Question on Cauchy Sequence definition?

Question

Kaplansky defines a Cauchy sequence if for any $\epsilon > 0$ there exists sufficiently large $i, j$ such that $D(x_i, x_j) < \epsilon$ for some sequence $\{x_n\}$ in a metric space. The sequence need not converge to some specified limit point, just that $x_i$ and $x_j$ become arbitrarily close. However, if $D(x_i, x_j) < \epsilon$, doesn't that mean $x_i \to x_j$?

Answer 1

What you know is that $\lim\limits_{i,j\to\infty}D(x_i,x_j)=0$. But you know nothing about the limits $\lim\limits_{i\to\infty}D(x_i,x_j)$, $\lim\limits_{j\to\infty}D(x_i,x_j)$. As an example, take $x_i=i^{-1}$. Then $\lim\limits_{i,j\to\infty}|i^{-1}-j^{-1}|=0$ yet $\lim\limits_{j\to\infty}D(x_i,x_j)=i^{-1}$ and $\lim\limits_{j\to\infty}D(x_i,x_j)=j^{-1}$.

Answer 2

Your definition is wrong or at least sloppily worded. "$(x_n)$ is Cauchy" means that for any $\epsilon > 0$, there exists $N \equiv N(\epsilon)$ such that if $i, j \in \mathbb{N}^+$ with $i \geq N$ and $j \geq N$, then $D(x_i,x_j) < \epsilon$. Roughly speaking, the terms in the sequence are getting closer and closer to each other. In some (but not all) metric spaces, any Cauchy sequence must converge. If the sequence does converge (to $x$), $x$ is not necessarily one of the terms in the sequence.