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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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Please don't post orders. If you have a question, pose it as a question. –  Arturo Magidin Aug 18 '11 at 3:46
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What are your definitions of "open" and "closed"? For example, I would consider the definition of "closed" to be "complement of an open set", so this matters quite a bit. Also, I assume we are working in $\mathbb{R}$ with its usual topology? –  Zev Chonoles Aug 18 '11 at 3:46
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@Jichao: In most books, the fact that a set is closed if and only if its complement is open is not a theorem, it's the definition. So clearly, you have a different definition. Please provide the definition you have of closed sets when you edit the question to get rid of the imperative. Thanks. –  Arturo Magidin Aug 18 '11 at 3:47
    
@Arturo: I'd expect a book about metric spaces (or even analysis focused entirely on $\mathbb{R}^n$ - indeed anything preceding general topology) to give a different definition of open set, i.e. a set is said to be open if every point is an interior point. And a point is said to be an interior point of a set if there is an open ball centered at the point contained in the set. In that setting the fact that finite intersections of open sets are open is indeed a theorem. –  kahen Aug 18 '11 at 4:01
    
@kahen: granted: if it is a problem from real analysis, I can see that. But the label does say "general topology". All the topology books I have give the complement definition... In any case, if we don't have the definition being used in this particular instance, it seems very hard to help/obey Jochao's orders. –  Arturo Magidin Aug 18 '11 at 4:09

2 Answers 2

up vote 3 down vote accepted

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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How do you show 'If $x$ is a limit point of $\cup G_i$, it is a limit point of one of the $G_i$' –  jspecter Aug 18 '11 at 4:09
    
It is not obvious that a limit point of a union is a limit point of a single one of the $G_i$; you need to prove that (you know every neighborhood of the point intersects some $G_i$, but why would all of them necessarily intersect the same $G_i$?) –  Arturo Magidin Aug 18 '11 at 4:11
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Humor me; The question deals with first principles as is. –  jspecter Aug 18 '11 at 4:23
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but if $x$ is a limit point of $\cup G_i$ and not a limit point of any of the $G_1, \ldots G_n$, then we can find open sets $U_1, \ldots U_n$ so $x \in U_j$ and $U_j \cap G_j=\emptyset$. The set $\cap U_i$ is an open set containing $x$ that does not intersect $\cup G_i$. –  ShawnD Aug 18 '11 at 4:24
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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). –  Arturo Magidin Aug 18 '11 at 4:43

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.

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