Q20·Section II (Mathematics)·20 of 145Adding the limit point 0 makes {1/n}\{1/n\}{1/n} compactLet E={0}∪{1/n:n∈N}E=\{0\}\cup\{1/n:n\in\mathbb{N}\}E={0}∪{1/n:n∈N} and F={1/n:n∈N}F=\{1/n:n\in\mathbb{N}\}F={1/n:n∈N}, then: OptionsAE is a compact subset whereas F is not compactBE and F are compact subsets of R\mathbb{R}RCF is compact subset whereas E is not compactDNeither E nor F is compactPick an option to check your answer.Show answerShow solution Report a problem with this question← Q19All questionsQ21 →Keyboard: A–D answer · S solution · ←/→ previous/next