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If we let F be a non-principal ultrafilter on the Natural numbers, and define $a,b \in \mathbb R^{\mathbb N}$ to be real-valued sequences. Then the equivalence relation ~ can be defined as a~b if the set $\{j \in \mathbb N | a_j = b_j \} \in F$.

I.e. that two sequences are equivalent if the set of natural numbers on which their points agree is in our ultrafilter. Now if we take the quotient of $ \mathbb R^{\mathbb N}$ / ~ then this creates the Hyperreal numbers.

I don't understand what is happening when I take this quotient though. I'm guessing it is doing something like partitioning the real valued sequences into equivalence classes by using each of the sequencs that are inside the ultrafilter? I'm not really sure though. Any help, this step is confusing.

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  • $\begingroup$ Three notational fixups you should do: the superscript to $\mathbb{R}$ should be $\mathbb{N}$ not lowercase 'n'; and you should escape the curly braces (precede them with backslash) around the set comprehension: \{j \in \mathbb{N} \mid a_j = b_j\} $\endgroup$
    – BrianO
    Oct 8 '15 at 20:40
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Think of the ultrafilter as an "almost all" quantifier. Sets in the ultrafilter $F$ are "big", so two sequences that agree on a set that's in $F$ agree "almost everywhere"; hence they're identified, treated as "equivalent" (they're in the same equivalence class in the quotient): they define the same hyperreal.

The "nonprincipal" requirement on F eliminates the trivial case, which doesn't give anything new. If F is principal – say, all subsets containing 17 – then F-similarity is just: $a \sim_F b \iff a_{17} = b_{17}$, so the quotient is isomorphic to $\mathbb{R}$.

Because $F$ is nonprincipal, it contains all cofinite sets – every set whose complement is finite. (Clearly the converse is true too.) Two sequences that agree on a cofinite set are "eventually equal" – they're identical after some $n_0 \in \mathbb{N}$.

Examples:

  • if $a = (5, 5, 5, \dots )$, then all other sequences which are eventually constantly equal to $5$ will be equivalent to $a$. The equivalence class of $a$ is the hyperreal version of $5$.

  • If $b = (\pi, \pi, \pi, \dots )$, then $b$ is not equivalent to $a$: $b \not\sim a$. The equivalence class of $b$ is the hyperreal version of $\pi$.

  • If $c = (0, 1, 2, \dots, n, \dots) = (n)_{n \in \mathbb{N}}$ then $c$ is not equivalent to either $a$ or $b$, and in fact the hyperreal that $c$ represents is greater than every standard real (the hyperreals are a non-Archimedean ordered field).

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  • $\begingroup$ So I get that they the two sequences will be treated as equivaent based on what you've said. But surely then we could just pick another two sequences and they will be equivalent and represent the exact same Hyperreal number? But thinking like that, every two sequences would represent the exact same Hyperreal number. I'm thinking about the quotient something along the lines of there is a big set of real-valued sequences, and another set (the relation) which is dividing up the set of real-valued sequences into various groups. Is that intuition somewhat correct? $\endgroup$
    – user253919
    Oct 8 '15 at 20:54
  • $\begingroup$ @hazza777 I made my initial reply/comment part of my answer. The examples are intended to show that, no, not all sequences are equivalent :) Your intuition about the set of sequences and the similarity relation are correct. The $\sim$ relation is a (very nontrivial) equivalence relation, so it partitions the sequences into equivalence classes. $\endgroup$
    – BrianO
    Oct 8 '15 at 21:22
  • $\begingroup$ I didn't see those inital examples. Thanks that has helped. I'm not too sure what the equivalence classes actually are in this case; usually i'd be fine with this but since it's using an ultrafilter i'm getting confused. The issue of real-valued sequences not forming a field because of zero divisors being present, is fixed when Hyperreals are constructed since one of the sequences becomes equal to the 0 equivalence class and the other 1 equivalence class? e.g. a=1,01,0,... b=0,1,0,1,... then ab=0,0,0,0,... then when Hyperreals constructed a=[0] b=[1] where [] denotes equivalence class $\endgroup$
    – user253919
    Oct 10 '15 at 14:34
  • $\begingroup$ Yes, because it's an ultrafilter you can't explicitly write down, say, a set comprehension specifying all the sequences equivalent to, say, [0, 0, 0 ...]. To do that, you'd need to be able to say exactly what sets are in the ultrafilter, and, well, you can't. Your "a" and "b" examples are good ones. Not both of them can be equivalent to [0], because the ultrafilter is proper. Either the evens or the odds are in the the ultrafilter, but just which one is, you can't know -- so you can't know just which of $a, b$ is $\sim [0]$.But one, and only one, is. $\endgroup$
    – BrianO
    Oct 10 '15 at 17:04
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As BrianO said, you can think of an ultrafilter as a consistent way to say something is true for "most" or "almost all" elements of your set (in this case $\mathbb{N}$). Of course the real power of ultrafilters comes from the word "consistent" in that sentence, and the axioms are set up precisely to ensure a high level of consistency.

For the construction of hyperreals we say that two sequences are equivalent if they agree for "most" values.

Here's one use of this, which is a step in showing the hyperreals are a field: A sequence $a_n$ of reals is invertible as a hyperreal iff it is not equivalent to the sequence of all zeroes. One direction is clear, if "most" of the entries of $a_n$ are zero, then the same will be true of the product with any other sequence so it can't possibly have a multiplicative inverse. The other direction is also nice, just construct a multiplicative inverse by saying $b_n = 1/a_n$ if $a_n \ne 0$ and $b_n = 17$ otherwise. Since "most" of the entries of $a_n$ are not zero, then "most" of the entries of $a_nb_n$ will be 1.

The real (or should I say hyperreal) fun begins when you consider a sequence like $a_n = 1/n$ which is greater than zero all of the time, and less than any positive real number "most" of the time.

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  • $\begingroup$ ... greater than zero all of the time, and less than any positive real number "most" of the time -- great turn of phrase :) $\endgroup$
    – BrianO
    Oct 8 '15 at 21:48

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