A set � ⊂ R is open if ___.

a. it contains all of its boundary points

b. it contains none of its interior points

c. every point in � has a neighborhood entirely contained in A

d. every point in � has a boundary entirely outside �

The correct answer is:

**c. Every point in \( A \) has a neighborhood entirely contained in \( A \).**

### Explanation:
In topology, a set \( A \subseteq \mathbb{R} \) is defined to be **open** if for every point \( x \in A \), there exists a neighborhood of \( x \) (an interval around \( x \)) that is entirely contained within \( A \). This means that you can move a small amount in any direction from \( x \) without leaving the set \( A \).

This property ensures that no point in \( A \) is a boundary point of the set, which aligns with the idea of "openness" in a topological sense.

Would you like more details on open sets or any other related concepts?

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Here are 5 related questions to expand your understanding:
1. What is the difference between an open set and a closed set?
2. Can a set be both open and closed (clopen)? If so, provide an example.
3. How is the concept of an open set used in real analysis?
4. What is the definition of a boundary point of a set?
5. What does it mean for a set to be open in higher dimensions, such as \( \mathbb{R}^n \)?

**Tip:** The complement of an open set in \( \mathbb{R} \) is a closed set, and vice versa. Understanding this relationship can be very helpful when studying topology.

A set A ⊂ R is open if ___.

a. it contains all of its boundary points

b. it contains none of its interior points

c. every point in A has a neighborhood entirely contained in A

d. every point in A has a boundary entirely outside A

Explore the concept of open sets in topology, focusing on the problem: 'A set A ⊂ R is open if every point in A has a neighborhood entirely contained within A.' This problem is essential for students studying topology or real analysis, typically at the undergraduate level.

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