# What's wrong with the classical Cauchy construction of the reals?

I am reading Bishop's "Constructive Analysis" and he says that defining a real number to just be an equivalence class of Cauchy sequences of rationals would be wrong. Why is that?

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The construction is actually due to Cantor. – Asaf Karagila May 4 '14 at 6:20
Right, the issue Bishop has is with Cantor's construction. He for some reason thinks that it doesn't satisfy constructive methods, and proposes an alternate construction. – Andrew S May 4 '14 at 6:23

Bishop refers here to a constructive approach to the reals. Classically, a Cauchy sequence of rationals is a sequence such that for any specified rational distance $\varepsilon >0$, there exists a modulus $N\in \mathbb N$ such that from that index onwards all elements are at distance at most $\varepsilon$ from each other. Constructively, this is unacceptable since the modulus $N$ needs to be constructed from the given sequence and the $\varepsilon >0$. In other words, a Cauchy sequences needs to be a sequence together with a function $\mathbb Q_+\to \mathbb N$ which is a constructive function giving an appropriate modulus for every $\varepsilon >0$. Now, exactly what a constructive function from one infinite countable set to another means can be answered in different ways (in the constructive world, classically there is no doubt how to continue). This is where it gets complicated.

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Why is Bishops construction better? He says that a real number is a sequence $(x_n)$ of rationals such that $|x_{m}-x_{n}| \leq m^{-1} + n^{-1}$ All it does is give a specific function from the indices to the rationals that a sequence must satisfy to be a real number. – Andrew S May 4 '14 at 7:01
yes, so he specified a particular modulus. Classically, not every Cauchy sequence need satisfy the stated condition. If Bishop defines these as the reals or an equivalence class of such as the reals (I don't remember and I don't have the book any more), there may be less Bishop reals than there are classical reals. – Ittay Weiss May 4 '14 at 8:17
Bishop says that to define the reals as equivalence classes of such sequences would be unnecessary, which makes perfect sense. Correct me if I'm wrong, but what I understand you to say is that the classical definition allows Cauchy sequences which may not have a constructible function to compute the modulus. If that's the case, why can we just define Cauchy sequences in the manner you described and still define the reals as equivalence classes of them? – Andrew S May 4 '14 at 8:30
I don't quite understand why there is no need to take equivalence classes, but if Bishop says so.... As for the classical construction, if you have constructive objections, then you probably reject the Cantor construction. Classically, a Cauchy sequence has some modulus but that modulus need not depend on the argument in any constructive way. Constructivists don't accept the existence of things like that, so they produce more constructive entities. Does that answer the question? – Ittay Weiss May 4 '14 at 8:56

Bishop is the prototypical constructivist-which means he won't accept any mathematical construction,procedure or definition that doesn't result in a precise and clear method of giving at least one example. Since Cauchy's definition is in terms of limits of a sequence of rationals and doesn't explicitly establish a bound for the limit, Bishop considers it mathematically suspect.

It's really a philosophical problem he has with the definition and nothing more-mathematically,the 2 are really equivalent to each other.I really should point out whether or not they're equivalent depends on what version of constructivism you're working in. As Steven Stadnicki points out in his objection-there are countably infinite"Bishop reals",which form "equivalence classes" under Bishop's definition where an countable constructive bijection algorithm cannot be devised (the Halting problem). As a result, by the absence of such a map, the Bishop reals are trivially uncountable! In intuitionism-Brouwer 's version of constructivism-there's a diagonalization arguement similar to Cantor's that proves the Bishop reals are uncountable.

So the truth is this matter is really subtle and sticky. But for our purposes,it certainly would be fair to say logically ,the Bishop reals are uncountable. You really need to ask a logician or set theorist this question,though-as I said,it's subtle and most people outside of research in constructive mathematics really understand it.

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I'm pretty certain the two are far from identical - IIRC, classically there are only countably many 'Bishop reals' (though of course since there's no constructive mapping from the integers to the Bishop reals, they're still 'internally uncountable'). – Steven Stadnicki Dec 10 '14 at 23:54
I've modified my answer and hope it's more satisfactory now, Steven. – Mathemagician1234 Dec 11 '14 at 7:59