Do hypercontinuous fields exist?

Solution 1:

The "upwards" direction of the Lowenheim-Skolem theorem (this direction is really just the compactness theorem in disguise) gives a positive answer to a wide variety of questions of this form: roughly speaking, any first-order theory has arbitrarily large models (assuming it has at least one infinite model in the first place). This gives, for example, the existence of real closed fields of cardinality $>\vert\mathbb{R}\vert$ as an immediate corollary. Bringing in the downwards Lowenheim-Skolem theorem (which is a genuinely different result) lets us further control the cardinality of the structures we build, so e.g. we can prove "There is a real closed field of cardinality $\vert\mathbb{R}\vert^{++++++}$ exactly."

Things get a little more complicated if we add non-first-order desiderata, and indeed certain properties are not so malleable - for examlpe, there is no Archimedean ordered field of cardinality $>\vert\mathbb{R}\vert$, and indeed every Archimedean ordered field is isomorphic to a unique subfield of $\mathbb{R}$ - but for what you've specifically asked, the Lowenheim-Skolem theorem(s) is a convenient "nuke."

As an aside, your use of the terms "continuous" and "hypercontinuous" is incorrect - continuity is a property of functions, not a name for a cardinality, and there is sadly no snappy term for "set with cardinality $>\vert\mathbb{R}\vert$" or "set with cardinality $2^{\vert\mathbb{R}\vert}$" or similar.