How Do Nonempty Closed Sets Intersect in Compact Spaces?

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SUMMARY

In a compact topological space, any decreasing sequence of nonempty closed sets has a non-empty intersection. This conclusion is established through the application of the Bolzano-Weierstrass theorem, which asserts that every sequence in a compact space has a convergent subsequence. The discussion highlights the importance of compactness in topology and its implications for closed sets.

PREREQUISITES
  • Understanding of compact topological spaces
  • Familiarity with closed sets in topology
  • Knowledge of the Bolzano-Weierstrass theorem
  • Basic principles of sequences in mathematical analysis
NEXT STEPS
  • Study the properties of compact spaces in topology
  • Explore the implications of the Bolzano-Weierstrass theorem
  • Investigate the relationship between closed sets and convergence
  • Learn about other theorems related to intersections of sets in topology
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Mathematicians, students of topology, and anyone interested in the properties of compact spaces and closed sets in mathematical analysis.

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Here is this week's POTW:

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Prove that in a compact topological space, any decreasing sequence of nonempty closed sets has non-empty intersection.

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Congratulations to castor28 for his correct solution, which is as follows:
Let $E$ be a compact topological space and $\{A_i\}$ a decreasing sequence of non-empty closed sets; $A_i^c$ is therefore an increasing sequence of open proper subsets of $E$.

Assume that $\bigcap A_i= \emptyset$. $\bigcup A_i^c=E$, and the $A_i$ constitute an open cover of $E$. As $E$ is compact, $\{A_i^c\}$ contains a finite sub-cover $\{A_{i_1}^c,\ldots,A_{i_n}^c\}$ whose union is $E$. As the $A_i^c$ constitute an increasing sequence, we have $ A_{i_n}^c=E$, and this contradicts the fact that the $A_i^c$ are proper subsets (because the $A_i$ are non-empty sets).
 

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