I'm trying to solve this question
Let S be a non-empty set and consider the partially ordered set(P(S
), ⊆). Show that every subset of P(S ) has a least upper bound.
I'm not sure I agree with the premise of the question.
If I take S = N, then I won't have an upper bound for the set of all
natural numbers (since it isn't closed).
For example, if S = N, then I could consider the chain:
{ {1}, {1,2}, {1,2,3}, ...}
This doesn't have an upper bound since the union of this chain isn't closed
Am I thinking correctly?
Best Answer
You are confusing the order relation "less than" on the natural numbers with the order relation "is a subset of" on the power set.
The set of all subsets of a set is closed under union. The infinite chain in the question has union all of $ \mathbb{N} $, which is its least upper bound in the subset ordering.