In ch. 2 of Rudin's Principles of Math Analysis, definition 2.18 gives the definition of a closed set: $E$ is closed if every limit point of $E$ is an interior point of $E$. After that, theorem 2.20 and the following corollary clearly state that a finite point set has no limit points. Yet, the immediate example below clearly states that a nonempty finite set is closed. What exactly am I missing because this sounds like a clear contradiction?
Limit points in closed finite sets?
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real-analysis
limits
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0There is no contradiction. Let $E$ be a nonempty finite set. Then $$\{ \mbox{limit points of } E \} = \emptyset \subseteq E^°$$ so $E$ is closed. – 2017-01-15
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0Something is not right about this. Are you sure that your definition of closed set is correct? – 2017-01-15
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0Crostul, is E^° the closure of E? – 2017-01-15
1 Answers
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With the definition the way you write it, the interval $ [0,1] $ would not be closed, since $0$ is not interior. What 2.18 says is that $E $ is closed if every limit point belongs to $E $ (no "interior").
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0Yes, I noticed my error. Thank you for clarifying. – 2017-01-15