Formulation
Given a non-empty totally ordered set R with the following four properties:
- R does not have a least nor a greatest element;
- the order on R is dense (between any two elements there is another);
- the order on R is complete, in the sense that every non-empty bounded subset has a supremum and an infimum;
- every collection of mutually disjoint non-empty open intervals in R is countable (this is the countable chain condition, ccc).
Is R necessarily order-isomorphic to the real line R?
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 isomorphic to R.
Read more about this topic: Suslin's Problem
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