The countable chain condition is the & alefsym; 1-chain condition.
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This poset satisfies the countable chain condition.
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The real line also satisfies the countable chain condition : every collection of mutually independent of the standard axiomatic system of set theory known as ZFC.
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:: : I think that the real heart of this problem lies within the realm of general topology, and in particular, the countable chain condition.
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In a Polish group " G ", the ?-algebra of Haar null sets satisfies the countable chain condition if and only if " G " is locally compact.
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Knaster's condition implies a countable chain condition ( ccc ), and it is sometimes used in conjunction with a weaker form of Martin's axiom, where the ccc requirement is replaced with Knaster's condition.
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This is called the " countable chain condition " rather than the more logical term " countable antichain condition " for historical reasons related to certain chains of open sets in topological spaces and chains in complete Boolean algebras, where chain conditions sometimes happen to be equivalent to antichain conditions.
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\mathbb { P } satisfies the "'countable chain condition "'( c . c . c . ) if and only if every antichain in \ mathbb { P } is countable . ( The name, which is obviously inappropriate, is a holdover from older terminology.
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If the requirement for the countable chain condition is replaced with the requirement that " R " contains a countable dense subset ( i . e ., " R " is a separable space ) then the answer is indeed yes : any such set " R " is necessarily order-isomorphic to "'R "'( proved by Cantor ).
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:: : I am going to stop now in case I get too carried away; ) but these problems illustrate the diversity of general topology and how simple ideas such as the OP's question may lead to such interesting questions ( for instance, in question one, we know that every separable space satisfies the countable chain condition; the obvious question is whether there exists a non-separable space which satisfies the countable chain condition ( that is, does the converse hold ? ).
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