Question

(4) Suppose that an → a. Prove or disprove: (a) If an is an upper bound fora set S for all n, then a is also an upper bound for S. (b) If an € (0,1) for all n, then a € (0,1). (c) If an € [0,1] for all n, then a € [0, 1]. (d) If an is rational, then a is rational.

Answer 1

an a So, fo an n,ENN such thal Eo lan-alE Anno a-anat Since o and choice cenbitay bossible of E is chaose So we can as smal a s Lim an lima tlim an a Hencs ih bnoved thal f o is an utpa bound fon a set S, 'a' is also an baund fon S Statement is evamble: On -e, a=2 a-3,.. . an (o,) but, timan nEN 4 (0,1)

Cet an € [0,1] ¥ n En but a Co1 Since an a , on anoE Such that nho lon-ale a-Ea a+ nno an E (a-6,ate) Now we choose ço such that Ca-E,ate)n To Obsenvc that fon all nno is fon infinifely an E la-Eate) mony n which contradic tte fact taf an E Co n eN Hene an -C011] ¥ ne IN ae Co] d Let On( a2 .XnE IN, an is national But we know that, (in an e . The statement is cunon
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