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math771

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(Title should be Connectedness of...)

Hi. I'm trying to prove that if a set of Dedekind cuts is bounded, it has a least upper bound. We've defined a Dedekind cut, called E, to be a nonempty subset of Q (i) with no last point, (ii) an upper bound in Q, and (iii) the property that if x belongs to Q and y belongs to E, then x < y implies that x belongs to E. And we've defined A < B for Dedekind cuts to mean that A is a subset of B.

I have made some tentative observations without any idea of how the whole proof will look. Dedekind cuts are open because of properties (i) and (iii). The union of a collection of Dedekind cuts will also be open then. In fact, it seems that the union of a bounded collection of Dedekind cuts would itself be a Dedekind cut. Furthermore, no upper bound of a Dedekind cut belongs to it (because of its openness). The same is true for a collection of Dedekind cuts. The complement of any Dedekind cut (the set of upper bounds of that cut) will be closed.

I'm not sure how to proceed. Any advice would be much appreciated. Thanks!

Hi. I'm trying to prove that if a set of Dedekind cuts is bounded, it has a least upper bound. We've defined a Dedekind cut, called E, to be a nonempty subset of Q (i) with no last point, (ii) an upper bound in Q, and (iii) the property that if x belongs to Q and y belongs to E, then x < y implies that x belongs to E. And we've defined A < B for Dedekind cuts to mean that A is a subset of B.

I have made some tentative observations without any idea of how the whole proof will look. Dedekind cuts are open because of properties (i) and (iii). The union of a collection of Dedekind cuts will also be open then. In fact, it seems that the union of a bounded collection of Dedekind cuts would itself be a Dedekind cut. Furthermore, no upper bound of a Dedekind cut belongs to it (because of its openness). The same is true for a collection of Dedekind cuts. The complement of any Dedekind cut (the set of upper bounds of that cut) will be closed.

I'm not sure how to proceed. Any advice would be much appreciated. Thanks!

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