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 both injective (one-to-one) and surjective (onto).

One possible bijective function is:
\[
g(x) = 
\begin{cases}
x & \text{if } x \in [0, 1) \\
\frac{1}{2} & \text{if } x = 1.
\end{cases}
\]

### Step-by-Step Analysis

1. **Domain and Codomain**: The function \( g \) is defined for all \( x \in [0, 1] \) and maps to values in \( [0, 1) \).
  
2. **Definition of the Function**:
   - For \( x \in [0, 1) \), \( g(x) = x \).
   - For \( x = 1 \), \( g(1) = \frac{1}{2} \).
  
3. **Injectivity**:
   - For \( x, y \in [0, 1) \), \( g(x) = x \) and \( g(y) = y \). If \( g(x) = g(y) \), then \( x = y \), which preserves injectivity on \( [0, 1) \).
   - For \( x = 1 \), \( g(1) = \frac{1}{2} \), which does not overlap with any values \( g(x) = x \) for \( x \in [0, 1) \).
   - Therefore, \( g \) is injective.

4. **Surjectivity**:
   - Every \( y \in [0, 1) \) has a preimage in \( [0, 1] \): for \( y \in [0, 1) \), set \( x = y \), and then \( g(x) = x = y \).
   - Therefore, \( g \) is surjective onto \( [0, 1) \).

Since \( g \) is both injective and surjective, it is bijective.

This function \( g \) meets the requirements for a bijection between \( [0, 1] \) and \( [0, 1) \).

Would you like a deeper explanation or have any questions? 

### Related Questions

1. What are other examples of bijections between closed and half-open intervals?
2. How does the continuity of a function affect its bijectivity?
3. Can we define a bijection between \( (0, 1) \) and \( [0, 1] \)?
4. How would you define a bijection from \( [0, 1] \) to \( [a, b) \) for any \( a < b \)?
5. Can we extend this function \( g \) to work on all real numbers?

**Tip:** A function is bijective if it is both one-to-one (injective) and onto (surjective). Always check both conditions when verifying a bijection.

Explore how to construct a bijection between the interval [0,1] and [0,1), and learn step-by-step how to prove it is a bijective function. This problem covers essential mathematical concepts in set theory and functions, focusing on injectivity and surjectivity, suitable for undergraduate students.

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