Are real numbers defined as Banach spaces? I was listening to this lecture by Norman Wildberger, where he explains that real numbers are constructed as equivalence classes of Cauchy sequences. He never says that these equivalence classes are Banach spaces, but I was left wondering if that is what he means.
 A: This is unfortunately a nonsense question.
There is a set of Cauchy sequences. This set has a relation on it ($(a_n) \sim (b_n)$ if $\lim_n (a_n - b_n) = 0$). There is therefore a quotient set or set of equivalence classes.
None of this ever passes through anything with more structure than sets. (If you like, it passes through rings: the set of Cauchy sequences is a ring, because I know how to add and divide in it, and the relation respects these operations, so that the quotient inherits them.)
In fact "$V$ is a Banach space" does not even make sense until you invent the real numbers. A Banach space is a pair $(V, N)$, where $N: V \to \Bbb R_{\geq 0}$ is a function satisfying certain properties.
A: The real numbers can be defined by the process of completion with respect to the usual norm $\lvert\:\cdot\:\rvert:\Bbb{Q}\to \Bbb{Q}$. The norm $\lvert \:\cdot\:\rvert$ defines a metric on $\Bbb{Q}$ by $d(x,y)=\lvert x-y\rvert$. We note that $\Bbb{Q}$ is not complete with respect to this metric because there are Cauchy sequences which do not converge (this is the definition of not being complete): for instance the sequence
$$
3,3.1,3.14,3.141,3.1415,\ldots
$$
is "trying" to converge to $\pi= 3.1415...$ but $\pi$ is not rational so this does not make sense. [Note that one does not really know what $\pi$ is before defining $\Bbb{R}$ so this reasoning is sort of bad, but it is a good source of motivation - you could do the same thing with $\sqrt{2}$ or any infinite decimal expansion you wish.] The completion of $\Bbb{Q}$ with respect to this norm is called $\Bbb{R}$ and is defined to be the set of Cauchy sequences in $\Bbb{Q}$ modulo the relation that two Cauchy sequences $(a_n)$ and $(b_n)$ are called equivalent if
$$
\lvert a_n-b_n\rvert\to 0
$$
as $n\to\infty$. Maybe this is already familiar to you. Anyway, the sequences themselves are not Banach spaces in any reasonable sense, but it is true that $\Bbb{R}$ has a norm defined by extending the norm $\lvert\:\cdot\:\rvert$ on $\Bbb{Q}$ from before. It is also true that the completion of a normed space with respect to a norm is complete (in that all Cauchy sequences converge). In particular, $\lvert\:\cdot\:\rvert$ extended from $\Bbb{Q}$ endows $\Bbb{R}$ with the structure of a complete normed vector space where we add Cauchy sequences by
$$
(a_n)+(b_n)=(a_n+b_n)
$$
and define scaling in the obvious way. So, $(\Bbb{R},\lvert\:\cdot\:\rvert)$ is a Banach space.
