Question

Let \( X = (\mathbb{R}, T) \), where \( T \) is the smallest topology on \( \mathbb{R} \) in which all the singleton sets are closed. Then, which of the following statements are TRUE?

• A. \( [0, 1] \) is compact in \( X \)
• B. \( X \) is not first countable
• C. \( X \) is second countable
• D. \( X \) is first countable

Hint:

In the discrete topology, every subset is open and every set is compact. However, the space is not second countable, and in some cases, it is not first countable.

Answer 1

The Correct Option is A, B

We are given that \( X = (\mathbb{R}, T) \), where \( T \) is the smallest topology on \( \mathbb{R} \) in which all singleton sets are closed. This means the topology \( T \) is the discrete topology. Step 1: Compactness of \( [0, 1] \) in \( X \)
In the discrete topology, every subset is open, including \( [0, 1] \). A set is compact if every open cover has a finite subcover. Since \( [0, 1] \) is a finite set in the discrete topology, it is trivially compact. Therefore, statement (A) is true. Step 2: First Countability of \( X \)
A space is first countable if every point has a countable local base. In the discrete topology, every point has a local base consisting of a single point, which means the space is first countable. Hence, statement (D) is false, and (B) is true, as \( X \) is not first countable in the sense that every set is open, and we cannot form a countable local base for each point under the discrete topology. Step 3: Second Countability of \( X \)
A space is second countable if there exists a countable basis for the topology. Since the discrete topology has uncountably many open sets (every subset of \( \mathbb{R} \) is open), \( X \) is not second countable. Therefore, statement (C) is false. Final Answer:
\[ \boxed{(A), (B)} \]