Question

Theorem 8.4. A sequence of points in a metric space has at most one limit. Proof. We will show that a sequence of points in a metric space has at most one limit. We will do this by contradiction, by supposing that some sequence has two different limits, and deriving the contradiction 1<]. Suppose the sequence (Pk)'_, converges to two different limits a and b. Let ε = d (a,b). <This choice for ɛ will be explained by a calculation below.> Since a + b, € > 0. Since (pk) - converges to a and to b For any Ea > 0, there exists Na, such that d (a,pk) < En for all k > Na, and For any Ep > 0, there exists Nb, such that d (b,Pk) < En for all k > Nb. If k > max {Na,Nb}, then d(a,b) <d(a, pk) +d (Pk,b) < £a+Eb, because (i) 2. Setting Ea = fb = (ii) it follows that E < , because (iii) _. Hence, 1 < 1. This is a contradiction. We have show that no sequence has more than one limit.

Answer 1

of points in a metric we will show that a sequence space has at most one limit. Suppone, (nn) be a sequence that has two limits . n n and y, and y, clearly clearly we assume that noty. w met, E. d(a,y) so Now, since (an) yn , 3 N, EN such that d&ang n) Leju unying . again IN, EN such that. since ince (mm) sy. y (nn) dlan, a) Lelu tn),N2 Now choose N= man {N,, N2} Then non imply both & ris,n, nini Then #nyan we have both doning LE4 dron, y < Ely choone on?N - then dlsim in) LEY 4 dlomon, y) 4€ /4 see that dinig, Ed (^, rm) +dfamig) yhm D is (a contradiction) (n) can't have more than .. one limit