By Ziemer W.
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E. , v = w). If domain h = V , then domain h is an initial segment of V and range h = W and the existence of h−1 in this case establishes v > w. 3. Theorem. The class of ordinal numbers is well-ordered. Proof. Let S be a nonempty set of ordinal numbers. Let α ∈ S and set T = {β ∈ S : β < α}. If T = ∅, then α is the least element of S. If T = ∅, let W be a well-ordered set such that α= ord(W ). 3. ORDINAL NUMBERS 35 ord(Wβ ), and there is a unique xβ ∈ W such that Wβ is isomorphic to the initial segment W (xβ ) of W .
1. TOPOLOGICAL SPACES 41 b are arbitrary negative real numbers, are open in the induced topology on Y , but none of them is open in the topology on X. However, all intervals J of the form J = {x = (x1 , x2 ) : x1 = 0, a ≤ x2 ≤ b} are closed both in the relative topology and the topology on X. 1. Theorem. Let (X, T ) be a topological space. Then (i) The union of an arbitrary collection of open sets is open. (ii) The intersection of a finite number of open sets is open. (iii) The union of a finite number of closed sets is closed.
I=1 44 3. 4. Definition. A family {Eα : α ∈ I} of subsets of a set X is said to have the finite intersection property if for each finite subset F ⊂ I Eα = ∅. 1. Lemma. A topological space X is compact if and only if every family of closed subsets of X having the finite intersection property has a nonempty intersection. Proof. First assume that X is compact and let {Cα } be a family of closed sets with the finite intersection property. Then {Uα } := {X \Cα } is a family, F, of open Cα were empty, then F would form an open covering of X and therefore sets.