Test1 - 1 Russell’s Paradox If S is a set there are two possibilities S ∈ S or S ∉ S Let G = S S is a set& S ∉ S Let B = S S is a

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Unformatted text preview: 1 Russell’s Paradox If S is a set, there are two possibilities. S ∈ S or S ∉ S . Let G = { S ∶ S is a set & S ∉ S } . Let B = { S ∶ S is a set & S ∈ S } C1: If G ∈ G then G is a set & G ∉ G C2: If G ∉ G , then G ∈ G 2 Functions A function f ∶ A → B ∋ ∀ x ∈ A → f ( x ) ∈ B . f maps to f ( x ) . f ( x ) is the image of x under f . x must map to a single f ( x ) . Definition 1 . A function f ∶ A → B is injective (or one-to-one) If, ∀ x,y ∈ A, f ( x ) = f ( y ) → x = y Definition 2 . A function f ∶ A → B is surjective (onto) If, range ( f ) = B 3 Logic/Truth Tables False implies True. 4 Definitions Definition 3 . A function f is continuous at a point x if, ∀ ǫ > , ∃ δ > 0, such that ∀ y , if divides.alt0 x − y divides.alt0 < δ then divides.alt0 f ( x ) − f ( y )divides.alt0 < ǫ . Definition 4 . A function f is not continuous at x when, ∃ ǫ > , ∀ δ > , ∃ y such that divides.alt0 x − y divides.alt0 < δ , but that ǫ ≤ divides.alt0≤ divides....
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This note was uploaded on 01/23/2012 for the course MATH 3012 taught by Professor Costello during the Fall '08 term at Georgia Institute of Technology.

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