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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 $x_i \to x_j$?

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What does $x_i \to x_j$ mean? –  njguliyev Oct 20 '13 at 15:57
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$x_i,x_j$ are terms of the sequence. For every $\varepsilon > 0$, you have a $k_\varepsilon \in \mathbb{N}$ such that for all $i,j \geqslant k_\varepsilon$ you have $D(x_i,x_j) < \varepsilon$. But $x_j$ is not something fixed. –  Daniel Fischer Oct 20 '13 at 15:57
    
well it has to be, for example $D(x_{34785634756872395}, x_{34785634756872395 + 1}) < \epsilon$. But of course, that can always be bigger... –  Don Larynx Oct 20 '13 at 16:01

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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 $\lim\limits_{j\to\infty}D(x_i,x_j)=j^{-1}$.

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And you subtract $\frac{1}{j}$ from $\frac{1}{i}$ to get something smaller than $\epsilon$? –  Don Larynx Oct 20 '13 at 16:03
    
@DonLarynx What do you mean? –  Pedro Tamaroff Oct 20 '13 at 16:05

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.

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