Keepin' it Hyperreal

I haven't posted anything here in a while, so let's do something cool... like construct $^*\mathbb{R}$

Swag.

The hyperreals, $^*\mathbb{R}$, (Abraham Robinson's world of nonstandard analysis) are a neat way to rigorously develop the notion of infinitesimal numbers in analysis.  Because we want our numbers to form a field, that means we'll also pick up a new description of 'infinitely large' numbers (quite distinct from the cardinals/ordinals though).

This will be more about the crazy set theory shenanigans that go into building $^*\mathbb{R}$, rather than actually working in  $^*\mathbb{R}$... maybe I'll cover the practical side in another blogpost...
You remember building $\mathbb{R}$ from the set of all cauchy sequences in $\mathbb{Q}$?  Well, to construct $^*\mathbb{R}$ from $\mathbb{R}$, we start with the set of all sequences in $\mathbb{R}$ (no restrictions at all) and call it $\mathbb{R}^\mathbb{N}$

Now we need to put some sort of equivalence relation of these sequences.  Intuitively, we're going to say two sequences, $\{a_i\}_{i \in \mathbb{N}}, \{b_i\}_{i \in \mathbb{N}} \in \mathbb{R}^\mathbb{N}$ are equivalent $\left( \mbox{i.e. }\{a_i\}_{i \in \mathbb{N}} \sim \{b_i\}_{i \in \mathbb{N}}\right)$ if they agree on a 'large' subset of $\mathbb{N}$: i.e. if the set $\{ i \in \mathbb{N}: a_i = b_i\}$ is 'large'.

It's a curious way to define an equivalence.  On one hand, a pair of cauchy sequences like $\{0.9, 0.99, 0.999 ...\}$ and $\{0, 0.9, 0.99 ...\}$, while very similar, do not precisely equal each other at any term, and thus won't be equal.  And sequences like $\{0,1,0,1 ...\}$ look like they're going to give us a headache.  Is it $1$ or is it $0$?  It ought to be one of them... but it's not clear which one...

Let's write out some nice properties for 'large' subsets of $\mathbb{N}$ to have, and see where that takes us.

$1.$ $\mathbb{N}$ itself is large (this way, a sequence is at least equivalent to itself)
$2.$ $\emptyset$ is not large (this way, not every pair of sequences in $\mathbb{R}^\mathbb{N}$ are equivalent)
$3.$ If $A$ is large and $A \subset B$, then $B$ is large
$4.$ If $A$ and $B$ are both large, then $A \cap B$ is large.

We can define a notion of 'largeness' by simply specifying what the collection of all 'large' sets is, some $\mathcal{F} \subset \mathcal{P}(\mathbb{N})$.  A good example (and also a good starting point) is the collection of all cofinite subsets (sets whose complement in $\mathbb{N}$ is finite).  Generally speaking, any $\mathcal{F}$ that fits these four criteria  is called a proper filter on $\mathbb{N}$.  As the notation would suggest, if $\emptyset \in \mathcal{F}$, we would call $\mathcal{F} = \mathcal{P}(\mathbb{N})$ the improper filter on $\mathbb{N}$.

Put those Filters on the backburner for now.  There's one more (tentative) property of 'large sets' that makes all the numerical magic happen:

$5.$  For any $A \subset \mathbb{N}$, $A$ is a large set if and only if $A^C$ is not a large set,

This lets us put a strict order relation $\prec$ on the equivalence classes in $\mathbb{R}^\mathbb{N}$.  We can say $\{a_i\}_{i \in \mathbb{N}} \prec \{b_i\}_{i \in \mathbb{N}}$ iff  $\{i \in \mathbb{N}: a_i < b_i\}$ is large.  By the usual trichotomy of $<$ and property #$4$ of large sets, precisely one of these sets will be large for any pair $\{a_i\}_{i \in \mathbb{N}}, \{b_i\}_{i \in \mathbb{N}} \in \mathbb{R}^\mathbb{N}$:

$\{i \in \mathbb{N}: a_i < b_i\}$, i.e. $\{a_i\}_{i \in \mathbb{N}} \prec \{b_i\}_{i \in \mathbb{N}}$

$\{i \in \mathbb{N}: a_i = b_i\}$, i.e. $\{a_i\}_{i \in \mathbb{N}} \sim \{b_i\}_{i \in \mathbb{N}}$

$\{i \in \mathbb{N}: a_i  > b_i\}$, i.e. $\{a_i\}_{i \in \mathbb{N}} \succ \{b_i\}_{i \in \mathbb{N}}$

Which gives the same trichotomy with $\prec$.  We also get transitivity pretty easily:

If $\{a_i\}_{i \in \mathbb{N}} \prec \{b_i\}_{i \in \mathbb{N}} \prec \{c_i\}_{i \in \mathbb{N}}$,

$\{i \in \mathbb{N}: a_i < b_i\}$ and $\{i \in \mathbb{N}: b_i < c_i\}$ are both large sets

Thus by #$3$ & #$4$,

$\{i \in \mathbb{N}: a_i < b_i\}\cap \{i \in \mathbb{N}: b_i < c_i\} \subset \{i \in \mathbb{N}: a_i < c_i\}$ is a large set
$\therefore\{a_i\} \prec \{b_i\} \prec \{c_i\} \Rightarrow \{a_i\} \prec \{c_i\}$

Neat!  These wonky criteria ensure that any definition of 'large sets' will have a natural ordering on the equilvance classes of $\mathbb{R}^\mathbb{N}$ it defines.  We even get a field structure when we use component-wise addition and multiplication:  If $\{a_i\}_{i \in \mathbb{N}} \nsim \{0\}_{i \in \mathbb{N}}$ (i.e. a 'large' subset of $\{a_i\}_{i \in \mathbb{N}}$ is nonzero) then $\{a_i\}_{i \in \mathbb{N}}$ has a multiplicative inverse: define $\{b_i\}_{i \in \mathbb{N}}$ such that

$b_i = \left\{ \begin{array}{l} a_i^{-1} \mbox{ when } a_i \neq 0 \\ 0 \mbox{ when } a_i = 0 \end{array} \right.$

Thus we have that $\{a_i\}_{i \in \mathbb{N}} \times \{b_i\}_{i \in \mathbb{N}} = \{a_i \times b_i\}_{i \in \mathbb{N}}$ agrees with $\{1\}_{i \in \mathbb{N}}$ on a 'large' set.

All right.  So a classification of the elements of $\mathcal{P}(\mathbb{N})$ into 'large' and 'not large' elements is guaranteed to produce a simply ordered field from $\mathbb{R}^\mathbb{N}$ that is 'bigger' than $\mathbb{R}$.

So when do we get a good definition of 'large' already?

And now the magic happens.

Let's look at those filters again.  Note that when $\mathcal{F}_1 \subset \mathcal{F}_2$, we can consider $\mathcal{F}_2$ a 'refinement' of $\mathcal{F}_1$.  In this sense, $\subset$ is a partial order relation on the set of filters over $\mathbb{N}$.

Theorem: A proper filter $\mathcal{F}$ is maximal iff $\forall A \in \mathcal{P}(\mathbb{N}), \, A \in \mathcal{F} \Leftrightarrow A^C \notin \mathcal{F}$

What does this mean?  The filters that satisfy condition #$5$ are precisely the maximal ones.
Given their significance, maximal proper filters are also called ultrafilters.

Proof:
Part one: "maximal $\Rightarrow$ #$5$" by way of  "$\neg$#5 $\Rightarrow \neg$ maximal"

If $A, A^C \notin \mathcal{F}$, we must show there exists a filter $\mathcal{F}'$ such that $\mathcal{F}\subsetneq  \mathcal{F}'\subsetneq \mathcal{P}(\mathbb{N})$

Let $\mathcal{F}' = \{C \in  \mathcal{P}(\mathbb{N}): \exists B \in \mathcal{F}\mbox{ for which } A \cap B \subseteq C\}$

(Now how do you show $\mathcal{F}'$ is a proper filter? Run through that list of conditions...)

Part two: "#$5$ $\Rightarrow$ maximal"

Suppose $\mathcal{F}'$ satisfies $\mathcal{F}\subsetneq  \mathcal{F}'$
Thus $\exists A \in \mathcal{F}' : A \notin \mathcal{F}$
By #$5$, $A^C \in \mathcal{F} \subsetneq \mathcal{F}'$
Because $\mathcal{F}'$ is a filter,  $A, A^C \in \mathcal{F}' \Rightarrow A \cap A^C = \emptyset \in \mathcal{F}'$
$\therefore \neg \exists \mathcal{F}': \mathcal{F}\subsetneq  \mathcal{F}'\subsetneq \mathcal{P}(\mathbb{N})$

 $\Box$

But where does this ultrafilter on $\mathbb{N}$ come from?  We use Zorn's Lemma to pick an ultrafilter that contains the (proper, but definitely not maximal) filter of cofinite sets in $\mathbb{N}$.

(Zorn's Lemma does apply: every ascending chain of filters in $\mathcal{P}(\mathbb{N})$ has a maximal element, as an arbitrary union of nested proper filters is still a proper filter)

So this is where $^*\mathbb{R}$ starts looking pretty crazy... given that we need the axiom of choice to pick our ultrafilter for us.  But a canny reader may want to point out further: shouldn't there be a lot of different ultrafilters that we could get in this way?  Wouldn't a different ultrafilter produce a different $^*\mathbb{R}$?

This is when even more crazy stuff happens.  Proving that all such 'versions' of $^*\mathbb{R}$ are isomorphic requires the Continuum Hypothesis.

Again, crazy set theory shenanigans.  The notion that you would need the continuum hypothesis to construct a system of numbers besides the ordinals... still boggles my mind.

Source: Robert Goldblatt's Lectures on the Hyperreals