Let's analyze the problem step by step.

The statement to be proved is: $\forall x \in (0, 1), \exists y \in (0, 1) \text{ such that } x < y.$

The proof provided follows these steps:

1. Let $x \in (0, 1)$ be given.

   The proof chooses $y = \frac{x + 1}{2}$.

2. Check that $y \in (0, 1)$:

   • First, verify that $\frac{1}{2} < y \leq 1$. To see this:
     • Since $0 < x < 1$, we have $\frac{1}{2} < \frac{x + 1}{2} \leq 1$.
     • Hence, $y \in (0, 1)$.

3. Verify that $x < y$:

   • $y = \frac{x + 1}{2}$
   • Clearly, $\frac{x}{2} < \frac{x + 1}{2}$, so $x < y$.

The dropdown selections should be as follows based on the analysis:

1. $\frac{x + 1}{2}$ is correctly chosen.
2. $\frac{1}{2} \leq y < 1$, so choose $\frac{1}{2} \leq y \leq 1$.
3. $\frac{x}{2} + \frac{1}{2}$, so the correct value is $x < y$.

This completes the proof as intended.

Would you like me to go into more details or have any specific questions?

Here are 5 related questions to expand this topic:

1. What is the significance of choosing $y = \frac{x + 1}{2}$ in the proof?
2. Can you choose a different expression for $y$ such that $x < y$?
3. How does this proof technique change if the interval was $(0, 2)$?
4. How can you generalize this proof for $x \in (a, b)$ for arbitrary $a$ and $b$?
5. How does this problem connect with concepts of limits in calculus?

Tip

In proofs involving inequalities, always check boundary cases carefully to ensure your selected values meet all given conditions.