Exercise · ZW-COMPACT-SPACE-008 · ★★☆ Intermediate
To prove that \(A=(0,1)\) is not compact it suffices to exhibit:
Choose one option.
Show answer
- A. an open cover of \(A\) with no finite subcover
Explanation
Compactness demands that every open cover has a finite subcover; to deny it one failing cover is enough, e.g. \(\{(1/n,1)\}_{n\ge2}\).
Review the concept: Compact space
A topological space \(X\) is compact if every open cover \(\mathcal{U}\) of \(X\) has a finite subcover.
\[X=\bigcup_{U\in\mathcal{U}}U\ \Rightarrow\ X=U_1\cup\dots\cup U_n,\ U_i\in\mathcal{U}\]
More practice problems: compact space
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- ZW-COMPACT-SPACE-005 Which subset of ℝ is compact?
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- ZW-COMPACT-SPACE-009 Let K_1,K_2 be compact subsets of a space X. Which statement is true?