- #1

jaci55555

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**uniqueness of limits: please check which answer i should use :)**

Prove that in a metric space (X, d) limits are unique. [x

_{n}] -> x and x

_{n}->y then x = y

By contradiction:

Assume x [STRIKE]=[/STRIKE]y. let |x - y|/3 (Can I just make a random assumption like this?)

|x - y| = |f(x) - x|+|f(x) - y|

defn of a limit: for all e>0 there exists d>0 st 0<|x-a|<d so that |f(x) - L|<e

thus |f(x) - x|+|f(x) - y|<= 2e

|x-y| <=2e

|x-y| +e <= 3e

|x-y| +e <= 3e(|x - y|/3)

Thus |x-y| + e<= |x-y| contradiction. therefore x=y

OR

by defn: for all e>0 there exists d>0 st 0<|x-a|<d so that |f(x) - L|<e

so 0<|x

_{n}-x|<d

0<|x

_{n}-y|<d

if you subtract them from each other then:

0<y-x<0

thus y-x = 0 and x = y