By Bilusyak N. I., Ptashnyk B. I.

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**Additional resources for A Boundary-Value Problem for Weakly Nonlinear Hyperbolic Equations with Data on the Entire Boundary of a Domain**

**Sample text**

Hence either a = x, or a ∈ x, or x ∈ a. But by lemma 29, applied to the element chain a ∈ b ∈ x, we must have a ∈ x. Proposition 32 A transitive proper subset c of an ordinal d is an element of that ordinal. Proof Since d is founded, there is y ∈ d − c with y ∩ (d − c) = ∅. We claim that c = y. Since d is transitive, y ⊂ d, and by construction of y, y ⊂ c. Conversely, let b ∈ c, then b ∈ d. Since d is alternative, either b ∈ y or b = y or y ∈ b. But b = y implies b ∈ d − c, a contradiction. Further, y ∈ b and b ∈ c imply y ∈ c by transitivity of c.

In fact, d ∩ (p − f (d)) = (q ∪ f (d)) ∩ (p − f (d)) = (q∩(p−f (d)))∪(f (d)∩(p−f (d))) = ∅∪∅ = ∅ because we suppose p∩q = ∅. So we have a disjoint union q ∪ p = d ∪ (p − f (d)). Now, we have a bijection ∼ ∼ f : d → f (d) and a bijection (the identity) p − f (d) → p − f (d). Therefore by sorite 20 (ii), we obtain the required bijection. This implies a famous theorem: Proposition 22 (Bernstein-Schröder) Let a, b, c be three sets such that there exist two injections f : a → b and g : b → c. If a and c are equipollent, then all three sets are equipollent.

V) and (vi) are immediate from the definitions. (vii) in view of (vi), w = y + z is a solution. For any two solutions w + y = w + y, one has w = w + y + y = w + y + y = w . y z y z x x · (y + z) x x·y +x·z Fig. 3. Distributivity of · over +. Remark 3 This structure will later be discussed as the crucial algebraic structure of a commutative ring, see chapter 15. Exercise 2 Let a = {r , s, t} as in exercise 1. Calculate the solution of w + y = z within 2a for y = {r , s} and z = {s, t}. Exercise 3 Let a = {∅}.

### A Boundary-Value Problem for Weakly Nonlinear Hyperbolic Equations with Data on the Entire Boundary of a Domain by Bilusyak N. I., Ptashnyk B. I.

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