# Why does every cauchy sequence of rational numbers converge?

In the construction of real numbers from cauchy sequence, we define real number to be the set of equivalence classes of cauchy sequences of rational numbers. But how can we ensure that all cauchy sequences of rational numbers converges (in Q or in R), what if there exists a cauchy sequence of rational number that diverges? A divergent cauchy sequence of rational would still represent a real number by our definition, but what kind of “number” can a divergent sequence represent? Or is this some assumption we all have to agree on?

• Who assumes that all Cauchy sequences of rational numbers converges? Jan 1 at 22:01
• They don't. We identify the Cauchy sequences with the real numbers. Jan 1 at 22:01
• Cauchy sequences of rationals (or rather, equivalence classes of these) are defined to be real numbers. Hence they all converge in $\mathbb R$ as a matter of definition. Of course, one does have to prove that this construction makes sense, and that's not an easy business.
– lulu
Jan 1 at 22:06
• Should add: other constructions of $\mathbb R$ are possible and if you are working with one of those, you can demonstrate the convergence as a theorem about that system. But you'd need to specify how you are defining $\mathbb R$ to say much more.
– lulu
Jan 1 at 22:11
• What's your mental image of a "divergent" Cauchy sequence? Are you sure the thing you're picturing is Cauchy?
– Karl
Jan 1 at 22:48