What makes the reals a small set? The following is more of a soft question than a concrete problem, but it haunts me for quite a while now.
The reals can be defined as the maximal Archimedean field. They are tiny in comparison to other sets; there are quite a lot of large sets. I find it counterintuitive that there are no Archimedean fields of arbitrary large cardinality. Why is this the case?
Note: I am not really asking for a proof of this fact. The ones I have seen (e.g. here) were perfectly fine, but they did not really give an intuitive reason, why larger sets cannot arise.
Thank you for your time.
 A: If an ordered field $F$ has cardinality $>|\mathbb{R}|$, then $\mathbb{N}$ has an upper bound in it.
Indeed, assume the contrary. Then $\mathbb{Q}$ is order-dense in $F$. So for $x \in F$, there is a sequence $(q_n)_{n \in \mathbb{N}} \in \mathbb{Q}^{\mathbb{N}}$ with $|x-q_n|<2^{-n}$ for all $n \in \mathbb{N}$. No other element $y$ of $F$ can satisfy this with respect to $(q_n)_{n \in \mathbb{N}}$ (take $n$ with $2^{-n}<\frac{|x-y|}{2}$). So this yields a surjective map $\mathbb{Q}^{\mathbb{N}} \rightarrow F$, sending any irrelevant sequence to $0$ and sending sequences like $(q_n)_{n \in \mathbb{N}}$ to their limit. Since $|\mathbb{Q}^{\mathbb{N}}|=|\mathbb{R}|$, this contradicts the assumption.
Now since $\mathbb{N}$ has an upper bound in $F$, it should have a least upper bound in $F$ if $F$ were complete. But for any upper bound $M$, we have $M-1>\mathbb{N}$, so there is no such supremum.
So to sum up: if $\mathbb{N}$ is bounded then the field is not complete, and if $\mathbb{N}$ is unbounded then the field is small.
A: If an ordered field is archimedean and complete, then all its points are limits of points in $\mathbb{Q}$. This gives you a bound on the cardinality of the field.
In a sense, filling the gaps between the points of $\mathbb{Q}$ can only create so many new points.
