Hitting the buffers with Least Upper Bound

I hit the buffers in my study of Least Upper Bounds this weekend. I was getting on ok until I looked more deeply at the strategy for determining the least upper bound of a set (Strategy 4.1, page 28 AA1).

The Strategy reads as follows:-

"Strategy 4.1. Given a subset E of ℜ, to show that M is the least upper bound, or supremum, of E, check that:

1. x≤M, for all x∈E
2. if M'<M, then there is some x∈E such that x>M' "

I had trouble with the second part.

Suppose we want to prove that the least upper bound of [0,2) is 2. The first part of the strategy is easy enough. We know that M=2 is an upper bound of [0,2) because

x≤2 for all x∈[0,2)

The second part had me bothered. The text says:-

"Suppose that M'<2. We must find an element x in [0,2) which is greater than M'. But by the Density Property, there is a real number x that is less than 2 and greater than both M' and 0. Thus x∈[0,2) and x>M', which shows that M' is not an upper bound of [0,2). Hence M=2 is the least upper bound of [0,2)."

My problem was when I started thinking about why 3 couldn't be the least upper bound according to this strategy. For example, if M'=1 (which is less than 3), then there is some x∈E such that x>M', namely x=1.5. It took ages for me to get my head round this.

The solution, of course, is that the second point of the strategy should be read as:

2. For all M'<M, then there is some x∈E such that x>M'

That 'for all' I think is implied in the fact that M' is a variable quantity. In the case where the upper bound is M=3, then step 2 isn't true when say M'=2.5 as we can't find any x in [0,2) such that x>2.5. Phew!!It definitely had me thrashing about for a few hours.

I also had to pick my way carefully through the proof of the Least Upper Bound Property of ℜ. It took some real concentration and trying it with an example before I got the gist of what was going on.

So, I am off the buffers now. AA1 is behind me and I am onto sequences in AA2.