• anonymous
This is a really stupid question, but is the set \(\{0\}\) compact?
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
I think it is, since the only sequence in it would be \((0,0,0,...)\) and its subsequences would be the same.
  • JamesJ
In the real numbers, any set that is bounded and closed is compact. Hence any finite set is compact.
  • anonymous
Just to give you a hint for further research: The theorem JamesJ is using is called the "Heine Borel" theorem if I'm not mistaken. You could also argue that the topology induced by the usual topology on R on a singleton set (like \(\{0\}\)) is the trivial topology consisting only of the point itself and the empty set. Therefore the only open cover of the singleton subspace is the singleton itself which is countable. This proves compactness directly by its definition: Any open cover of a singleton set has the singleton set itself as a countable subcover.

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