[FOM] disguised set theory "DST"
Randall Holmes
m.randall.holmes at gmail.com
Tue Oct 4 21:43:47 EDT 2011
Dear FOMers,
In this note I will explain why Frode Bjordal has not succeeded in
proving Infinity
or inconsistency in Zuhair al-Johar's proposed set theory DST
(disguised set theory).
I should say that I am not arguing as an advocate of this very strange
theory, but as someone
who has been trying for some time to show either that it is
inconsistent (as it rather appears
it ought to be) or to discover what a model might look like.
The simplest exposition of DST is as follows: it is a first order
theory with primitive
relations e (private membership) and = (equality).
Axiom I: (Az.z e x <-> z e y) -> x=y (extensionality for private membership)
Axiom II: trans(a) means (Axy.x e y and y e a -> a e a). x c= y
means (Az, z e x -> z e y).
For any x there is a set
TC(x) (unique by axiom I) such that for all y, y e TC(x) iff y e A
for every set A such that trans(A) and x c= A.
Definition: x E y means x e y and not y e TC(x).
The E relation is called the public membership relation.
Axiom III: For every formula phi mentioning only E and =, there is a unique set
{x|phi} such that for all x, x e {x|phi} iff phi. Note that it is the
*private* extension
of {x|phi} that is determined by the formula phi: this means for
example that set brackets
cannot appear except in parameters in an instance of Axiom III.
Always remember that TC(x) is the transitive closure of x with respect
to the *private* membership relation.
There are variations. I have suggested including a full class theory
(so that Axiom II
would be impredicative class comprehension as in Kelley-Morse; Axiom
III would have an additional
restriction in only allowing sets (e-elements) as parameters, and an
object witnessing
Axiom III would be a set in the sense of being an e-element. I also
note that in a class
treatment there is no reason that TC(x) might not be a proper class
for some sets. I definitely
think that a useful version of this theory (something whose existence
I doubt) would need
some kind of induction principle to support reasoning about transitive
closures; the full
class theory would handle this neatly.
I have considered not allowing = in instances of axiom III. I have
considered using the supertransitive closure
(which includes subsets of its elements as well as elements of its
elements) instead of the transitive
closure in the definition of E.
For the discussion of Bjordal's posts, the variations are not needed.
It is important to note that while e is nonwellfounded because it has
cycles (for example {x | x = x} = V
is the universal set for e and so V e V), the relation E, while also
non-well-founded
(a proof of this would take a little work) is acyclic. In a chain x1
E x2 E x3 E ... E xn
we cannot have xn = x1, because we would then have each xi in the
transitive closure of each other xj,
as the relation x e TC(y) is transitive, and x[n+1] e TC(xn) is
excluded for each n by
the definition of E.
We turn to Bjordal's proposals for an argument for Infinity, which are
similar to arguments
that I have attempted. Define 0 as {x | x =/= x}, the empty set.
Note that this is *the*
empty set only in relation to private membership. The set {V} for
example whose only private
element is the universe, is readily seen to also have no public elements.
Bjordal wants to define ordinal successor intuitively by x' = {u | u E
x or u = x}. He
is aware that x' defined thus cannot appear in an instance of Axiom
III, so he suggests
a contextual definition of x' E y as "there is a w E y such that the
public elements of w
are exactly the public elements of x and x itself". He then defines N
as {n | (forall I.0 E I
and (forall m.m E I -> m' E I) -> n E I}.
This exact definition fails for a technical reason having to do with
the choice of quantifier
in the contextual definition. There is an even simpler failure:
it happens that not all objects have any successors at all (there is
no V' for example)
but Im willing to restrict the notion of inductive set to say that
elements of inductive sets
which have a successor have some successor in the inductive set, as a
friendly amendment).
It happens that 0 has more than one "successor". The set {0}
exists and its public extension is the same as its private extension.
Define W = {x | x E V}
(this is not the same as V: its private elements are the public
elements of V, that is, the
sets which do not have V in their transitive closure). {0,W} has the
same public extension
as {0}, and also {0,W} E V is true ({0,V} has the same public extension as {0}
but is not a public member of anything). Observe that V-{{0}} and
V-{{0,W}} are both inductive
sets -- anything which has a successor and is in either of these sets
has a successor in that set.
But this means that {0} is not a public element of N. In fact, for
any successor 1 of
0, V-{1} is inductive in the sense required by Bjordal's definition of
N, so 1 is not in N.
It rather appears that Bjordal's N is just {0}. This set is not
closed under successor, which
shows directly that his claims about consequences of his definition are false.
So, I make a further friendly amendment: change the contextual
definition of x' E y to
"for every w such that the public elements of w are exactly the public
elements of x
and x itself, and such that w is a public element of some set, w E y".
This set is in my opinion defined correctly. But it does not have
nice properties. Clearly
0 e N, and since N is obviously not in TC(0), we have 0 E N as well.
Now suppose x E N.
We can deduce that x e N, and so that x belongs to every public
extension which contains 0
and is closed under successor in our precise sense, and so any
successor x' of x also belongs
to all these public extensions, so for any successor x' of x, we have
x' e N. At this point
in Bjordal's argument, he forgets the distinction between public and
private membership -- he jumps
to x' E N. Unfortunately this does not follow. We know that N is not
in the private extension of TC(x)
because x E N: so we know that N is not in the transitive closure of
any of the public elements
of x'. But x' might have private elements (private because they have
x' in their TC) which
also have N in their TC. One could protest -- but what about the
specific x' defined
by {u | u E x or u = x}? This has exactly the right private members
and we can see that
N is not in its TC. Unfortunately, it might be the case that the
public extension of this "x'"
might not be the same as its private extension: if x' e TC(x), this
will happen. This implies
that x is a very bad set; unfortunately there does not seem to be any
way to exclude such bad
sets from our inductive sets (the conditions that define them involve
e essentially and cannot
in any obvious way be expressed in terms of E and =).
It should be noted of course that any model of DST (if there are any)
is externally infinite; the question
here is whether DST contains a set which it can tell is infinite in
internal terms.
In a second post, Bjordal presented an argument for inconsistency of DST:
Here is Bjordal's argument, with my comments in brackets:
Let ordered pairs be defined a la Kuratowski
[<x,y> = {{x},{x,y}} can be defined as a private extension;
if <x,y> is in the transitive closure of {x,y} it will not have the
correct public extension; there is
no way to specify the privately defined <x,y> using just E and =, but
of course one can write a formula
expressing the idea that a set p is a Kuratowski ordered pair of x and
y; such p's will not be unique,
but can be used to represent relations definable in terms of E and =,
as long as the objects related
are capable of being thrice iterated public elements of sets]
Let Uz signify the union set of z
[the public membership relation is nonextensional so this may not
describe a single object,
and in terms of the public membership
there is no guarantee that a set *has* a union, though I do not have a
counterexample handy. We can
write a formula which says "U is a union set of x" in terms of the
public membership]
Bjordal continues to use the notations for empty set and ordinal
successor from above.
Let t(a) be the set provided by the comprehension
{x | (forall y.<0,a> E y and (forall uv.<u,v> E y -> <u',U(v)> E y) -> x E y}}
[because of failures of notation due to nonuniqueness of objects
mentioned, this is difficult,
but in fact something like this set can be defined. The intention is
to produce a function taking
each natural number n to the nth iterated union of a. The problems
with induction evidenced above
suggest to us that this is not going to work perfectly, but as will be
seen I'm willing to grant for
the sake of argument that this construction does exactly what is wanted.]
Let X be given by the comprehension {x | (forall ny.<n,y> E tc(x) -> not y E x)}
[X is supposed to be the set of all x such that x is not an element of
the transitive closure of x
with respect to public membership. Bjordal does not realize that
public membership is acyclic in all cases,
as the subsequent discussion reveals. It is not permissible to use
tc(x) in an instance of comprehension
in this way, because tc(x) is a set with a specified *private*
extension. However, I am willing to
grant that this whole construction works as intended, because the set
of all x such that the public transitive
closure of x does not contain x is indeed provided by Axiom III in a
different way: it is the universe!!!]
Clearly x e X iff not X e TC(x) [no, x E X iff not X e TC(x), where X = V].
For if xeX and XeTC(x) then xeTC(x),
and x would be cyclic. [I *think* that Bjordal is here confusing
public and private membership.
He thinks that membership in X will exclude x e TC(x). It doesn't.
The definition of X, if taken as
succeeding, excludes members with cycles in public membership,
which don't exist anyway, as I showed earlier in this note;
it does not exclude cycles in private membership, whose existence is
quite hard to describe in terms of E and
= (quite probably impossible). If there is a formula in E and = which
is equivalent to x e TC(x)
we DO get a paradox in DST; Bjordal is in many ways on the right track.]
As xEX iff xeX and not XeTC(x), we have that
xEX iff xeX. [No, as in the actual case X=V we have x e X true in all
cases but x E X true
only when X not in TC(x).]
Suppose first that X is cyclic, i.e. XEX or XE(2)X or…
Then we derive in a finite number of steps that X is not cyclic.
[As I have shown above, there are no sets which are cyclic in the
sense of public membership.]
Suppose next that X is not cyclic.
[the only possibility]
Then X fulfills the comprehension
condition for X and we derive that XEX.
[Just as at the end of his infinity argument,
Bjordal forgets that comprehension gives *private* extensions. We can
only conclude
X e X -- in the case X=V which actually obtains, we have V e V but we
do not have V E V
because V e TC(V) = V.]
So X is cyclic iff X is not
cyclic according to the suggested set up.[As we see above, the
argument fails in various ways.]
Conclusions: This theory is very strange. I would like to see its
status settled: it seemed to me
at first blush that it ought to be inconsistent and relatively easy to
show inconsistent, but it seems not
to be the case (a great deal of information is concealed by the
restriction of formulas defining sets
to those using public membership). It is also clear that the world of
DST (if there is such a world) is
quite weird. If some FOMer can come up with a model or an
inconsistency proof I will be very interested,
but do remember that some very subtle distinctions are at play here.
The fact that extensions for one membership
relation are defined in terms of another membership relation is
reminiscent of the situation in Kisielewicz's
"double extension set theory", two versions of which I showed to be
inconsistent in a quite elaborate way,
though the consistency of the weakest system he proposed remains an
open problem. Unlike Zuhair, I do
not see this theory as a place to actually do mathematics -- it is too
strange, even if it is consistent.
I'm happy to discuss the other things I know or think I know about
this theory with anyone who is interested. I state
for the record that I have made a fair number of mistakes myself
trying to think about DST; it is extremely tricky.
--
Sincerely, Randall Holmes
Any opinions expressed above are not the
official opinions of any person or institution.
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