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Question
Math
Posted 9 months ago
Let X be a topological space, and let A, B, and C be nonempty subsets of X such that A c B c C; that is, A is a proper subset of B, and B is a proper subset of C. Which of the following statements is (are) true? I. If A and C are connected, then B is connected. II. If A and C are compact, then B is compact. III. If A and C are Hausdorff, then B is Hausdorff.
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Answer from Sia
Posted 9 months ago
Solution by Steps
step 1
Consider the given subsets A A , B B , and C C of a topological space X X with ABC A \subset B \subset C
step 2
Analyze statement I: If A A and C C are connected, then B B is connected
step 3
Since A A is connected and AB A \subset B , any disconnection of B B would lead to a disconnection of A A , which contradicts the assumption that A A is connected
step 4
Similarly, since BC B \subset C and C C is connected, B B cannot be disconnected without disconnecting C C
step 5
Therefore, B B must be connected if A A and C C are connected
step 6
Analyze statement II: If A A and C C are compact, then B B is compact
step 7
Compactness is a topological property that is hereditary for closed subsets
step 8
If B B is a closed subset of C C , then B B is compact because C C is compact. However, without additional information about the closedness of B B , we cannot conclude that B B is compact
step 9
Analyze statement III: If A A and C C are Hausdorff, then B B is Hausdorff
step 10
The Hausdorff property is hereditary for all subsets, so if C C is Hausdorff, then any subset of C C , including B B , is also Hausdorff
Answer
I and III are true. II cannot be determined to be true without additional information about the closedness of B B .
Key Concept
Topological properties and hereditary nature in subsets
Explanation
Connectedness is preserved between nested subsets if the outer sets are connected. The Hausdorff property is hereditary for all subsets. Compactness is hereditary for closed subsets, but without knowing if B B is closed in C C , we cannot conclude B B is compact.

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