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Homework Statement
9. Show that the least upper bound axiom also holds in Z (i.e., each nonempty subset of Z with
upper bound in Z has a least upper bound in Z), but that it fails to hold in Q.
http://gyazo.com/4c0b79cbb1d15cd5edf0c96ec612a55c
Homework Equations
I'll split the question into 2 parts.
The Attempt at a Solution
(a). We want to show the sup axiom holds for ##\mathbb{Z}##.
Suppose ##A \subseteq \mathbb{Z}## is nonempty and bounded above.
Let ##M = max(A)## so that ##a ≤ M, \forall a \in A## so that M is an upper bound for A.
Now I have a quick question about this before I continue. I've shown that ##M## is an upper bound, but now I have to show that if ##U \in \mathbb{Z}## is any upper bound then ##M ≤ U##.
The question is how to go about this. Should I consider the set ##-A## or is there perhaps a more straightforward alternative.
(b) I'm guessing I should assume the contrary that the sup axiom holds for ##\mathbb{Q}## and then arrive at a contradiction?