Question

Math

Posted 4 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 4 months ago

Solution by Steps

step 1

Consider the given subsets $A$, $B$, and $C$ of a topological space $X$ with $A \subset B \subset C$

step 2

Analyze statement I: If $A$ and $C$ are connected, then $B$ is connected

step 3

Since $A$ is connected and $A \subset B$, any disconnection of $B$ would lead to a disconnection of $A$, which contradicts the assumption that $A$ is connected

step 4

Similarly, since $B \subset C$ and $C$ is connected, $B$ cannot be disconnected without disconnecting $C$

step 5

Therefore, $B$ must be connected if $A$ and $C$ are connected

step 6

Analyze statement II: If $A$ and $C$ are compact, then $B$ is compact

step 7

Compactness is a topological property that is hereditary for closed subsets

step 8

If $B$ is a closed subset of $C$, then $B$ is compact because $C$ is compact. However, without additional information about the closedness of $B$, we cannot conclude that $B$ is compact

step 9

Analyze statement III: If $A$ and $C$ are Hausdorff, then $B$ is Hausdorff

step 10

The Hausdorff property is hereditary for all subsets, so if $C$ is Hausdorff, then any subset of $C$, including $B$, is also Hausdorff

Answer

I and III are true. II cannot be determined to be true without additional information about the closedness of $B$.

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 $B$ is closed in $C$, we cannot conclude $B$ is compact.

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