Question

Let \( \{(a, b) : a, b \in {R, a<b \} }\) be a basis for a topology \( \tau \) on \( {R} \). Which of the following is/are correct?

• A. Every \( (a, b) \) with \( a<b \) is an open set in \( ({R}, \tau) \)
• B. Every \( [a, b] \) with \( a<b \) is a compact set in \( ({R}, \tau) \)
• C. \( ({R}, \tau) \) is a first-countable space
• D. \( ({R}, \tau) \) is a second-countable space

Hint:

For topology problems, verify properties like openness, compactness, and countability based on the basis of the topology.

Answer 1

The Correct Option is A

Step 1: Verifying open sets. The intervals \( (a, b) \) are part of the basis for the topology \( \tau \), so they are open in \( ({R}, \tau) \). 

Step 2: Compactness of \( [a, b] \). In general, \( [a, b] \) may not be compact in the topology \( \tau \), as compactness depends on the specific topology induced. 

Step 3: First-countability. \( ({R}, \tau) \) is first-countable because at each point \( x \in {R} \), a countable basis of open sets can be constructed using \( (x - \frac{1}{n}, x + \frac{1}{n}) \). 

Step 4: Second-countability. Second-countability is not guaranteed unless \( \tau \) has a countable basis for all open sets, which is not specified here. 

Step 5: Conclusion. The correct answers are \( {(1), (3)} \).