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Unformatted text preview: 54 23. November 29th: Continuity and Topology. 23.1 . Verify formally that lim x → 1 x 2 = 1. Answer: We need to verify that ∀ > , ∃ δ > 0 : 0 <  x 1  < δ = ⇒  x 2 1  < . Given , let δ = min( 3 , 1). If  x 1  < min 3 , 1 then in particular  x 1  < 1, so 0 < x < 2, and it follows that 1 < x + 1 < 3: so  x + 1  < 3 . But then we have  x 1  < 3 and  x + 1  < 3, and hence  x 2 1  =  x 1  ·  x + 1  < 3 · 3 = , as needed. 23.2 . Prove Lemma 23.1. Answer: We have to verify that if ( † ) ∀ > , ∃ δ > 0 : 0 <  x c  < δ = ⇒  f ( x ) L  < and ( ‡ ) ∀ > , ∃ δ > 0 : 0 <  x c  < δ = ⇒  f ( x ) L  < then L = L . Informally: Statement ( † ) says that f ( x ) is as close as we want to L for all x close enough to c ; statement ( ‡ ) says that f ( x ) is as close as we want to L for the same x . But if L and L are not the same number, f ( x ) cannot be arbitrarily close to both of them at the same time. Therefore necessarily L = L ....
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 Fall '11
 Aluffi
 Topology, Empty set, Metric space, Topological space, Bijections, R. Thus

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