1st December 2004

Munkres §10

Ex. 10.1. If a subset of a well-ordered set has an upper bound, the smallest upper bound is a least upper bound (supremum) for the set. (This proof is a tautology!)

Ex. 10.2.

(a). The smallest successor x_{+} of any element x is the immediate successor. (The iterated successors of x has the order type of a section of Z_{+}.)

(b). Z.

Ex. 10.4.

(a). Let A be a simply ordered set containing a subset with the order type of Z_{−}. Then this subset does not have a smallest element so A is not well-ordered. Conversely, let A be simply ordered set containing a nonempty subset B with no smallest element. Let b_{1} be any element of B. Since b_{1} is not a smallest element of B there is some element b_{2} of B such that b_{2} < b_{1}. Continuing inductively we obtain an infinite descending chain ⋯ < b_{n+1} < b_{n} < ⋯ < b_{2} < b_{1} forming a subset of the same order type as Z_{−}.

(b). A does not contain a subset with the order type of Z_{−}.

Ex. 10.6.

(a). For any element α of S_{Ω}, the set {x ∈ S_{Ω} | x ≤ α} = S_{α} ∪ {α} is countable but S_{Ω} itself is uncountable [Lemma 10.2].

(b). For any element α ∈ S_{Ω}, the set S_{α} ∪ {α} is countable so its complement, {x ∈ S_{Ω} | x > α} = (α, +∞), in the uncountable set S_{Ω}, is uncountable [Lemma 10.2, Thm 7.5].

(c). We show the stronger statement [Thm 10.3] that X_{0} is not bounded from above. We do this by assuming that X_{0} has an upper bound α and find a contradiction. The (non-empty) simply ordered set (α, +∞) is well-ordered [p. 63], it has no largest element by (a), and each element of (α, +∞), except the smallest element, has an immediate predecessor. Thus (α, +∞) has the order type of Z_{+}, in particular (α, +∞) is countable, contradicting (b). (Let x be any element of (α, +∞). Since (α, +∞) does not contain an infinite descending chain [Ex 10.4], α is an iterated immediate predecessor of x and x is an iterated immediate successor of α.)

Ex. 10.7. We show the contrapositive. Let J_{0} be any subset of J that is not everything. Let α be the smallest element of the complement J − J_{0}, the smallest element outside J_{0}. This means that α ∉ J_{0} and that any element smaller than α is in J_{0}, i.e. S_{α} ⊂ J_{0}. Thus J_{0} is not inductive.

References