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Suppose $G_i$ is closed set, how to prove $\cup_{i=1}^n G_i$ is closed.

Please do not use $(\cup_{i=1}^n G_i)^c = \cap_{i=1}^n ({G_i}^c)$. That is you could not use the theorem that a set is closed if and only if the complement of the set is open.

The question is derived from Rudin's PMA. The definitions are

(a) If every point of a set is interior point, then the set is open.
(b) If every limit point of a set is in the set, then the set is closed.

Becase the caption of the chapter is topology, so I use the tag general-topology. However, The book have not define topology or order topology.

Sorry for my bad English, I didn't mean to offend you by saying that.

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    @Jichao: By the way, it is not true that $\cup_{i=1}^n G_i = \cap_{i=1}^n G_i^c$; what *is* true is that $(\cup_{i=1}^n G_i)^c = \cap_{i=1}^n(G_i^c)$.2011-08-18

2 Answers 2

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Assuming your definition is a set is closed if it contains all its limit points, you can argue as follows.

First we show that if $x$ is a limit point of $\cup_{i=1}^n G_i$, then $x$ is a limit point of at least one of the $G_i$'s. If this were not the case, then we could find open sets $U_1, \ldots U_n$ with $x \in U_j$ and $U_j \cap G_j = \emptyset$ for all $j$. Then the open set $\cap_{i=1}^n U_i$ is an open set containing $x$ that has empty intersection with $\cup_{i=1}^n G_i$. This contradicts the fact that $x$ is a limit point of $\cup_{i=1}^n G_i$ and thus the assumption that $x$ was not a limit point of any of the $G_i$'s is wrong.

Now if $x$ is a limit point of $\cup G_i,$ as we just saw, it is a limit point of one of the $G_i$'s. Since $G_i$ is closed, it contains all its limit points and thus $x \in G_i \subset \cup_{j=1}^n G_j.$ So $\cup G_i$ contains all its limit points and as a result is closed.

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    @user14467: You might want to add this to the body of your question; then perhaps some of the downvoters will retract those votes (they cannot if the question has not been edited since they downvoted).2011-08-18
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Let's take the following definition: a subset $S$ of ${\bf R}^n$ is called closed if every sequence of points in $S$ that converges in ${\bf R}^n$ converges to something in $S$. Now let $S$ be the union of a finite number of closed sets, and suppose you have a sequence of points of $S$ converging in ${\bf R}^n$. Then at least one of those finitely many closed sets contains infinitely many terms in the sequence. By standard stuff that infinite subsequence converges, and converges to the same point as the original sequence; since the subsequence is all in one of the closed sets, it converges to something in that set, and, thus, to something in $S$, QED.