So, lets say we have this maximal set T of true statements (truths, for short). {t

_{1}, . . . , t

_{i}, t

_{i + 1}, . . .}. In particular, let's consider {t

_{1},t

_{2}}. The first question we need to answer: Is this subset itself a truth? If this subset is not a truth, then the entire argument from the creation of the power set is meaningless, because the power set does not consist of truths, but of collections of truths, and there is no reason to presume the cardinality of all collections of truths would be the same as the cardinality of all truths. In fact, since the proof relies on looking at elements of P(T) as if there were in T, for Dr. Vallicella's argument to be cogent, we need to apply the standard that {t

_{1}, t

_{2}} is itself a truth. We are not given a definition for this truth, unfortunately, and how it relates to t

_{1}or t

_{2}, possibly via some sort of truth table. At least, since we are using sets, we know that {t

_{1}} = {t

_{1}, t

_{1}} = {t

_{1}, t

_{1}, t

_{1}}.

However, that leads us to another area of fuzzy definition. Is {t

_{1}} the same truth as t

_{1}? Is {t

_{1}} the same truth as {{t

_{1}}}? Will {t

_{1},t

_{2}} be the same truth as {t

_{1},t

_{2},{t

_{1},t

_{2}}}? Basically, can we remove all internal braces (except for the empty set)?

If we we allow the removal of all internal braces, then the proof falls apart, because all of the elements of P(T) will already be elements of T, after removing the internal braces and reducing the duplications. For example, let's look at a world of one atomic truth (that is, truths that not sets of other truths). T = {{}, t

_{1}}. Then P(T) = {{},{{}},{ t

_{1}},{{},t

_{1}}} = {{},{}, t

_{1},{},t

_{1}} = {{}, t

_{1}} = T.

So let us consider the construction where we can not remove internal braces. Now, since we have a valid method of creating a new truth from previously existing truths, by inclusion in sets, that means for any set Q of atomic truths, we find the power set of that set of truths, and the power set of that first power set of truths, and the power set of that second power set of truths, etc., already in T. How far can we go? Do we allow for there to be (loosely speaking) an infinite number of brace levels? If we do not allow that, then we know either |T| (the cardinality of T) = |{1, 2, 3, 4, ...}|, that is, T is countable, whenever Q is finite or countable, otherwise |T| = |Q|. However, this has the defect of removing much of P(T) from being eligible to be in T, because P(T) will include elements with an infinite number of braces.

In fact, if we place any limit M at all on the number of braces, we find that |Q| being less than or equal to M means |T| = M, otherwise |T| = |Q|, and either way P(T) will have elements that are not capable of being in T. So, the only way around this is place no restriction on the number of levels of inclusion. This has the side effect of making T a proper class even when |Q| = 1, so P(T) does not even exist.

So, it would seem regardless of set-up we are left with a choice of P(T) = T, P(T) having elements that do not qualify to be in T, or P(T) not existing. Regardless, the attempted proof fails.

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