Let (a, b),(x, y) ∈ R × R and define ≺ as follows:
(a, b) ≺ (c, d) iff
a < c or a = c and b < d:
Define (a, b) ≼ (c, d) if and only if (a, b) = (c, d) or (a, b) ≺ (c, d).
Show that there is a subset of R R that has a lower bound but no greatest lower bound.
Let S = {(x, y) ∈ (Q+)x (Z+) | P ∈ irrationals and Z ∈ positive integers {x > 1, y = 1}
Claim: (0, 0) is a lower bound for S because 0 < x because x ∈ P+.
Thus (0, 0) ≼ (x, y) by the given order.
Assume that (b, 1) is the greatest lower bound for S.
I know I need to prove there is no greatest lower bound in the set, but I'm not sure how to. Any guidance would be appreciated.
First of all, note that a point $(x,y)$ will be a lower bound if and only if $x\leq 1$ (why?).
With that in mind, suppose that $(x,y)$ is a lower bound. What can we say about $(x,y+1)$?
For future reference, this way of ordering a set of numbers is called Lexicographical order.