Corollary 1.16

If $$x_n$$ is a convergent sequence in a metric space, then it cannot converge to two distinct points.

baruch_shahi
Take a sequence $$(x_n)$$ in $$X$$, and suppose it converges to two distinct points $$x$$ and $$y$$. That is, assume that $$\lim\limits_{n\to\infty}{x_n}=x$$ and $$\lim\limits_{n\to\infty}{x_n}=y$$.

From Proposition 1.15, there exists $$\epsilon >0$$ such that $$B(x,\epsilon)\cap B(y,\epsilon)=\varnothing$$, since $$x$$ and $$y$$ are distinct.

By the definition of convergence, there exists $$N'\in\mathbb{N}$$ such that $$x_n\in B(x,\epsilon)$$ for all $$ n\ge N'$$. Similarly, there exists $$N\in\mathbb{N}$$ such that $$x_n\in B(y,\epsilon)$$ for all $$ n\ge N$$. Set $$N=\max{\{N',N''\}}$$.

Observe that for any $$n\ge N$$, we have $$x_n\in B(x,\epsilon)$$ and $$x_n\in B(y,\epsilon)$$. That is, for $$n\ge N$$, we have $$ x_n\in B(x,\epsilon)\cap B(y,\epsilon)$$. This is a contradiction by our choice of $$\epsilon$$.

Therefore, $$x=y$$. $$\blacksquare$$

baruch_shahi 13:24, March 3, 2011 (UTC -5)

'''Looks alright to me! --Steven.clontz 02:36, March 11, 2011 (UTC)'''