Question

Consider the following topologies on the set \( \mathbb{R} \) of all real numbers:
\( T_1 = \{ U \subset \mathbb{R} : 0 \notin U \text{ or } U = \mathbb{R} \} \),
\( T_2 = \{ U \subset \mathbb{R} : 0 \in U \text{ or } U = \emptyset \} \),
\( T_3 = T_1 \cap T_2 \).
Then the closure of the set \( \{1\} \) in \( (\mathbb{R}, T_3) \) is

• A. \( \{1\} \)
• B. \( \{0, 1\} \)
• C. \( \mathbb{R} \)
• D. \( \mathbb{R} \setminus \{0\} \)

Hint:

In topologies, the closure of a set includes all points that cannot be separated from the set by open sets. In particular, sets in the intersection of topologies may have different closures.

Answer 1

The Correct Option is C

The topology \( T_3 \) is the intersection of \( T_1 \) and \( T_2 \), so it consists of sets that either contain 0 or are the entire real line. The closure of a set in a given topology is the smallest closed set containing that set. In \( T_3 \), since \( 1 \) is contained in every open set that contains 0, its closure will include all points except 0. Thus, the closure of \( \{1\} \) in \( (\mathbb{R}, T_3) \) is \( \mathbb{R} \setminus \{0\} \), corresponding to option (D).