Proving Q is a Lattice but Not a (sigma)-Lattice

[beeftrax]

I'm reading "A Course in Advanced Calculus" by Robert Borden, and one of the problems begins as follows:

"Prove that the field Q is a lattice, but not a (sigma)-lattice, under the usual order" (pg.25)

Q is of course the rational numbers.

However, Q doesn't seem to be a lattice, since the supremum of, say, [0,1] doesn't exist, since given any upper bound eg 1.1, a smaller upper bound eg 1.01 that is still in Q can be found.

So is Q not in fact a lattice, or am I missing something?

I apologize if this is in the wrong forum.

 

[Hurkyl]

1 is a least upper bound of [0, 1]

 

[beeftrax]

It is, isn't it. I feel silly. At the risk of getting another simple answer to a stupid question, what about an interval between 0 and a positive irrational number, say sqrt(2). Does the supremum of such an interval lie within Q?

 

[beeftrax]

On further thought, I'll answer my own question (or try to). The subset I described isn't finite, so it's lack of a supremum doesn't mean that Q isn't a lattice.

 

[Hurkyl]

Exactly right.