Re: Skolem's Paradox and why is math the way it is?
From: KRamsay (kramsay_at_aol.com)
Date: 11/28/04
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Date: 28 Nov 2004 17:44:44 GMT
Sorry about the lag; I think I've been trying to do too many things
all together. I'll probably have to taper off my contribution soon.
In article <39d6e584.0411201743.513fc@posting.google.com>,
troubled6man@yahoo.com (J.E.) writes:
|> Most mathematics, incidentally, uses only a relatively uncontroversial
|> portion of set theory. People deal with things like sets of real
|> numbers all the time, but not so often the parts that depend on
|> (say) the axiom of replacement.
|
|Since we haven't proven that ALL the set theory axioms taken together
|are consistent, then you'd expect that if a smaller subset works for
|physics, that someone would have tried to prove that that smaller set
|of axioms was consistent. Is there such a proof?
There is indeed a proof that the axioms of set theory, without
replacement, are consistent. However! The proof uses the axiom
of replacement. By Goedel's second incompleteness theorem, a system
like Z (the system without the axiom of replacement) can't prove
its own consistency, so some additional axiom is needed, of course.
The most familiar proof works by showing the existence of a model
of Z. The most obvious model of Z is V_{omega+omega}, which is the
first part of the cumulative hierarchy: Let V_0 be the empty set,
and by induction let V_{n+1} be the set of all subsets of V_n for
n=0,1,2,.... These are all finite sets, which can be written as
finite strings if need be. They're known as the hereditarily finite
sets, since not only are they finite, but their members are finite,
the members of their members are finite, and so on.
Then let V_{omega} be the set of all hereditarily finite sets. This
is a countably infinite set. It requires the axiom of infinity to
prove that it exists, but its elements can be put into correspondence
with the natural numbers. Incidentally V_{omega} is a model of all the
axioms of ZFC except for the axiom of infinity.
Define V_{omega+(n+1)} to be the set of all subsets of V_{omega+n}
inductively. Then finally let V_{omega+omega} be the union of all
of the V_{omega+n} for natural numbers n. That last step is the only
step in the proof where the axiom of replacement is needed.
[...]
|> I don't think there is a be-all theory.
|
|Isn't being a "be all" theory part and parcel of the "standard
|interpretation" that everything that could exist for anyone that is
|"small" enough to be a set, is a actually a set?
No. It seems as though you're confusing an interpretation with a
theory. One can believe that one has a coherent interpretation in
which when one says "set", it really means "any kind of set". That
doesn't mean that one has a "be all" theory of what they're like.
It's like the difference between believing that you know what an
apple is, and knowing everything that there is to know about apples.
|The only reason to
|insist on this rather than just that enough sets exist to satisfy the
|axioms is because one wants to pretend like one can have "an
|everything", where the universe of discourse of standard
|interpretation set theory is superior to all other universe of
|discourses.
No. There are larger domains of discourse. Set theory is merely a
relatively large one.
|Otherwise why is that interpretation consider necissary or even
|standard?
It seems to me that you're putting the cart before the horse again.
I consider the primary purpose of the axioms to be to investigate a
domain of inquiry systematically, *not* to define the limits of the
domain of inquiry.
Unless you have a good reason to think that it's impossible to talk
about "all real numbers" (and I don't think you do), it's a very
strange suggestion that we should talk only about some of them.
|> |But IF
|> |logic avoids having infinite regresses into higher-order logics so
|> |that we CAN sit down and discuss how you make theories, so isn't that
|> |worth considering?
|>
|> What infinite regress into higher-order logic is there for anyone?
|
|Like you say in your previous post, you need SO set theory to define
|the strongly inaccessible cardinal, in order to get a faithful model
|of set theory, but once you introduce SO set theory, people will want
|the "other sets" too, because the whole POINT of introducing that
|cardinal was to get "all the sets" that were missing in previous
|models. You aren't "succeeding" at getting all the sets.
I don't think I said you needed second-order set theory to define a
strongly inaccessible cardinal. It only needs a first-order definition
in terms of the "epsilon" relation. As usual, of course, to say that
an element of a model satisfies this definition as relativized to the
model doesn't mean the same thing as saying that its a cardinal
satisfying the definition.
I think you need to distinguish between various senses of "get", here,
pertaining to the scope of variables, the language as a whole, and the
axioms.
If I say that all real numbers are either <0, >0 or =0, then my
statement implicitly contains a quantifier for a variable ranging
over all the real numbers. The statement succeeds in "getting" all
the reals in the sense that it quantifies over them.
If I say that I can define any arbitrary algebraic number, then I've
"gotten" them all in second sense. This sense is relative to my language,
since what definitions I can provide depend on how rich my language is.
If all I have are the elementary school operations of +,-,*,/ and maybe
simple roots x^(1/n), then my language is too weak to define all of them.
There is no language (with finite expressions) strong enough to "get"
all the real numbers in this sense.
If I say that I can prove the existence of a weak inaccessible cardinal,
then I have "gotten" it in a third sense, which depends on what can be
proven (which depends on which axioms are accepted).
These three senses are in order of increasing narrowness. In order to
prove that something exists, I need to be able to describe it. In order
to describe it, I need to have (implicitly at least) variables that
range over a domain that includes it.
When you say "get", you often seem to be sliding between these senses.
You seem often to be trying to treat them as if they were the same
thing. You don't seem to see any problem with treating the narrowest
sense (what we can prove to exist) as if it "should" somehow be the same
as the range of our quantifiers. I don't see any point in doing that.
If we have been talking about real numbers, I don't see any point in
deciding to assume that we are *always* talking about some subset of
definable or provably existing real numbers instead. The only reason I
can think of for wanting to do that, not just sometimes but generally,
would be if there was somehow a serious problem with the concept of
"real number", some genuine ambiguity or incoherence.
[...]
|> You can't say that a graph has three connected components, in it.
|
|Do you have a citation for that result, or better yet can you state
|your definition of graph and connected component?
I don't have a citation for it offhand. There's nothing special
about three, by the way. I used that because I was thinking I could
go on to point out that "This graph has at least three connected
components" as well as "this graph has at least four connected
components" were both expressible in IF logic, but not "this graph
has (exactly) three connected components".
Connectedness is a familiar example of a non-first-order property of
a structure. It's not expressible in IF logic because IF logic
extends first-order logic by permitting an existential quantifier
over subsets of the structure (in effect).
Those are the sigma-1-1 properties. But connectedness is a pi-1-1
property, which is how it escapes being first-order definable. It's
usually defined as meaning that any two vertices are joined by a path.
That's equivalent to the nonexistence of a way to divide the graph
into two nonempty disjoint subsets, with no edges connecting any
vertex in the one with any vertex in the other.
Lemme see if I can sketch a proof that it's not also sigma-1-1.
Suppose there is a game (associated to a sentence in IF-logic)
with a winning strategy for the verifier on an infinite connected
graph where the degree of the vertices is bounded above by some
natural number n. For simplicity, we can take a set of vertices
indexed by the integers (including negative integers) where the
edges connect adjacent vertices. The winning strategy consists of
a collection of functions f(a1,...,a_m) for different values of m.
I claim that there exist disconnected graphs where the verifier
also has a winning strategy.
First, a simple example. I'm pretty sure that the graph consisting
of two disjoint copies of the original graph is an example. Take
the points (x,y) in the plane where y=0,1 and x is an integer, and
join the points (x,y) and (x+1,y) by edges to form the graph. I'm
suffering a little writer's block on the proof, though.
Second, pulling out the big guns. The original graph has just two
relations on it, xEy meaning that x and y are joined by an edge,
and x=y. Now augment the structure by adding the functions f that
correspond to the verifier's winning strategy. That's now a model
of the first-order sentence saying that the verifier wins the game
regardless of what the falsifier plays.
The upward Lowenheim-Skolem theorem says (as a special case) that if
a first-order sentence has an infinite model, it also has an uncountable
model. So the sentence saying the functions f are a winning strategy
for the verifier, and that all of the vertices have degree 2, also holds
true for some functions f' on a graph with uncountably many vertices.
Since a connected components of a graph whose vertices have degree 2
is always countable, this graph with uncountably many vertices is
disconnected.
The same proof works just as well for the sentence, "this graph has
three connected components". Any game associated with a sentence of
IF logic that can be won on a graph with three components can always
be won on a graph with more components.
To me this just reveals something missing in IF logic. We can understand
nearly as easily what it means to be able to win the following game:
the refuter picks two vertices, and to win the verifier has to present
vertices one at a time, each connected to the previous one, and starting
with the first vertex given by the refuter get to the other one. It's true
that this involves the notion of a finite sequence of moves, but I don't
see how that can be much worse than the kind of arbitrary strategy
allowed the players in a game associated with an IF logic sentence.
IF logic just is so limited that we can't say it. We can even say what I
would call the REAL negation of the claim that the graph is connected:
that it can be divided into two nonempty parts that aren't connected to
each other.
[...]
|> |Every model of set theory lacks a set that should exist as much as the
|> |alleged "uncounted real" should exist.
|>
|> If by "model" you mean a set with an epsilon relation on it, then
|> this is correct, but people often mean by "model" either a set *or*
|> a proper class with an epsilon relation on it. The cumulative
|> hierarchy does not "lack" a set that "should" exist-- it consists
|> by definition in all the well-founded pure sets.
|
|This is really hard to discuss non-circularly. The words structure,
|class, function, set, collection, relation all have definitions "in
|the theory" and to use the same words outside of the theory is begging
|for confusion.
I think trying to force them to be theory-dependent in an inconsistent
way is begging for deeper confusion.
I don't think there is such a thing as a different definition of
function, for example, "in the theory". It's possible that you are
alluding to relativizing some of these concepts to models, but you
need to distinguish theories from models.
|What do you want to take as given?
I'm pretty flexible about what to take as given, so long as we're
consistent with it.
We can start the formal development by taking "set" as an undefined
term, either believing that we know a definite meaning for it, or by
merely proceeding as though it does. To remain consistent with such
a starting point, however, makes a lot of statements nonsensical, like
saying that this domain of sets lacks a set that "should exist".
If by model, you mean a set having an epsilon relation and so on, then
it's circular to try to define "set" relative to "model", because this
sense of "model" depends on the concept of "set" already. "Set" needs
to have a meaning that doesn't depend on models. We do not need to start
out with a "model" in this sense of some theory-- it's circular to try
to start that way. It's true of this kind of model that there are sets
that are not members of it; it does not contain itself, for instance, and
it is by definition a set.
If by model you mean something broader, like what philosophers sometimes
call a domain of discourse, then we can, if you like, call the domain of
sets that we start out with a "model", but then there is no longer any
sense in saying that our starting model is missing any sets. We have just
defined the domain of discourse to consist of all such objects that we
are going to be calling "sets".
When you asked questions about deciding to "use" the minimal model,
since the minimal model is defined in terms of the concept of "set",
whatever you meant by "minimal model" was dependent on some prior
concept of "set". Certainly if you want to use such a model for physics
there isn't an inconsistency, but you are still stuck with the fact that
you started out with one brand of set theory, and then created a second
kind that depends on the original kind. Most people figure that they
are better off just sticking to whatever kind of set they started out
with.
| We could have a
|third person in the game, and have the third person start talking,
|saying "a in M, aea in A, a' in M, a'ea' in A, aea' in E, a'ea in A,
|a'' in M, a''ea'' in A, a''ea' in A, a''ea in A, a'ea'' in A, aea'' in
|A, a''' in M, a'''ea''' in A, a'''ea in A, a'''ea'' in A, a'''ea'' in
|A, aea''' in A, a'ea''' in A, a''ea''' in A, ..." and but where the
|third person chooses freely whether to say "xey in A" or "xey in E",
These little scenarios where you describe some outside source generating
a structure incrementally strike me as having so little about them that
is analogous to the way mathematics actually works or how we talk about
it, that I can hardly think of anything to say about them.
[...]
|I don't understand your claim about the cumulative hierarchy, once you
|"finish" the model, someone can take the standard interpretation and
|say that some sets are missing,
Why?
|isn't V=L considered "restrictive" by
|mathematicians?
The best guess I can come up with here is that you're confusing two
definitions of hierarchies here, the definition of the cumulative
hierarchy (whose members constitute V) and the definition of the
constructive hierarchy (whose members constitute L).
Keith Ramsay
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