What do you mean by "dense" here? I am familar with the property of being denseinsome set but what set do you intend here? No the set of real numbers or rational numbers- that's not true.

It might help to prove first: If A and B are cuts then one and only one is true: A= B, A is a proper subset of B, or B is a proper subset of A.(b)Prove that theunionof two cuts is always a cut.

(c)Let be any collection of cuts. Prove that their union

is either a cut or else is equal to .

[/quote](d)[Define what it means for a linearly ordered set to be "complete". (I know this,if every subset of X which is bounded above has the least upper bound, X is complete) ]

Recall that the set of real numbers isdefined as the set of all cuts. Prove that is complete.

(Jason: I know I have to use the previous part some how. Every subset of is a set of cuts and these are all bounded above, so is the proof just that these cuts have the least upper bound in ? And if so, how are they cuts?!)p[./qu0te]

If A is such a set of cuts, then, as above, the union of all cuts in A is itself a cut. (You need the upper bound on A to show this union is not the set of all rational numbers. Show that that union is the least upper bound.

Well, that's straight forward. Just show that the three properties you listed originally are true for this set.(e)For a rational number r with the followingdefinition

I need to prove that is acut.

No, it's not the empty set. because r> 0, and contains every rational number less than 0, the intersection contains every rational number less than 0.(f)Let denotethe set of all positive rational numbers. Describe theintersection:

is it a cut? (Jason: I thought this might be something similar to the empty set? In which case it's not a cut.)

Yes. If a is a lower bound for set A then -a is an upper bound for {-x| x in A}.(g)Prove thatevery setthathas a lower bound, also has a greatest lower bound. (Jason: here I think because iscompleteand then take the negative of allcuts?)

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Help on any of this ismuchappreciated.