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The standard procedure for completing a metric space is adding a limit for every Cauchy sequence, thus making the space Cauchy-complete. When elementary analysis is first taught, however, the completeness of the real numbers is usually introduced using the axiom of completeness, which asserts that the real numbers are Dedekind complete, that is, every Dedekind cut is generated by a real number (or equivalently, every upper-bounded nonempty subset has a least upper bound). For the real numbers, the two definitions coincide; but for a general metric space, the Dedekind definition is stronger.

I'm struggling to understand, or to give an intuitive explanation, for why the Cauchy definition truly implies "completeness", in the sense that any "place" (or "hole") in the space will have a point in it. I can rationalize the Dedekind definition: it basically implies that wherever you "cut" the line, you will find a number there; thus there are no "holes". If this definition was provable from the Cauchy one, I would not have complained; However the Cauchy definition is strictly weaker, and thus I'm struggling to see why does Cauchy completeness truly counts as "completeness", in the intuitive or geometric sense of "continuousness". Can anyone find a sort of intuitive or graphical explanation for that?

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    see [here](http://faculty.uml.edu/jpropp/dedekind.pdf) for a related discussion – 2017-01-31
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    You need an ordering on the space for the concept of Dedekind completeness to make sense. General metric spaces aren't ordered. – 2017-01-31
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    If you build real numbers as the Cauchy completion of rational, you can prove that Dedekind criterion also holds. So the Cauchy definition is not weaker than Dedekind, and it allows to complete the spaces where the order of elements is not defined. – 2017-01-31
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    @Wolfram Cauchy completeness is weaker in the sense that for an ordered field, Dedekind completeness is equivalent to Cauchy completeness plus the field being Archimedian. A non-Archimedian ordered field can be Cauchy complete, but not Dedekind complete. – 2017-01-31
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    @DanielFischer That is true, but still doesn't explain why Cauchy completeness is related to the intuitive notion of "continuousness". – 2017-01-31
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    I don't know what your "intuitive" notion of "continuousness" is, so I can't say much about that. Perhaps that if there were a hole (a missing point), you could approximate that arbitrarily well with points that are still there, and that would give rise to a Cauchy sequence (in sufficiently nice spaces, a Cauchy filter in more general spaces) that doesn't converge. – 2017-01-31
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    @DanielFischer That definitely helps me understand, thank you. It has now occurred to me, though, that Cantor's lemma is also a nice explanation for "continuousness", since you would like every "converging" sequence of intervals to converge to a point and not a "hole". Is Cantor's lemma equivalent to Cauchy completeness? – 2017-01-31
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    "Cantor's lemma" being that a sequence of nested closed intervals whose length shrinks to $0$ has nonempty intersection? In an Archimedian setting, that's equivalent to Cauchy completeness. In a non-Archimedian setting, it isn't, since the Cauchy criterion is defined using the Archimedian $\mathbb{Q}$ or $\mathbb{R}$, while the length of a nondegenerate closed interval could be infinitesimal, so a Cauchy sequence need not generate a sequence of intervals whose length shrinks to zero. Hence the intersection property need not guarantee a limit of the Cauchy sequence then. – 2017-01-31
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    @DanielFischer I'm actually slightly confused now, since Wikipedia says the properties are indeed equivalent: https://en.wikipedia.org/wiki/Cantor's_intersection_theorem (see section "Variant in complete metric spaces") Does wikipedia define the Cauchy criterion different than you? – 2017-01-31
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    A metric space is an Archimedian setting, no two distinct points have an infinitesimal distance. – 2017-01-31
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    @DanielFischer Of course, now I understand... thank you so much. – 2017-01-31

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