How do you prove that a topological space X with topology T is compact if it has the finite intersection property?
The finite intersection property (FIP) states that any finite intersection of open sets in X is non-empty. To prove that X is compact if it has the FIP, we use the following steps:
- Assume that X is not compact. Then, there exists an open cover of X that has no finite subcover.
- Let C be such an open cover. Since X is not compact, there exists a point x ∈ X such that every open set in C that contains x also contains another point y ∈ X that is not in x.
- For each open set O ∈ C that contains x, let O' be an open set that contains y but not x.
- The collection {O': O ∈ C} is a finite intersection of open sets in X. However, this intersection is empty since no open set in C can contain both x and y.
- This contradicts the FIP, so our assumption that X is not compact must be false. Therefore, X must be compact.
Related Questions
- What is the definition of a topological space?
- What is the definition of the finite intersection property?
- Why is the FIP important for proving compactness?
- Can a topological space be compact without having the FIP?
- What are some examples of topological spaces that have the FIP?
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