# Uniqueness of limits: please check which answer i should use

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

Prove that in a metric space (X, d) limits are unique. [xn] -> x and xn ->y then x = y

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<|xn-x|<d
0<|xn-y|<d
if you subtract them from each other then:
0<y-x<0
thus y-x = 0 and x = y

## Answers and Replies

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I think i have it:
I know that if x(n) -> x then d(x(n), x) -> 0
and also that: d(x, y) <= d(x(n), x)+ d(x(n), y)
and distace is always greater than 0 therefore:
0<=d(x, y)<=0
therefore x =y