# Non-foundationalist foundational thinking

Vaughan Pratt pratt at cs.stanford.edu
Thu Jun 9 03:10:32 EDT 2022

```HF writes

>
> In particular, I have never seen an account of what mathematical objects
> "really are" that is even remotely convincing.
>

Actually he has, in a talk I gave in Toronto's Fields Institute a few years
ago, that I claim would be convincing to many if not to Harvey himself.

Compared to the idea that every mathematical object is a set with a
well-defined rank such that the empty set is of rank zero and so on, I find
the category theory notion of an object more convincing, even if Harvey
doesn't

In any given category one can look at the arrows into a given object and
those out of that object.  The ratio of arrows in to arrows out, suitably
weighted according to the other ends of those arrows, gives a much better
idea of the "flexibility" of the object in the context of the category
housing it than does merely representing it as a set of some rank.

The more arrows out relative to arrows in, the more flexible the object.

In the case of finite dimensional vector spaces, that ratio is 1.  As such,
those vector spaces have what one might think of as ideal flexibility,
neither too much nor too little.

Let X be any fixed finite set of cardinality m.  Finite sets Y of
cardinality n have n^m arrows in from X and m^n arrows out to X.  Most sets
Y being larger than X, i.e. n >> m, we can compare n^m and m^n and we find
that the arrows into Y from X  grow with increasing Y as a polynomial of
degree m while those from Y to X grow exponentially with base m.

So regardless of what X is, all sets much larger than X have far more
arrows out than in.  Sets are insanely flexible, which might be why
Poincare was hoping the interest in them would die down.

Now consider finite Boolean algebras.  These are precisely the dual of
finite sets: they have far more arrows in than out, with m and n
interchanged in the above.  As such Boolean algebras are insanely stiff.

Sticking to finite objects, does this generalize beyond vector spaces,
sets, and Boolean algebras?

Yes, it generalizes to all finite matrices over any given set K.  These are
the finite members of the category Chu(Set,K).  A great variety of
categories of finite objects can be embedded both fully and faithfully in
Chu(Set,K) for sufficiently large K.

Taking K = {0,1} is already enough to encompass a great variety of
categories of interest to lattice theorists and other students of partial
orders.  And K = {0,1,2} extends this enormously.

There may be even better accounts of objects showing what they "really
are".  If there are better accounts than objects as Chu spaces I'm all ears.

Vaughan Pratt
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