Let be a metric space. If is a sequence in which converges to both and , prove that
Suppose x were not equal to y. Then d(x,y) is not 0. Apply the definition of convergence to both x and y with to get a contradiction:
Since converges to a, there exist such that if n> , then . Since converges to b, there exist such that if n> then . Use the triangle inequality to show that those are impossible.
Suppose to the contrary that converges to two distinct limit points a and b in the metric space (X, d).
Then by the the axiom of triangle inequality in the metric space.
By definition of a convergent sequence, there exists integers N1 and N2 such that if n >= N1, then and if n >= N2, then . Let N= max (N1, N2).
Then, if n >= N,
.
Let .
Then,
, which leads a contradiction since d(a,b) < d(a,b) is not true.
Thus, a sequence in a metric space cannot converge to more than one limit.
There is an intuitive way to consider this problem.
If then almost all of the terms of are in any ball that contains .
So if then there are two disjoint balls, one containing a the other containing b.
BUT THAT IS IMPOSSIBLE. Almost all of the terms of are in any ball that contains .
This may be similar to HallsofIvy, but it is slightly different and he/she seems to have LaTeX mistakes.. All you need is the triangle inequality
Choose such that if then
Also, choose such that if then
Now choose then for we have that .
Since was arbitrary this implies that which by the axioms of metric spaces is only true if