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The question is asking to prove the statement:

\[
\forall x \in [0,1), \exists y \in (0,1) \text{ such that } x < y.
\]

The goal of the proof is to show that for every \( x \in [0, 1) \), there exists a \( y \in (0, 1) \) such that \( x < y \).

Let's fill in the blanks step by step:

1. **Select \( y \)**: Since we need \( y \) to be greater than \( x \) but still in the interval \( (0,1) \), we can define \( y \) as a value that depends on \( x \) and ensures \( y > x \). A common choice is to set \( y = \frac{x + 1}{2} \), which lies between \( x \) and 1.

   So, let \( y = \frac{x + 1}{2} \).

2. **Since \( 0 \leq x < 1 \)**: Now, since \( x \in [0, 1) \), it follows that \( 0 \leq x < 1 \), and we need to show that \( y = \frac{x + 1}{2} \) is still in the interval \( (0, 1) \).

   We know \( 0 < \frac{x + 1}{2} < 1 \) because \( x + 1 > x \), and dividing by 2 keeps the value within \( (0,1) \). Thus, \( y \in (0, 1) \).

3. **Thus, we have shown \( y \in (0, 1) \)**: At this point, we've established that \( y = \frac{x + 1}{2} \) is between 0 and 1.

4. **Since \( 0 \leq x < 1 \), we have \( y = \frac{x}{2} + \frac{1}{2} \)**: This step provides an alternate form of \( y \). The expression \( \frac{x + 1}{2} \) can be written as \( \frac{x}{2} + \frac{1}{2} \). 

5. **Therefore, \( x < y \)**: Finally, because \( \frac{x + 1}{2} > x \), we have shown that \( x < y \).

Thus, the proof is complete, and we have demonstrated that for every \( x \in [0, 1) \), there exists a \( y \in (0, 1) \) such that \( x < y \).

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Would you like more details on any part of this proof? Here are some related questions that might help you explore further:

1. Why is \( y = \frac{x + 1}{2} \) a good choice for proving \( x < y \)?
2. Can you use any other values for \( y \) to prove \( x < y \)?
3. What happens if \( x = 1 \)? Why is \( x \in [0, 1) \) and not \( [0, 1] \)?
4. How can this type of proof be extended to other intervals or inequalities?
5. Could you modify this proof for a stricter inequality, such as \( x \leq y \)?

**Tip:** When choosing \( y \) in a proof like this, ensure that the value is not only greater than \( x \) but also lies within the specified bounds (in this case, the interval \( (0,1) \)).

Prove the statement: ∀x ∈ [0,1), ∃y ∈ (0,1) such that x < y.

This proof shows that for any x in the interval [0, 1), there exists a y in (0, 1) such that x < y. The proof involves selecting a y that lies strictly between x and 1 and verifying that it belongs to the interval (0, 1). Key mathematical concepts include real analysis, inequalities, and interval properties.

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