Find a bijection g : [0, 1] −→ [0, 1), and make sure to prove it is a bijection.

To construct a bijection \( g: [0, 1] \to [0, 1) \), we need a function \( g \) that is:

1. **Injective**: Every element in \([0, 1]\) has a unique image in \([0, 1)\).
2. **Surjective**: Every element in \([0, 1)\) has a preimage in \([0, 1]\).

### Step 1: Define the function \( g \)

Let's define the function \( g \) as follows:
\[
g(x) = 
\begin{cases} 
x & \text{if } x \in [0, 1) \\
\frac{1}{2} & \text{if } x = 1 
\end{cases}
\]

This definition essentially maps every point in \([0, 1)\) to itself, while specifically mapping \(1\) in the closed interval \([0, 1]\) to a point in \([0, 1)\), namely \(\frac{1}{2}\).

### Step 2: Prove Injectivity of \( g \)

To prove injectivity, assume that for any \( x_1, x_2 \in [0, 1] \), if \( g(x_1) = g(x_2) \), then \( x_1 = x_2 \).

1. If \( x_1, x_2 \in [0, 1) \), then \( g(x_1) = x_1 \) and \( g(x_2) = x_2 \), so \( g(x_1) = g(x_2) \) implies \( x_1 = x_2 \).
2. If \( x_1 = 1 \) and \( g(x_1) = \frac{1}{2} \), there is no \( x_2 \in [0, 1) \) such that \( g(x_2) = \frac{1}{2} \). Thus, \( g(x) \) maps \(1\) uniquely to \(\frac{1}{2}\) without overlap.

Therefore, \( g \) is injective.

### Step 3: Prove Surjectivity of \( g \)

To show surjectivity, we need to confirm that for every \( y \in [0, 1) \), there exists an \( x \in [0, 1] \) such that \( g(x) = y \).

1. For \( y \in [0, 1) \), if \( y \neq \frac{1}{2} \), we can let \( x = y \), which is in \([0, 1)\) and satisfies \( g(x) = x = y \).
2. For \( y = \frac{1}{2} \), we take \( x = 1 \), which maps under \( g \) to \( g(1) = \frac{1}{2} \).

Thus, every \( y \in [0, 1) \) has a corresponding \( x \in [0, 1] \) such that \( g(x) = y \), proving that \( g \) is surjective.

### Conclusion

Since \( g \) is both injective and surjective, it is a bijection from \([0, 1]\) to \([0, 1)\).

Would you like more details or have any questions?

Here are 5 related questions that explore this topic further:
1. What is an example of a bijection from \((0, 1]\) to \((0, 1)\)?
2. Can we create a bijection from \([0, 1]\) to \((0, 1]\)?
3. How can we construct a bijection from any closed interval \([a, b]\) to any open interval \((c, d)\)?
4. What properties of a function guarantee it can be bijective on given intervals?
5. Can a bijective function between intervals be differentiable?

**Tip:** To check if a function is bijective, verify it meets both the injective and surjective criteria.

Explore the construction of a bijective function g: [0, 1] −→ [0, 1) by defining g(x) with specific mappings. This problem delves into set theory, function mapping, and proofs of injectivity and surjectivity, suitable for undergraduate students in mathematics.

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