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The following is the definition of an open set from Rudin's Real and Complex Analysis (3rd edition):

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A closed set would be an open set's complement. From this definition, why would a closed set contain its limit points?

J. W. Tanner
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Incognito
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  • Presumably the book goes on to redefine limit points and convergence in topological-space terms (ie no metric) and start probing theorems equivalent ent to the ones for metric spaces? (Possibly including an explanation of this somewhere, if it's recognisable.) – timtfj Feb 04 '19 at 11:08
  • Topologically limit points are nearly useless. Why is something of so little importance given a central roll in metric space theory? I consider that misleading. – William Elliot Feb 04 '19 at 21:21

2 Answers2

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Let $X$ be a topological space and $S$ a closed subset of $X$.

If $x$ is a limit point of $S$, then each open set $U$ with $x \in U$ has the property that $ U \cap (S \setminus \{x\}) \ne \emptyset.$

Now suppose that $x \notin S$. Then $U:= X \setminus S $ is open and $x \in U.$

Then we have $ U \cap (S \setminus \{x\}) \ne \emptyset$ , on the other hand we have $ U \cap (S \setminus \{x\}) = \emptyset$, a contradiction.

So: $x \in S.$

Fred
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    This would be easier to follow for novices like me if an open set $U$ with $x\in U$ was called an open neighbourhood of $x$. – timtfj Feb 04 '19 at 11:28
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You have to use the topological notion of a limit: A sequence $(x_{n})$ in $X$ converges to some $x\in X$ if and only if for all open sets $V\subset X$ containing $x$ there exists an $N\in\mathbb{N}$ such that for all $n>N$ we have $x_{n}\in V$.

Now let $C$ a closed set, suppose there is some sequence $(x_{n})$ in $C$ converging to some $x\in X\setminus C$. By definition $X\setminus C$ is an open set so there exists an $N\in \mathbb{N}$ such that for all $n>N$ we have $x_{n}\in X\setminus C$, which is a contradiction.

William Elliot
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Floris Claassens
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    The question is not about limits. It is about limit points. – William Elliot Feb 04 '19 at 11:03
  • You could let x be a limit point of the closed set and construct a sequence within the closed set, that converges to x, so x is in the closed set. But that proof will fail for spaces that are not 1st countable. – William Elliot Feb 04 '19 at 11:12
  • True, or you could just work with nets follow the same method and get the result that way. Given it is an elementary question I thought it worthwhile to avoid using nets. – Floris Claassens Feb 04 '19 at 11:17