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## Introductory Analysis

by: Mrs. Lenore King

35

0

2

# Introductory Analysis MTH 4101

Mrs. Lenore King
Florida Tech
GPA 3.91

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Mrs. Lenore King on Monday October 12, 2015. The Class Notes belongs to MTH 4101 at Florida Institute of Technology taught by Staff in Fall. Since its upload, it has received 35 views. For similar materials see /class/221672/mth-4101-florida-institute-of-technology in Mathematics (M) at Florida Institute of Technology.

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Date Created: 10/12/15
MTH 41015101 Introduction to Analysis Fall 2008 Practice problems Vll 11162008 1 If f is a continuous mapping of a metric space X into a metric space Y prove that fE C fE for every set E C X Show by an example fE can be a proper subset of Solution If fE a then there is nothing to prove Suppose that fE 31 1 Then there exits a y E Thus there exists a 1 E E such that f1 1 Since E E U E we have that 1 E E or 1 E E Also note that fE fE U fE Now let i1 E E Then f1 E fE C ii H1 6 E we have for every 6 gt 0 there exists a z E E N51 z 7amp1 Since f is continuous on X we have for every 6 gt 0 there exists a 60 gt 0 such that dX1q lt 60 implies dyf1fq lt 6 Thus we have for 6 60 dyf1fz lt 6 So f E N5f1 Since 6 is arbitrary we have f1 E fE Hence f1 E Hence the result To show that fE can be a proper subset of fE consider f 000 Y R E Z1 fE fE E Z where as fE o u an e Z Let F and 9 be continuous mappings of a metric space X into a metric space Y and let E be a dense subset of X Prove that fE is dense in fX lf f1 91 for all 1 E E prove that f1 91 for all X Solutionzi To show that fE is dense in fX we need to show that every point of fX is either a point of fE or a limit point of fE or both Suppose that y E fX Then there exists a point 1 E X such that f1 9 Since E is dense in X 1 6 EU E H1 6 E then 9 E fE and we are done H1 6 E and 1 E then ther exists a sequence 1 E E such that 1 a1 We then have f n a f1 That is f1 E Now suppose f1 91 for all 1 E E Let x E XE Since E is dense in X we have a sequence qn E E such that 9 a x So f flirn q lirn fqn lirn9qn 9lirnqn Thus f 9 for all z E X If f is de ned on E the graph of f is the set of points x for z E E Suppose E is compact prove that f is continuous on E if and only if its graph is compact Solution Let G E i Suppose f is continuous Then since E is cornact fE is also cornpact We claim that thr product of nitely many compact sets is compact Hence E gtlt fE is compact ii Suppose that f is not continuous at some point 1 E E Then there exists a sequence pn E E such that 1 a 1 but f n does not converge to 1 906 A 4 V A U V A CT fp That is7 there exists a E gt 07 such that dfpnfp 2 E for all n Thus7 we have pn a p but dfpnfp gt 6 Put7 Pn 10 7 fpn a sequence in G Since G is compact7 there exists a subsequence PM that convergees to some point p7q in G Since p7q is a point in the graph of f we must have fp q But7 pm a p and dfpnkfp 2 6 This is a contradiction Let f be a real uniformly continuous function on a bounded set E in R Prove that f is bounded on E Solution Suppose that f is uniformly continuous on a bounded set E and that f is not bounded on E That is7 for every M there exists a zM E E such that lfMl 2 M Since the sequence of real numbers xM is bounded7 there exists a convergent subsequence Thus7 MM is Cauchy implying that fsz is Cauchy see the problem below This is a contradiction since fsz is unbounded Hence the result Suppose f is a uniformly continuous mapping of a metric space X into a metric space Y and prove that is a Cauchy sequence in Y for every Cauchy sequence in X Solution Let E gt 0 be given Since7 f is uniformly continuous7 there exists a 6 gt 0 such that dfx7 lt 6 for all my 6 X such that dxy lt 6 Let be a Cauchy sequence in X Then7 for the 6 above7 there exists a N such that dxnxm lt 6 for all mm 2 NThis means7 since dfn7 lt E for all mm 2 N Hence7 fzn is Cauchy Let 01 be the closed interval Suppose f is a continuous mapping of I into I Prove that f z for at least one x E I Solution Let 9 f 7 x If 90 0 or 91 0 then the result is true Suppose7 90 31 0 and 91 31 0 Then7 f0 31 0 and f1 31 1 Since7 f 01 a 01 f0 gt 0 and f1 lt1 Thus7 90 gt 0 and 91 lt 0 Since 9 is a continuous function7 by intermediate value theorem7 Theorem 423 9 0 for some z E I Hence7 fx z for some z E I

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