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

To define a bijection \( g: [0,1] \to [0,1) \), we need a function that is both injective (one-to-one) and surjective (onto). The function must map each element in the closed interval \([0,1]\) to a unique element in the half-open interval \([0,1)\), and every element in \([0,1)\) must have a preimage in \([0,1]\).

### Step 1: Define the function

Let's define the following function:

\[
g(x) =
\begin{cases}
\frac{x}{2}, & \text{if } x \in [0, 1) \\
\frac{1}{2}, & \text{if } x = 1
\end{cases}
\]

- For all \( x \in [0,1) \), the function maps \( x \) to \( \frac{x}{2} \), which stays in the interval \([0,1)\).
- At \( x = 1 \), we map it to \( \frac{1}{2} \), which is also in the interval \([0,1)\).

### Step 2: Prove that \( g \) is injective (one-to-one)

To prove injectivity, we need to show that if \( g(x_1) = g(x_2) \), then \( x_1 = x_2 \).

- For \( x_1, x_2 \in [0,1) \), if \( g(x_1) = g(x_2) \), this means \( \frac{x_1}{2} = \frac{x_2}{2} \). Hence, \( x_1 = x_2 \).
- For \( x_1 = 1 \), the only \( x \) that maps to \( \frac{1}{2} \) is 1 itself by definition, so no other element in \( [0,1) \) can map to \( \frac{1}{2} \).

Thus, \( g \) is injective.

### Step 3: Prove that \( g \) is surjective (onto)

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

- For \( y \in [0,1/2) \), we can take \( x = 2y \). Since \( y \in [0,1) \), the corresponding \( x \in [0,1) \), and \( g(x) = \frac{x}{2} = y \).
- For \( y = \frac{1}{2} \), we have \( g(1) = \frac{1}{2} \), so there exists \( x = 1 \) that maps to \( \frac{1}{2} \).

Thus, \( g \) is surjective.

### Conclusion

The function \( g(x) = \frac{x}{2} \) for \( x \in [0,1) \), and \( g(1) = \frac{1}{2} \), is both injective and surjective, so it is a bijection from \( [0,1] \) to \( [0,1) \).

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Would you like further details or have any questions?

Here are five related questions to explore further:
1. Can you construct another bijection from \( [0,1] \) to \( [0,1) \)?
2. How would the proof change if you wanted to find a bijection from \( (0,1] \) to \( (0,1) \)?
3. Can you find a bijection between \( [0,1] \) and \( (0,1) \)?
4. How would you extend the concept of a bijection to higher dimensions, e.g., \( [0,1]^n \)?
5. What are examples of bijections between unbounded intervals, such as \( [0,\infty) \) and \( (0,\infty) \)?

**Tip**: When proving injectivity, remember to focus on showing that different inputs map to different outputs. For surjectivity, ensure every output has a corresponding input.

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

Explore how to construct a bijection from the interval [0,1] to [0,1) with a step-by-step proof. This problem involves defining an injective and surjective function, applying bijection properties, and verifying injectivity and surjectivity. Ideal for students studying undergraduate mathematics.

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