So I'm going through Velleman's book "how to prove it: a structured approach", I'm kinda confused about exercise 22 at section 4.4 It goes like this:
"Suppose that R is a partial order on A,
$B_1$⊆ A, $B_2$⊆ A,
$x_1$ is the least
upper bound of $B_1$,
$x_2$ is the least upper bound of $B_2$.
Prove that if
$B_1$ ⊆ $B_2$ then {$x_1$,$x_2$}∈ R ."
Now since I can think of an example when this is not true I suspect I must have gotten something wrong, maybe what an upper bound is?
My example:
let A={1, 2, 3, 4, 5, 132, 264, 420, 2940}
Let R={(x,y) ∈ A × A| x divides y }
then since R is reflexive, transitive and antisimmetric we can conclude that R is a partial order on A.
Now let $B_1$={ 2,3,4 } and $B_2$ = { 2,3,4,5 },
clearly $B_1$ ⊆A , $B_2$ ⊆ A and $B_1$⊆$B_2$.
Now I believe 132 is an upper bound of $B_1$
that is:
132∈A and ∀ z if z ∈ $B_1$ then (z,132) ∈ R
now let H ={h| h is an upper bound of $B_1$ } it seems to me that 132 is the smallest element of H and I think this is what "least upper bound of $B_1$" means.
Then the same line of reasoning would apply to 420 so I think 420 is the least upper bound of $B_2$
But (132,420) ∉R
I'm probably making the most stupid mistake but could someone point out what I got wrong?