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Just want to attempt to check if my understanding/intuition for the construction of the Hyperreal numbers via an ultraproduct is correct. Appreciate any corrections or help.

So Hyperreals are constructed by using a non-principal ultrafilter. Which is essentially a tool that obeys several properties the most important (for being non-prinipal) of which is no finite subsets are contained in the ultrafilter.

It can also be shown that a filter can be extended to an ultrafilter (via Zorn's lemma) and can therefore be turned into an non-principal ultrafilter if we simply take the filter consisting of all cofinite sets and extend that to be an ultrafilter.

using this non-prinicpal ultrafilter an equivalence relation is constructed on real-valued sequences. (The equivalence relation being the "size of two real-valued sequences agree and belong to the non-prinicpal ultrafilter"). This will essentially allow the set to be partitioned into pieces. Since we have an equivalence relation it is then easy to get the various equivalence classes.

Once we have the equivalence relation to get the Hypperreal numbers we take the quotient of the set of real-valued sequences with the equivalence relation. (is this similar to a quotient set, apart from the real-valued sequences form a ring, so it's a quotient ring; not sure on the name for it?).

Thanks for reading and pointing out any holes if there are any.

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  • $\begingroup$ Reasonable sketch of the construction, if we use say the natural numbers as the index set. The parenthesized sentence beginning with "(The equivalence relation being" is either vague or not right. Yes, the ultrapower is a quotient structure, it is a field and much more. It contains a natural copy of the reals, via the equivalence classes of constant sequences. The key properties of the ultrapower have not been mentioned, that it is (isomorphic to) an elementary extension of the reals, over whatever first-order language we may choose. $\endgroup$ – André Nicolas Jul 30 '15 at 17:13
  • $\begingroup$ Andre, Yes I'm assuming the natural numbers (can we use other number systems for the index set?). What i mean was say there was set a={1,2,3,4} and b={1,5,7,4} then their agreement set would be {1,4} as thoese are the indicies that would match one another. Regarding the properties what would be the technical term for the quotient structure? And isn't it only isomorphic if you assume the continuum hypothesis? $\endgroup$ – user253919 Jul 30 '15 at 17:17
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    $\begingroup$ The term "construction" is a little misleading, since the proof of existence of a non-principal ultrafilter on the naturals is non-constructive. One might add that we get something really different from the reals, since for example the equivalence class of the sequence $(1/(n+1))$ is different from $0$ but "smaller" than the equivalence class of any positive constant sequence. $\endgroup$ – André Nicolas Jul 30 '15 at 17:20
  • $\begingroup$ This is what I need to tackle next. The arithmetic of Hyperreals... It's a really interesting topic so far. $\endgroup$ – user253919 Jul 30 '15 at 17:25
  • $\begingroup$ There is not much interesting connected with the continuum hypothesis. Any infinite set will do as the index set. There is a technical additional restriction on the ultrafilter, that comes into play only with certain cardinals so large that we can consistently assume they do not exist (measurable cardinals). The quotient structure is called an ultrapower. In the definition of the equivalence relation, it looks as if you have the right idea, but it is imprecisely expressed. "Size" has nothing to do with it, unless we decide to call sets in the ultrafilter size $1$, and sets not in size $0$. $\endgroup$ – André Nicolas Jul 30 '15 at 17:28
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The equivalence relation is that the set of integers on which the sequences agree is an element of the ultrafilter.

If $A$ is a commutative ring, and $I$ is an ideal of $A$ then you can define the quotient ring $A / I$ as $\{a + I \mid a \in A\}$, and $(x + I) + (y + I) = (x+y) + I$, $(x + I)(y + I) = xy + I$. And $I$ is maximal iff $A / I$ is a field.

Here the ideal $I$ is $\{u \in {\mathbb{R}}^{\mathbb{N}} \mid u^{-1}(\{0\}) \in U\}$ where $U$ is the ultrafilter. It is an ideal because

-$\forall u,v \in I$, $u^{-1}(\{0\}) \cap v^{-1}(\{0\}) \subset (u - v)^{-1}(\{0\})$ and $U$ is a filter.

-$\forall (u,v) \in I \times A$, $u^{-1}(\{0\}) \subset (uv)^{-1}(\{0\})$ and $U$ is a filter.

It is maximal because if $J$ is an ideal and $I \subsetneq J$, then for $u \in J - I$, $\mathbb{N} - u^{-1}(\{0\}) \in U$ because $U$ is an ultrafilter. Therefore, $\chi_{u^{-1}(\{0\})}$ is in $I$, in $J$. If $v = n \mapsto \frac{1}{u(n)},0$ whether $u(n) \neq 0$ or $u(n) = 0$, we have $uv + \chi_{u^{-1}(\{0\})} = 1_A$ is in $J$ too, so $J = A$.

($\chi_E$ is the caracteristic function of $E$)


$A / I$ is a ring iff $U$ is a filter, it is a field iff $U$ is a maximal filter (= ultrafilter), and it is isomorphic to the field of real numbers iff $U$ is principal. (I have not proven these claims so you can try to do it, it may be a little tedious but useful and not very hard)


This is what I can say if I don't want to talk about model-theoritic structures, but ultraproducts really get interesting when one starts to talk about theses structures.

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  • $\begingroup$ Thanks nombre. Will have a go at trying to prove those claims (not sure how i'll do as relatively new to ring theory). I would be interested in learning more about ultraproducts, but at the moment think I'd be overwhelmed. $\endgroup$ – user253919 Jul 30 '15 at 17:44
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It has been shown that the Continuum Hypothesis implies that if R(1) and R(2) are hyper-real fields defined by free (non-principal) ultrafilters U(1) and U(2) on the natural numbers (or on any countably infinite set),then R(1) and R(2) are isomorphic ordered fields.(Sorry I've forgotten the reference). The negation of the Continuum Hypothesis along with some extra hypotheses that are known to be consistent with modern set theory ("ZFC") will produce some non-isomorphic R(1) and R(2).

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