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DEFINITIONS Math 355 1. For a real number x, the absolute value of x, denoted by |x|, is defined by ½ |x| = x, if x > o −x, if x ≤ 0 2. The absolute value function has the following properties: (a) |x| ≥ 0 ∀x ∈ R. (b) |x| = 0 if and only if x = 0. (c) For every real number x, | − x| = |x|. (d) For all real numbers x and y, |xy| = |x||y|. (e) Let a > 0. Then |x| ≤ a ⇐⇒ −a ≤ x ≤ a and |x| < a ⇐⇒ −a < x < a. (f) For every real number x, −|x| ≤ x ≤ |x|. (g) Triangle Inequality: For all real numbers x and y, |x + y| ≤ |x| + |y| (h) Reverse Triangle Inequality: For all real numbers x and y, ||x| − |y|| ≤ |x − y|. 3. Let X be a nonempty set and d be a metric on X. Then the ordered pair (X, d) is called a metric space. The function d is also called distance function or simply distance. Often d is omitted and one just writes X for a metric space if it is clear from the context what metric is used. 4. For the set R define d : R × R −→ R by d(x, y) = |x − y|. This is called the usual metric on R and the phrase ”the metric space R” will always refer to this metric space. p 5. For the set R2 define d : R2 × R2 −→ R by d((x1 , y1 ), (x2 , y2 )) = (x1 − x2 )2 + (y1 − y2 )2 . This is called the usual metric on R2 and the phrase ”the metric space R2 ” will always refer to this metric space. p 6. For the set R3 define d : R3 ×R3 −→ R by d((x1 , y1 , z1 ), (x2 , y2 , z2 )) = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 . This is called the usual metric on R3 and the phrase ”the metric space R3 ” will always refer to this metric space. 7. Let X be a nonempty set. Define d : X × X −→ R by ½ d(x, y) = 0 1 if x = y if x = 6 y Then d is a metric on X and is called the discrete metric on X and (X,d) is called the discrete metric space. 8. CAUCHY-SCHWARZ INEQUALITY: ∀x, y ∈ Rn |x.y| ≤ |x||y| 9. The Triangle Inequality in Rn : ∀x, y ∈ Rn |x + y| ≤ |x| + |y|. Let x, y ∈ Rn . Then |x + y|2 = (x + y) · (x + y) = x · x + y · y + 2x · y ≤ |x|2 + |y|2 + 2|x.y| ≤ |x|2 + |y|2 + 2|x||y| = (|x| + |y|)2 . Take the square root of both side to have |x + y| ≤ |x| + |y|. Verification of (M4 ): If P1 = (x1 ,p y1 ), P2 = (x2 , y2 ), and P3 = (x3 , y3 ). p Then |P1 − P2 | = p (x1 − x2 )2 + (y1 − y2 )2 , |P1 −P3 | = (x1 − x3 )2 + (y1 − p y3 )2 and |P2 −P3 | = (x2 − x3 )2 + (y2 − y3 )2 . 2 + (y − y )2 = |P −P | = Now using the triangle inequality, |x+y| ≤ |x|+|y|, we have (x1 − x3 )p 1 3 1 3 2 + (y − y )2 + |P − P − P | = |(P − P ) + (P − P )| ≤ |P − P | + |P − P | = (x − x ) 2 3 1 2 2 3 1 2 2 3 1 2 1 2 p1 (x2 − x3 )2 + (y2 − y3 )2 NOTE: For this class, we will assume the following theorem: The sum of the lengths of two sides of a triangle is greater than or equal to the length of the third side. (Equality holds when the three vertices lie on a line; that is, the triangle is degenerate).