Return to Answer

One problem is that the set of formal Laurent series is not a real closed field (an ordered field where every positive element has a square root, and every polynomial of odd degree has a root). That particular problem would be fixed if one considered instead the real closure of the field of Laurent series, which is the field of Puiseux series (Laurent series involving powers of $x^{1/n}$ where $n$ is allowed to vary). So the question would become: is the field of hypperreals isomorphic to the field of Puiseux series?

The answer is "certainly not"; the hyperreals are much "bigger", or it would be better to say "much more saturated" than Puiseux series. As you probably know, the hyperreals do not form a complete ordered field (only the standard real numbers do that), but they do have properties that approximate completeness. Namely, there is the $\eta_1$ property: given countable sets $L$ and $U$ such that every element of $L$ is less than every element of $U$, there exists an element that is an upper bound of all elements of $L$ and a lower bound of all elements of $U$. So for example: in the field of Puiseux series (and taking $x$ to be an infinitesimal element), there is a pretty big gap between all the constants (forming an $L$) and the $x^{-1/n}$ (forming a $U$). The hyperreals fill in such gaps.

(Added on request.) The hyperreals thus form an ordered field extension of the field of Puiseux series, and admits elements such as $\sin \varepsilon$ where $\varepsilon$ is an infinitesimal.

One problem is that the set of formal Laurent series is not a real closed field (an ordered field where every positive element has a square root, and every polynomial of odd degree has a root). That particular problem would be fixed if one considered instead the real closure of the field of Laurent series, which is the field of Puiseux series (Laurent series involving powers of $x^{1/n}$ where $n$ is allowed to vary). So the question would become: is the field of hypperreals isomorphic to the field of Puiseux series?

The answer is "certainly not"; the hyperreals are much "bigger", or it would be better to say "much more saturated" than Puiseux series. As you probably know, the hyperreals do not form a complete ordered field (only the standard real numbers do that), but they do have properties that approximate completeness. Namely, there is the $\eta_1$ property: given countable sets $L$ and $U$ such that every element of $L$ is less than every element of $U$, there exists an element that is an upper bound of all elements of $L$ and a lower bound of all elements of $U$. So for example: in the field of Puiseux series (and taking $x$ to be an infinitesimal element), there is a pretty big gap between all the rational constants (forming an $L$) and the $x^{-1/n}$ (forming a $U$). The hyperreals fill in such gaps.

(Added on request.) The hyperreals thus form an ordered field extension of the field of Puiseux series, and admits elements such as $\sin \varepsilon$ where $\varepsilon$ is an infinitesimal.

One problem is that the set of formal Laurent series is not a real closed field (an ordered field where every positive element has a square root, and every polynomial of odd degree has a root). That particular problem would be fixed if one considered instead the real closure of the field of Laurent series, which is the field of Puiseux series (Laurent series involving powers of $x^{1/n}$ where $n$ is allowed to vary). So the question would become: is the field of hypperreals isomorphic to the field of Puiseux series?

The answer is "certainly not"; the hyperreals are much "bigger", or it would be better to say "much more saturated" than Puiseux series. As you probably know, the hyperreals do not form a complete ordered field (only the standard real numbers do that), but they do have properties that approximate completeness. Namely, there is the $\eta_1$ property: given countable sets $L$ and $U$ such that every element of $L$ is less than every element of $U$, there exists an element that is an upper bound of all elements of $L$ and a lower bound of all elements of $U$. So for example: in the field of Puiseux series (and taking $x$ to be an infinitesimal element), there is a pretty big gap between all the constants (forming an $L$) and the $x^{-1/n}$ (forming a $U$). The hyperreals fill in such gaps.

(Added on request.) The hyperreals thus form an ordered field extension of the field of Puiseux series, and admits elements such as $\sin \omega$ where $\omega$ is an infinitesimal.

One problem is that the set of formal Laurent series is not a real closed field (an ordered field where every positive element has a square root, and every polynomial of odd degree has a root). That particular problem would be fixed if one considered instead the real closure of the field of Laurent series, which is the field of Puiseux series (Laurent series involving powers of $x^{1/n}$ where $n$ is allowed to vary). So the question would become: is the field of hypperreals isomorphic to the field of Puiseux series?

The answer is "certainly not"; the hyperreals are much "bigger", or it would be better to say "much more saturated" than Puiseux series. As you probably know, the hyperreals do not form a complete ordered field (only the standard real numbers do that), but they do have properties that approximate completeness. Namely, there is the $\eta_1$ property: given countable sets $L$ and $U$ such that every element of $L$ is less than every element of $U$, there exists an element that is an upper bound of all elements of $L$ and a lower bound of all elements of $U$. So for example: in the field of Puiseux series (and taking $x$ to be an infinitesimal element), there is a pretty big gap between all the constants (forming an $L$) and the $x^{-1/n}$ (forming a $U$). The hyperreals fill in such gaps.

(Added on request.) The hyperreals thus form an ordered field extension of the field of Puiseux series, and admits elements such as $\sin \varepsilon$ where $\varepsilon$ is an infinitesimal.

One problem is that the set of formal Laurent series is not a real closed field (an ordered field where every positive element has a square root, and every polynomial of odd degree has a root). That particular problem would be fixed if one considered instead the real closure of the field of Laurent series, which is the field of Puiseux series (Laurent series involving powers of $x^{1/n}$ where $n$ is allowed to vary). So the question would become: is the field of hypperreals isomorphic to the field of Puiseux series?

The answer is "certainly not"; the hyperreals are much "bigger", or it would be better to say "much more saturated" than Puiseux series. As you probably know, the hyperreals do not form a complete ordered field (only the standard real numbers do that), but they do have properties that approximate completeness. Namely, there is the $\eta_1$ property: given countable sets $L$ and $U$ such that every element of $L$ is less than every element of $U$, there exists an element that is an upper bound of all elements of $L$ and a lower bound of all elements of $U$. So for example: in the field of Puiseux series (and taking $x$ to be an infinite element), there is a pretty big gap between all the constants (forming an $L$) and the $x^{1/n}$ (forming a $U$). The hyperreals fill in such gaps.

(Added on request.) The hyperreals thus form an ordered field extension of the field of Puiseux series, and admits elements such as $\sin \omega$ where $\omega$ is an infinitesimal.

One problem is that the set of formal Laurent series is not a real closed field (an ordered field where every positive element has a square root, and every polynomial of odd degree has a root). That particular problem would be fixed if one considered instead the real closure of the field of Laurent series, which is the field of Puiseux series (Laurent series involving powers of $x^{1/n}$ where $n$ is allowed to vary). So the question would become: is the field of hypperreals isomorphic to the field of Puiseux series?

The answer is "certainly not"; the hyperreals are much "bigger", or it would be better to say "much more saturated" than Puiseux series. As you probably know, the hyperreals do not form a complete ordered field (only the standard real numbers do that), but they do have properties that approximate completeness. Namely, there is the $\eta_1$ property: given countable sets $L$ and $U$ such that every element of $L$ is less than every element of $U$, there exists an element that is an upper bound of all elements of $L$ and a lower bound of all elements of $U$. So for example: in the field of Puiseux series (and taking $x$ to be an infinitesimal element), there is a pretty big gap between all the constants (forming an $L$) and the $x^{-1/n}$ (forming a $U$). The hyperreals fill in such gaps.

(Added on request.) The hyperreals thus form an ordered field extension of the field of Puiseux series, and admits elements such as $\sin \omega$ where $\omega$ is an infinitesimal.