Show that if A is a non-empty set with lowest upper bound L and A is a subset of

Question

Show that if A is a non-empty set with lowest upper bound L and A is a subset of B and B has a lowest upper bound M, then L <= M

Explanation / Answer

This is surely pretty straightforward. All we have to do is demonstrate a subset of Q that has an upper bound but no least upper bound (in Q). [Since Q is a subset of R the set will obviously have a least upper bound in R: i.e. it has to have an irrational least upper bound, as a subset of R.] So, the most obvious example is X = {x ? Q: x^2 < 2} - i.e. the set of rationals between -v2 and v2, but expressed in a way that doesn't require irrational numbers. This obviously has an upper bound, 2 is an easy one: if x > 2 then x^2 > 4 so x ? X. So 2 is an upper bound for X. However, it doesn't have a least upper bound in Q. (Obviously we know the l.u.b. in R is v2.) We have to prove this. Suppose z is an upper bound for this set. Then first of all z is positive and z^2 > 2 (since for any rational less than v2 we know we can find another one in between it and v2; I'm omitting the tedious algebraic details here). So we want to find another rational number y such that y^2 > 2 and 0 2 and y>0 is enough to show y is an upper bound.) I'll borrow Wikipedia's example here rather than trying to recreate one: y = z - [(z^2-2) / (z+2)]= 2 (z+1) / (z+2). Since z is rational and positive, so is y; from the first expression we have that y 0 and z+2 > 0. It remains to show that y^2 > 2. Now y^2 = 4 (z+1)^2 / (z+2) ^2 so y^2 - 2 = [4z^2 + 8z + 4 - 2z^2 - 8z - 8] / (z+2)^2 = (2z^2 - 4) / (z+2)^2 = 2 (z^2 - 2) / (z+2)^2 Since z^2 > 2, it follows that y^2 - 2 > 0 and hence y^2 > 2. So for any rational upper bound z we can find a smaller rational upper bound y. Hence X does not have a least upper bound in Q, and so Q does not have the least upper bound property.

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