Return to Answer

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.

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.

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 $n^*\leq N$.

So $U$ is a nonprincipal ultrafilter on $\mathbb{N}$.

Basically, the ultrafilter 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.)

Thus, in a model of ZF with no nonprincipal ultrafilter on $\mathbb{N}$, there is no structure of the hyperreals satisfying the transfer principle.

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.

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

• 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 $n\in X^*=Y^*=(X\cap Y)^*$ 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 $n^*\leq N$. So

Basically, the ultrafilter 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.)

Thus, in a model of ZF with no nonprincipal ultrafilter on $\mathbb{N}$, there is no structure of the hyperreals satisfying the transfer principle.

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 $n^*\leq N$.

So $U$ is a nonprincipal ultrafilter on $\mathbb{N}$.

Basically, the ultrafilter 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.)

Thus, in a model of ZF with no nonprincipal ultrafilter on $\mathbb{N}$, there is no structure of the hyperreals satisfying the transfer principle.