By C.A. Reiter, W.R. Jones

This publication could be of curiosity to arithmetic scientists operating within the components of linear algebra, summary algebra, quantity conception, numerical research, operations examine and mathematical modelling

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**Example text**

Special binary relations . . . . . . . . . . . . . . . . . 5. Equivalence relations and partitions . . . . . . . . . . . . . 6. Ordered sets . . . . . . . . . . . . . . . . . . . . 7. An introduction to cardinality . . . . . . . . . . . . . . . 8. Isomorphisms of ordered sets. Ordinal numbers . . . . . . . . . . 9. Application: relational databases . . . . . . . . . . . . . . 80 Set theory is one of the most fundamental parts of modern mathematics.

Thus the number of ways to form the subset A is 2 × 2 × × · · · × 2 = 2n . 8 implies that if U is a universal set, then ∪, ∩ and ′ are operations on (U). ) Thus, we obtain an algebraic structure ⟨???? (U); ∪‚ ∩ ‚ ′ ‚ ∅‚ U⟩, called the powerset Boolean algebra of U. 9 (Properties of the Cartesian product) For any sets A, B and C: 1. A × (B ∪ C) = (A × B) ∪ (A × C), (B ∪ C) × A = (B × A) ∪ (C × A) 2. A × (B ∩ C) = (A × B) ∩ (A × C), (B ∩ C) × A = (B × A) ∩ (C × A) 3. A × (B − C) = (A × B) − (A × C) Proof: We prove the identity 1.

3. In particular, the composition of bijections is a bijection. 4. If gf is injective then f is injective. 5. If gf is surjective then g is surjective. Proof: 1. Suppose f and g are injective and that a1 ‚ a2 ∈ A such that gf (a1 ) = gf (a2 ). We must show that a1 = a2 . Thus g(f (a1 )) = g(f (a2 )). However, since g is injective this means that f (a1 ) = f (a2 ) which in turn means that a1 = a2 since f is injective. 2. Exercise. 3. Follows immediately from 1 and 2. 4. Exercise. 5. Suppose that gf is surjective and that c ∈ C.