In mathematics a topological space is called countably compact if every countable open cover has a finite subcover.

Equivalent definitions

A topological space X is called countably compact if it satisfies any of the following equivalent conditions: (1) \Rightarrow (2): Suppose (1) holds and A is an infinite subset of X without \omega-accumulation point. By taking a subset of A if necessary, we can assume that A is countable. Every x\in X has an open neighbourhood O_x such that O_x\cap A is finite (possibly empty), since x is not an ω-accumulation point. For every finite subset F of A define. Every O_x is a subset of one of the O_F, so the O_F cover X. Since there are countably many of them, the O_F form a countable open cover of X. But every O_F intersect A in a finite subset (namely F), so finitely many of them cannot cover A, let alone X. This contradiction proves (2). (2) \Rightarrow (3): Suppose (2) holds, and let (x_n)_n be a sequence in X. If the sequence has a value x that occurs infinitely many times, that value is an accumulation point of the sequence. Otherwise, every value in the sequence occurs only finitely many times and the set is infinite and so has an ω-accumulation point x. That x is then an accumulation point of the sequence, as is easily checked. (3) \Rightarrow (1): Suppose (3) holds and is a countable open cover without a finite subcover. Then for each n we can choose a point x_n\in X that is not in. The sequence (x_n)_n has an accumulation point x and that x is in some O_k. But then O_k is a neighborhood of x that does not contain any of the x_n with n>k, so x is not an accumulation point of the sequence after all. This contradiction proves (1). (4) (1): Conditions (1) and (4) are easily seen to be equivalent by taking complements.

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