Question

Let \( X \) be an uncountable set. Let the topology on \( X \) be defined by declaring a subset \( U \subset X \) to be open if \( X - U \) is either empty or finite or countable, and the empty set to be open. Then, which of the following is/are TRUE?

• A. Every compact subset of \( X \) is closed
• B. Every closed subset of \( X \) is compact
• C. \( X \) is \( T_1 \) (singleton subsets are closed) but not \( T_2 \) (Hausdorff)
• D. \( X \) is \( T_2 \) (Hausdorff)

Hint:

In the cofinite topology, compact sets are always closed because their complement is cofinite. Also, a space is \( T_1 \) if all singletons are closed, but a space is Hausdorff if any two distinct points can be separated by disjoint open sets. In the cofinite topology, this is not the case.

Answer 1

The Correct Option is A, C

Step 1: Understanding the Topology on \( X \)
The topology on \( X \) is defined such that a subset \( U \subset X \) is open if its complement \( X - U \) is either empty, finite, or countable. This is a specific topology called the cofinite topology. In this topology:
Every cofinite subset of \( X \) is open.
Singleton sets are closed because their complement is cofinite (i.e., countable).
A set is compact in this topology if every open cover has a finite subcover. Since the open sets are cofinite, compact sets are closed.

Step 2: Compact Subsets and Closedness (Option A)
In the cofinite topology, every compact subset of \( X \) is closed. This is because the complement of a compact set is cofinite, and the complement of a cofinite set is finite or countable, which is the condition for closed sets in this topology.

Step 3: Closed Subsets and Compactness (Option B)
Every closed subset of \( X \) is not necessarily compact. A closed set can be uncountable, and uncountable sets in this topology do not satisfy the compactness condition. Thus, Option B is not true.

Step 4: \( T_1 \) and \( T_2 \) Conditions (Option C and D)
The space is \( T_1 \) because singleton sets are closed. However, it is not Hausdorff because two distinct points cannot be separated by disjoint open sets in this topology. Thus, Option C is true, and Option D is false.

Step 5: Conclusion
Thus, the correct answers are \( \boxed{A} \) and \( \boxed{C} \).

Final Answer
\[ \boxed{A \quad \text{Every compact subset of } X \text{ is closed}} \] \[ \boxed{C \quad X \text{ is } T_1 \text{ (singleton subsets are closed) but not } T_2 \text{ (Hausdorff)}} \]