Countably Compact

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Homework Statement


Show that ##X## is countably compact if and only if every nested sequence ##C_1 \supset C_2 \supset ...## of closed nonempty sets of ##X## has a nonempty intersection.

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The Attempt at a Solution



Suppose that ##X## is not countably compact. Then there exists a countable open cover ##\{U_n\}## of ##X## that has no finite subcover. Consider the collection of closed sets ##C_n = X - (U_1 \cup ... \cup U_n)##; If ##x \in C_{n+1}##, then ##x \in X-U_i## for every ##i=1,....,n+1##, so in particular ##x \in X-U_i## for ##i=1,...,n##. Moreover, this collection consists entirely of nonempty sets, for if ##C_n =X - (U_1 \cup ... \cup U_n)## were empty, then we would have a finite subcover. Hence, ##\{C_n\}## is a nested sequence of closed nonempty sets. By way of contradiction, suppose that the intersection is nonempty. Then ##x \in X - (U_1 \cup ... \cup U_n)## for every ##n##. Since ##\{U_n\}## is a cover, ##x## must be in ##\bigcup U_i##, which implies there exists a ##k## such that ##x \in U_k \subseteq U_1 \cup ... \cup U_k##. This contradicts the fact that ##x \in X - (U_1 \cup ... \cup U_k)##. Hence, the intersection has to be empty.


Now, suppose that there exists a nested sequence ##\{C_i\}## of nonempty closed sets whose intersection is empty. Then ##U_i = X-C_i## forms a collection of open sets, and since ##\bigcup U_i = \bigcup (X-C_i) = X - \cap C_i = X - \emptyset = X##, we see moreover that it is an open cover. Now, if there were to exist a finite subcover, say ##\{U_{k_1},...,U_{k_n} \}##, where ##k_1 \le ... \le k_n##, then ##X \subseteq \bigcup U_{k_i} = X - \bigcap C_{k_i} = X - \bigcap C_{k_i} = X - C_{k_n}##, implying that ##C_{k_n} = \emptyset##, which is a contradiction. Hence, there cannot be a finite subcover.


How does this sound?
 

Answers and Replies

  • #2
andrewkirk
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Yes that proof looks sound to me, and nicely intuitive too.
 

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