Consider the following topologies on the set \( \mathbb{R} \) of all real numbers.
\[
T_1 \text{ is the upper limit topology having all sets } (a, b) \text{ as basis.}
\]
\[
T_2 = \{ U \subset \mathbb{R}: U \text{ is finite} \} \cup \{\emptyset\}.
\]
\[
T_3 \text{ is the standard topology having all sets } (a, b) \text{ as basis.}
\]
Then:
Show Hint
- \( T_2 \subset T_3 \subset T_1 \)
- \( T_1 \subset T_2 \subset T_3 \)
- \( T_3 \subset T_2 \subset T_1 \)
- \( T_2 \subset T_1 \subset T_3 \)
The Correct Option is A
Solution and Explanation
- \( T_3 \) is the standard topology on \( \mathbb{R} \), which also has all open intervals \( (a, b) \) as its basis.
The standard topology \( T_3 \) is finer than the upper limit topology \( T_2 \) because every open set in \( T_2 \) is also an open set in \( T_3 \), but not the reverse. Since \( T_3 \) is finer than \( T_2 \), and \( T_1 \) is the coarser topology that allows open sets in the form of \( (a, b) \), the correct relation is \( T_2 \subset T_3 \subset T_1 \). Final Answer: (A) \( T_2 \subset T_3 \subset T_1 \)
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Then
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