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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?

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  • $\begingroup$ There is no contradiction. Let $E$ be a nonempty finite set. Then $$\{ \mbox{limit points of } E \} = \emptyset \subseteq E^°$$ so $E$ is closed. $\endgroup$
    – Crostul
    Jan 15, 2017 at 10:43
  • $\begingroup$ Something is not right about this. Are you sure that your definition of closed set is correct? $\endgroup$
    – BigbearZzz
    Jan 15, 2017 at 10:54
  • $\begingroup$ Crostul, is E^° the closure of E? $\endgroup$
    – user143074
    Jan 15, 2017 at 11:01

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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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  • $\begingroup$ Yes, I noticed my error. Thank you for clarifying. $\endgroup$
    – user143074
    Jan 15, 2017 at 11:15

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