Countably compact space

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) formula_1 (2): Suppose (1) holds and "A" is an infinite subset of "X" without formula_2-accumulation point. By taking a subset of "A" if necessary, we can assume that "A" is countable.
Every formula_3 has an open neighbourhood formula_4 such that formula_5 is finite (possibly empty), since "x" is "not" an ω-accumulation point. For every finite subset "F" of "A" define formula_6. Every formula_4 is a subset of one of the formula_8, so the formula_8 cover "X". Since there are countably many of them, the formula_8 form a countable open cover of "X". But every formula_8 intersect "A" in a finite subset (namely "F"), so finitely many of them cannot cover "A", let alone "X". This contradiction proves (2).
(2) formula_1 (3): Suppose (2) holds, and let formula_13 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 formula_14 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) formula_1 (1): Suppose (3) holds and formula_16 is a countable open cover without a finite subcover. Then for each formula_17 we can choose a point formula_18 that is "not" in formula_19. The sequence formula_13 has an accumulation point "x" and that "x" is in some formula_21. But then formula_21 is a neighborhood of "x" that does not contain any of the formula_23 with formula_24, so "x" is not an accumulation point of the sequence after all. This contradiction proves (1).
(4) formula_25 (1): Conditions (1) and (4) are easily seen to be equivalent by taking complements.