Deriving X ∉ X through an axiomatic ∊ relation
No claim of originality is offered.
From a modern vantage point, we need not resort to a theory of types or an axiomatic ban on X ∊ X. We are not in the same position as Russell, Whitehead, Zermelo and other pioneers in wrestling with how to best ground mathematics in some overall system.
The element or membership relation ∊ pairs an element with a set. We adopt the convention that the left-hand part of the pair is an element and the right-hand part is a set. So x ∊ X is ∊< x,X >. X is defined by the collection of x's with which it is paired, whether intensionally (by rule) or extensionally (by particulars). We must beware regarding the left part of the relation as a set A and the right part as a set B, making the relation a subset of the cross product set A X B. For then we face infinite regress.
This I suppose could be handled by an infinite sequence of axioms. But better, I think, is to say that A and B are primitive sets (or collections), which are defined so as to not form a cross product. Nestedness is not permitted, as in making a pair into an element. That is < [ y, Y ], X > is impermissible. (Note the similarity to the axiom of foundation.)
But though we strive to guard against circularity, we face the skepticism of those who argue that circularity is inevitable in logic systems. Certainly we have that Goedel showed that most systems are elusive and elliptical, if not circular.
As long as we accept the basics of first order logic as usually taught today, perhaps to include an axiom of the relation of membership, we find that Russell's paradox evaporates without special exertions.
Consider first a proof that the null set is a subset of itself.
We use ordinary ("material") implication to obtain:
x ∊ ∅ --> x ∊ ∅
as the left-hand side is false, the implication is what we call vacuously true, since an assertion is true when a false proposition implies a false proposition.
Using the standard transform, we get x ∉ ∅ v x ∊ ∅
which is true.
So we have shown that ∅ ⊂ ∅
Note that we also have
x ∊ ∅ --> x ∉ ∅
Someone using "formal implication" would see this as a contradiction, but under "material implication," we use the transform and find no contradiction.
x ∉ ∅ v x ∉ ∅
which is true, but does us little good in proving the subset theorem.
Now consider the claim
∅ ∉ ∅ --> ∅ ∉ ∅
This transforms to a tautology
∅ ∊ ∅ v ∅ ∉ ∅
One alternate must be true, and we know that it is the right-hand proposition by definition of the null set.
Now we proceed to the non-null cases where
i) X ∊ X --> X ∊ X,
and
ii) X ∊ X --> X ∉ X
Case i) transforms to
X ∉ X v X ∊ X
while Case ii transforms to
X ∉ X v X ∉ X.
We may combine cases i) and ii) into a logical product, which then collapses to case i). One or the other of case i)'s alternates is true, but which?
The next step is to examine Russell's proof.
A set Y may or may not be a member of itself. Before Russell, there was no clear reason to debar a set belonging to itself.
So we write M for the set of sets that don't contain themselves as members and N for the set of sets that do so.
That is, Y ∊ M <--> Y ∉ N. The two sets are disjoint.
As M is a set, it must belong with either the sets that contain themselves or the sets that don't.
M ∊ M says that M contains itself. This in turn says that M is not in N, which means that M does not contain itself. That is, M ∊ M <--> M ∉ M.
Using the transform on the rightward arrow, we have M ∉ M v M ∉ M. For the leftward we get M ∊ M v M ∊ M. The biconditional reduces to
M ∊ M & M ∉ M,
a contradiction. An ordinary system (written here as a logical product)
ΠKi --> M ∊ M & M ∉ M
is inconsistent.
In other words, Russell proves that X ∊ X is only true in an inconsistent system. So, if we expect our system to be consistent, X ∊ X is false.
Thus, we have the proof that X ∉ X for any set X.
But we are of course taking for granted here other axioms found in standard logics and set theories.
Posted by Paul Conant. Ozark, Mo. Please let me know of any errors or oversights, major or minor.
From a modern vantage point, we need not resort to a theory of types or an axiomatic ban on X ∊ X. We are not in the same position as Russell, Whitehead, Zermelo and other pioneers in wrestling with how to best ground mathematics in some overall system.
The element or membership relation ∊ pairs an element with a set. We adopt the convention that the left-hand part of the pair is an element and the right-hand part is a set. So x ∊ X is ∊< x,X >. X is defined by the collection of x's with which it is paired, whether intensionally (by rule) or extensionally (by particulars). We must beware regarding the left part of the relation as a set A and the right part as a set B, making the relation a subset of the cross product set A X B. For then we face infinite regress.
This I suppose could be handled by an infinite sequence of axioms. But better, I think, is to say that A and B are primitive sets (or collections), which are defined so as to not form a cross product. Nestedness is not permitted, as in making a pair into an element. That is < [ y, Y ], X > is impermissible. (Note the similarity to the axiom of foundation.)
But though we strive to guard against circularity, we face the skepticism of those who argue that circularity is inevitable in logic systems. Certainly we have that Goedel showed that most systems are elusive and elliptical, if not circular.
As long as we accept the basics of first order logic as usually taught today, perhaps to include an axiom of the relation of membership, we find that Russell's paradox evaporates without special exertions.
Consider first a proof that the null set is a subset of itself.
We use ordinary ("material") implication to obtain:
x ∊ ∅ --> x ∊ ∅
as the left-hand side is false, the implication is what we call vacuously true, since an assertion is true when a false proposition implies a false proposition.
Using the standard transform, we get x ∉ ∅ v x ∊ ∅
which is true.
So we have shown that ∅ ⊂ ∅
Note that we also have
x ∊ ∅ --> x ∉ ∅
Someone using "formal implication" would see this as a contradiction, but under "material implication," we use the transform and find no contradiction.
x ∉ ∅ v x ∉ ∅
which is true, but does us little good in proving the subset theorem.
Now consider the claim
∅ ∉ ∅ --> ∅ ∉ ∅
This transforms to a tautology
∅ ∊ ∅ v ∅ ∉ ∅
One alternate must be true, and we know that it is the right-hand proposition by definition of the null set.
Now we proceed to the non-null cases where
i) X ∊ X --> X ∊ X,
and
ii) X ∊ X --> X ∉ X
Case i) transforms to
X ∉ X v X ∊ X
while Case ii transforms to
X ∉ X v X ∉ X.
We may combine cases i) and ii) into a logical product, which then collapses to case i). One or the other of case i)'s alternates is true, but which?
The next step is to examine Russell's proof.
A set Y may or may not be a member of itself. Before Russell, there was no clear reason to debar a set belonging to itself.
So we write M for the set of sets that don't contain themselves as members and N for the set of sets that do so.
That is, Y ∊ M <--> Y ∉ N. The two sets are disjoint.
As M is a set, it must belong with either the sets that contain themselves or the sets that don't.
M ∊ M says that M contains itself. This in turn says that M is not in N, which means that M does not contain itself. That is, M ∊ M <--> M ∉ M.
Using the transform on the rightward arrow, we have M ∉ M v M ∉ M. For the leftward we get M ∊ M v M ∊ M. The biconditional reduces to
M ∊ M & M ∉ M,
a contradiction. An ordinary system (written here as a logical product)
ΠKi --> M ∊ M & M ∉ M
is inconsistent.
In other words, Russell proves that X ∊ X is only true in an inconsistent system. So, if we expect our system to be consistent, X ∊ X is false.
Thus, we have the proof that X ∉ X for any set X.
But we are of course taking for granted here other axioms found in standard logics and set theories.
Posted by Paul Conant. Ozark, Mo. Please let me know of any errors or oversights, major or minor.
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