Is non-existence of the hyperreals consistent with ZF?

Question

I know that it is possible to construct the hyperreal number system in ZFC by using the axiom of choice to obtain a non-principal ultrafilter. Would the non-existence of a set of hyperreals be consistent with just ZF, without choice? Let me be conservative, and say that by a "set of hyperreals," I just mean a set together with some relations and functions such that the transfer principle holds, and there exists $\epsilon > 0$ smaller than any real positive real number.

Answer 1

The answer is yes, provided ZF itself is consistent. The reason is that the existence of the hyperreals, in a context with the transfer principle, implies that there is a nonprincipal ultrafilter on $\mathbb{N}$.

Specifically, if $N$ is any nonstandard (infinite) natural number, then let $U$ be the set of all $X\subset\mathbb{N}$ with $N\in X^*$. This is a nonprincipal ultrafilter on $\mathbb{N}$, since:

• If $X\in U$ and $X\subset Y$, then $N\in X^*\subset Y^*$, and so $Y\in U$.
• If $X,Y\in U$, then $N\in X^*\cap Y^*=(X\cap Y)^*$ and so $X\cap Y\in U$.
• If $X\subset\mathbb{N}$, then every number is in $X$ or in $\mathbb{N}-X$, and so either $N\in X^*$ or $N\in(\mathbb{N}-X)^*$ and thus $X\in U$ or $\mathbb{N}-X\in U$.
• For any particular standard natural number $n$, the set $X=\{m\in \mathbb{N}\mid n\leq m\}$ is in $U$, because $n^*\leq N$.
• The empty set $\emptyset$ is not in $U$, since $N\notin\emptyset=\emptyset^*$.

So $U$ is a nonprincipal ultrafilter on $\mathbb{N}$. The way that I think about $U$ is that it concentrates on sets that express all and only the properties held by the nonstandard number $N$. (See also my answer to A remark of Connes, where I make a similar point, and explain that, therefore, nonstandard analysis with the transfer property implies that there must be a non-measurable set of reals.)

Thus, in a model of ZF with no nonprincipal ultrafilter on $\mathbb{N}$ (and as Asaf mentions in the comments, there are indeed such models if there are any models of ZF at all), there is no structure of the hyperreals satisfying the transfer principle.

Answer 2

In response specifically to the title of the question: "Is non-existence of the hyperreals consistent with ZF?", technically speaking the answer is NO. Kanovei and Shelah constructed a definable model of the hyperreals in ZF; see http://arxiv.org/pdf/math/0311165.pdf

Therefore ZF is not "consistent with the nonexistence of the hyperreals". Of course, to prove any of their properties (such that that they are actually a proper extension, satisfy transfer, etc) one needs AC, but the same goes for many other crucial mathematical results (see below).

Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows.

Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF?

Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.