prove every real number in set (0,1) is an interior point of (0,1)

To prove that every real number in the set \( (0, 1) \) is an interior point of \( (0, 1) \), we can rely on the definition of an **interior point**.

### Definition of an Interior Point:
A point \( x \) is said to be an interior point of a set \( S \) if there exists an **open interval** around \( x \) that is entirely contained within \( S \). In other words, there is some \( \varepsilon > 0 \) such that the interval \( (x - \varepsilon, x + \varepsilon) \) is a subset of \( S \).

### Proof:

1. **Let \( x \in (0, 1) \):**
   - Consider any point \( x \in (0, 1) \). By definition, \( x \) is a real number such that \( 0 < x < 1 \).

2. **Find an \( \varepsilon \) such that the interval \( (x - \varepsilon, x + \varepsilon) \subset (0, 1) \):**
   - Since \( x \) is strictly between 0 and 1, we know that there is some positive distance between \( x \) and the endpoints of the interval 0 and 1.
   - Specifically, the distance from \( x \) to 0 is \( x - 0 = x \), and the distance from \( x \) to 1 is \( 1 - x \).
   - Let \( \varepsilon = \min(x, 1 - x) \). This ensures that \( \varepsilon \) is the smaller of the distances between \( x \) and the two endpoints of the interval (0 and 1).

3. **Verify that the interval \( (x - \varepsilon, x + \varepsilon) \) lies within \( (0, 1) \):**
   - By construction, \( \varepsilon \leq x \) and \( \varepsilon \leq 1 - x \).
   - Therefore:
     - \( x - \varepsilon \geq 0 \) (so the left endpoint of the interval is greater than or equal to 0).
     - \( x + \varepsilon \leq 1 \) (so the right endpoint of the interval is less than or equal to 1).
   - This shows that the open interval \( (x - \varepsilon, x + \varepsilon) \) is completely contained within \( (0, 1) \).

4. **Conclusion:**
   - Since for every \( x \in (0, 1) \), we can find an open interval \( (x - \varepsilon, x + \varepsilon) \subset (0, 1) \), we conclude that every point in \( (0, 1) \) is an interior point of \( (0, 1) \).

Thus, **every real number in \( (0, 1) \) is an interior point of \( (0, 1) \)**.

Would you like further details on any part of this explanation?

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Here are 5 related questions to expand on this topic:
1. How does the definition of an open set relate to interior points?
2. Can you prove whether \( 0 \) or \( 1 \) are interior points of \( [0, 1] \)?
3. How would you prove that a closed interval, like \( [0, 1] \), has no interior points at its boundary?
4. What is the relationship between interior points and boundary points of a set?
5. How does the concept of interior points apply to higher-dimensional spaces, such as in \( \mathbb{R}^2 \)?

**Tip:** When proving properties about sets and intervals, always start with the definitions and verify conditions step by step.

Prove every real number in set (0,1) is an interior point of (0,1).

Explore how to prove that every real number in the interval (0, 1) is an interior point. This proof relies on the definition of interior points and the concept of open intervals in topology. Suitable for students studying real analysis at the undergraduate level.

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