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

To solve this problem, let's analyze the provided options based on the topology defined on the set \( X \). The topology declares a subset \( U \subset X \) to be open if \( X - U \) (the complement of \( U \)) is either empty, finite, or countable, and the empty set is also open. 

1. Assessing the Option: Every compact subset of \( X \) is closed
   • A subset \( K \subset X \) is closed if \( X - K \) is open, meaning \( X - K \) is empty, finite, or countable.
   • A subset is compact if every open cover has a finite subcover. In this topology, compact subsets are characterized by being finite because any infinite subset won't be compact, given the requirement for finite subcovers.
   • Since finite subsets are automatically closed because their complements are open (being the entire set minus a finite part), this statement holds true.
2. Assessing the Option: Every closed subset of \( X \) is compact
   • For the given topology, a closed subset \( F \subset X \) means \( X - F \) is open (i.e., empty, finite, or countable).
   • A closed set is compact only if it satisfies the compactness criterion. However, a closed subset can be uncountable, exceeding the bounds of compactness criteria. Thus, not all closed sets in this topology are compact.
   • Therefore, this statement is false.
3. Assessing the Option: \( X \) is \( T_1 \) (singleton subsets are closed) but not \( T_2 \) (Hausdorff)
   • A topology is \( T_1 \) if all singleton sets are closed. In our definition, a singleton \( \{x\} \subset X \) is closed if \( X - \{x\} \) is open, i.e., empty, finite, or countable. Since \( X \setminus \{x\} \) is typically infinite but countable or finite, every singleton set is closed, confirming \( T_1 \) property.
   • Hausdorff spaces (\( T_2 \)) require any two distinct points to have disjoint neighborhoods. In this topology, it's not generally possible to find disjoint open sets for any two distinct elements, as many choices won't meet the open set criteria simultaneously. Hence, it's not \( T_2 \).
   • This statement is true.
4. Assessing the Option: \( X \) is \( T_2 \) (Hausdorff)
   • As concluded from the previous option, \( X \) does not satisfy the \( T_2 \) Hausdorff condition due to its coarse nature, where distinguishing points with disjoint open neighborhoods isn't feasible.
   • This statement is false.

Therefore, the correct answers are: Every compact subset of \( X \) is closed, and \( X \) is \( T_1 \) (singleton subsets are closed) but not \( T_2 \) (Hausdorff).