Understanding Real Analysis: Intervals, Bounds, and Problem-solving
Explore the concepts of intervals and bounds in real analysis, including open and closed intervals, semi-closed intervals, least upper bound, and greatest lower bound. Learn how to solve problems based on intervals and bounded sets through detailed explanations and examples.
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DEPARTMENT OF MATHEMATICS SGGSJ GOVERNMENT COLLEGE PONTA SAHIB
COURSE NAME : REAL ANALYSIS COURSE NAME : REAL ANALYSIS COURSE CODE : 201TH COURSE CODE : 201TH
CONTENTS CONTENTS Intervals Bounds : i. Greatest lower bound ii. Lowest Upper Bound Problems based on intervals and the bounded sets
INTERVALS Let a and b be the two distinct real numbers with a < b (say) then , (i) Open Interval : The set of all real numbers between a and b is said to form an open interval from a to b denoted by (a,b).In symbols (a,b) = {x:a<x<band x R }. (ii) Closed Interval : The set of all real numbers between aand bincluding the end points a and b is said to form a closed interval and is denoted by [a,b] . In symbols [a,b] = {x: a x band x R} .
(iii) Semi-closed (or semi-open) intervals : An interval in which one end point is included and other end point is excluded is called semi-closed interval. In symbols [a,b) = {x: a x<band x R} . (iv) Infinite interval : The set of all the real numbers x such that x>a forms an infinite set and is denoted by (a, ). In symbols (a, ) = {x: x>a and x R} .
BOUNDS LEAST UPPER BOUND : Let S be a non-empty subset of R which is bounded above. Then there exist a real number u which is the smallest of all the upper bounds of S . This number is called the least upper bound and is written as l.u.b of S =u . GREATEST LOWER BOUND : Let S be a non-empty subset of R which is bounded below. Then there exist a real number l which is the smallest of all the lower bounds of S . This number is called the greatest lower bound and is written as g.l.b of S =l.
Problems based on intervals and the bounded sets PROBLEM 1 Solve SOLUTION: Here Clearly , x-2 0 i,.e., x 2 Two cases arise : Case 1:x-2 >0 i,.e., x >2.Then ,
x + 3 < 5x 10 x > solution set is Case 2: x 2 < 0 i.e., x <2. Then , x + 3 > 5x 10 -4x > -13 But x < 2 x (- ,2)
x(-,2) solution set is (- ,2) Hence the solution set of the given inequality is (- ,2)
PROBLEM 2 Show that the following set is bounded. Also find their l.u.b and g.l.b : { sin x cos x }2 : 0 x } SOLUTION: Let S = { sin x cos x }2 : 0 x } Now { sin x cos x }2 = sin2x cos2x + 2sinxcosx = 1 +sin2x Also 0 x 0 2x 2 -1 sin2x 1 0 1+ sin2x 2 0 (sin x cos x )2 2 S is bounded and its l.u.b = 2 and g.l.b = 0
