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?


1 Answer 1


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.

  • $\begingroup$ I've edited the question with an example to illustrate my thinking. As I see it, I could take an infinite chain as subet. The chain doesn't have an upper bound since it isn't closed under union $\endgroup$
    – nz_21
    Commented May 18, 2019 at 15:27
  • 1
    $\begingroup$ @NZ_21 See my edit. $\endgroup$ Commented May 18, 2019 at 15:38

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