f(x_2) = x_2$. They are distinct if $x_1 \neq x_2$.
  • If one of them is $1$, $f(1) = 1 - \epsilon$ is distinct from all values in $[0, 1)$.

• Surjective: For any $y \in [0, 1)$:

  • If $y < 1$, then for $x = y$, we have $f(y) = y$.
  • Also, $f(1) = 1 - \epsilon$ can represent a number extremely close to $1$.

4. Verify and Summarize

We’ve established the function $f$ meets the criteria for being both injective and surjective, confirming that it is indeed a bijection from $[0, 1]$ to $[0, 1)$.

Final Answer

The explicit bijection between $[0, 1]$ and $[0, 1)$ can be defined as: $f(x) = \begin{cases} x & \text{if } 0 \leq x < 1 \\ 1 - \epsilon & \text{if } x = 1 \end{cases}$ where $\epsilon$ is an arbitrarily small positive number.