Exercise · ZW-COMPACT-SPACE-004 · ★☆☆ Basic
Let \(X\) be an infinite set with the discrete topology. Is it compact?
Choose one option.
Show answer
- D. No: the singletons form a cover with no finite subcover
Explanation
Each \(\{x\}\) is open and indispensable to cover \(x\); finitely many singletons cover only finitely many points.
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
- ZW-COMPACT-SPACE-001 Which of these subsets of ℝ is compact?
- ZW-COMPACT-SPACE-002 Is a finite topological space compact?
- ZW-COMPACT-SPACE-003 For integers n≥ 1, the intervals (-n,n) cover ℝ. What is the smallest n with [-5,5]⊆(-n,n)?
- ZW-COMPACT-SPACE-005 Which subset of ℝ is compact?
- ZW-COMPACT-SPACE-006 Let X=\a,b,c,d,e\ with the discrete topology and the cover \a,b\,\c,d\,\e\,\b,c\,\d,e\,\a,e\. What…
- ZW-COMPACT-SPACE-007 Let X be an infinite set with the cofinite topology (the open sets are ∅ and complements of finite…
- ZW-COMPACT-SPACE-008 To prove that A=(0,1) is not compact it suffices to exhibit:
- ZW-COMPACT-SPACE-009 Let K_1,K_2 be compact subsets of a space X. Which statement is true?