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I'm trying to understand the ultrapower construction of the hyperreal numbers as described in Wikipedia.

The motivation is given as the surprising ability to find a total order of sequences of real numbers. Apparently if we select a free ultrafilter $U$ of the natural numbers, then it can be used to determine which indices "matter" in the set and then (from the article):

We write $(a_0, a_1, a_2, ...) ≤ (b_0, b_1, b_2, ...)$ if and only if the set of natural numbers $\{ n : a_n ≤ b_n \}$ is in $U$.

This is where I don't follow. Are we still working component-wise and then "$\land$-ing" the resultant sequence together as mentioned previously in the article?

The "intuitive explanation" which follows briefly mentions a parallel to Cantor's construction of the reals ... did I miss a step where this became a Cauchy sequence by implication?

In short: why does choosing a special set $U$ of indices of sequences to "matter" create a total order relation?

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$U$ isn't a set of indices, it's a set of sets of indices. The idea is that two sequences are equal (or rather, name the same element) if they agree "most of the time" in the sense of $U$. Similarly, one sequence $A$ (names an element of the ultrapower which) is less than or equal to (the element named by) another sequence $B$ iff "most of the time"each entry of $A$ is $\le$ the corresponding entry of $B$. And so forth.

For example, consider the following sequences:

  • $A_1=(1,0,0,0,0,...)$

  • $A_2=(0,1,0,0,0,...)$

  • $A_n$ has a $0$ in every place except the $n$th place, where it has a $1$.

I claim that these sequences are all equal (or rather, name the same element) in the ultrapower! For every $i,j$, the set of $n$ such that the $n$th elemenet of $A_i$ is $=$ the $n$th element of $A_j$ is cofinite, and every cofinite set is in $U$ since $U$ is a nonprincipal ultrafilter, so we always have $[A_i]_U\le [A_j]_U$ (this is a bit different from what you wrote - I'm being careful to distinguish between the sequence $A_i$ and the equivalence class $[A_i]_U$ which it belongs to; remember that elements of the ultrapower are equivalence classes of sequences, not individual sequences).


How does this generate a total order?

Well, suppose I have two sequences $A=(a_n)_{n\in\mathbb{N}}$ and $B=(b_n)_{n\in\mathbb{N}}$. Let $$X=\{n: a_n\le b_n\},\quad Y=\{n: a_n>b_n\}.$$ Clearly $X\sqcup Y=\mathbb{N}$, so since $U$ is an ultrafilter exactly one of $X$ and $Y$ is in $U$. If $X\in U$, then $[A]_U\le [B]_U$; if $Y\in U$, then $[A]_U>[B]_U$.

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  • $\begingroup$ Awesome thank you, I can see I should spend some more time with ultrafilters and ultrapowers! $\endgroup$ – KCE Feb 12 at 7:18
  • $\begingroup$ I'm not sure how chat works, I thought I created a room that would invite you to it but it doesn't seem that way now. For a little more discussion on the topic I'd appreciate your attention to: chat.stackexchange.com/rooms/89602/… $\endgroup$ – KCE Feb 12 at 9:42

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