In addition to all the other problems this has (existence is seriously not unproblematic), the central argument against infinities is simply wrong, even if we admit the possibility of “brute distinguishability”.
If S is a set of these “brute distinguishables”, and T=S u {x}, where x is yet another brute distinguishable (one that isn’t in S), then S and T are distinguishable because T contains x and S doesn’t. Whether brutely or not, the fact is that we declared that x is distinguishable, so this is valid.
Now evidently you disagree with this last part; but if we instead accept your argument, we get other problems. For instance, the number two is impossible. Why? Well, say {x,y,z,w} is a set of four distinct brute distinguishables; then {x,y} and {z,w} are different; and yet they are indistinguishable, because the only distinguishing property they have is their cardinality, and thus we have a contradiction! This is exactly the same argument you used, but with “infinity” replaced by “two”. The only difference is that when you considered infinite sets, you considered S and T with S a subset of T, while here neither is a subset of the other. But this difference has no relevance to the reasoning. The assertion that “the only distinguishing property they have is their cardinality” is simply false. I could have used singletons instead of two-element sets, I was just afraid you’d miss the point and reply “But we assumed that x and y are distinguishable, thus so are {x} and {y}!” (Now wait. Have I proved that 2 doesn’t exist or that 4 doesn’t exist? More on that in a moment.)
Well, OK, there is one difference between the infinite case and the finite case—a finite set of brute distinguishables could maybe be said to have a defining property, while an infinite one certainly couldn’t. (Though honestly even the finite really couldn’t—it could only have a definition relative to your set/class of brute distinguishables.) But this is irrelevant; you don’t need defining properties to distinguish things; any property will do. You essentially admit this yourself when you consider the cardinality as a potentially distinguishing property, even though any two infinite sets of brute distinguishables would be equally undefinable, regardless of their cardinalities. You could admit only as distinguishing properties those that don’t make reference to particular brute distinguishables—but then you’re abandoning the assumption that they are brute distinguishables, because now given two brute distinguishables x and y, they have exactly the same properties. And if it was just undefinability you were after, why did you introduce this assumption of brute distinguishables in the first place? Undefinable sets exist anyway, if you want to make that the basis of your argument.
Edit Jan 1 2013: Oops! There’s a mistake in the above paragraph—I said any two such infinite sets would be equally undefinable, but (provided we accept the definabillity of finite ones, which as I detailed I don’t think we should), this is incorrect, because they could be cofinite. Thus, the above paragraph should really talk about sets of brute distinguishables that are both infinite and co-infinite. This makes essentially no difference to the point.
Even if we accept your argument this far, despite all the mistakes I’ve pointed out, all it shows is that there can’t be an infinite set of brute distinguishables, that there can be at most finitely many. (Except really it shows that there can’t even be 4, by the argument above, or even 2, by the version with singletons. At which point you no longer really have brute distinguishability at all.) It’s not an argument against infinite sets, just that a particular class of things must be finite.
Even if we accept your conclusion that there can’t be infinite sets, this still doesn’t do what you want. You aren’t just trying to argue against infinite sets, you’re trying to argue against “infinite quantities”. What the hell are “infinite quantities”? You seem to be under the mistaken impression that “infinite quantities” is a sensible unified notion, when in fact there are many different systems of infinities we use depending on what the situation calls for and that don’t fit together. Here, why don’t I just point you to the discussion article I wrote on the matter some time ago.
And really, you should know better than to base an argument on the existence of something just because it’s “conceptually possible”, i.e. you personally find it possible to conceive of. That really has no bearing on whether it’s actually possible. That is as silly as making arguments based on P-zombies being conceptually possible.
In addition to all the other problems this has (existence is seriously not unproblematic), the central argument against infinities is simply wrong, even if we admit the possibility of “brute distinguishability”.
If S is a set of these “brute distinguishables”, and T=S u {x}, where x is yet another brute distinguishable (one that isn’t in S), then S and T are distinguishable because T contains x and S doesn’t. Whether brutely or not, the fact is that we declared that x is distinguishable, so this is valid.
Now evidently you disagree with this last part; but if we instead accept your argument, we get other problems. For instance, the number two is impossible. Why? Well, say {x,y,z,w} is a set of four distinct brute distinguishables; then {x,y} and {z,w} are different; and yet they are indistinguishable, because the only distinguishing property they have is their cardinality, and thus we have a contradiction! This is exactly the same argument you used, but with “infinity” replaced by “two”. The only difference is that when you considered infinite sets, you considered S and T with S a subset of T, while here neither is a subset of the other. But this difference has no relevance to the reasoning. The assertion that “the only distinguishing property they have is their cardinality” is simply false. I could have used singletons instead of two-element sets, I was just afraid you’d miss the point and reply “But we assumed that x and y are distinguishable, thus so are {x} and {y}!” (Now wait. Have I proved that 2 doesn’t exist or that 4 doesn’t exist? More on that in a moment.)
Well, OK, there is one difference between the infinite case and the finite case—a finite set of brute distinguishables could maybe be said to have a defining property, while an infinite one certainly couldn’t. (Though honestly even the finite really couldn’t—it could only have a definition relative to your set/class of brute distinguishables.) But this is irrelevant; you don’t need defining properties to distinguish things; any property will do. You essentially admit this yourself when you consider the cardinality as a potentially distinguishing property, even though any two infinite sets of brute distinguishables would be equally undefinable, regardless of their cardinalities. You could admit only as distinguishing properties those that don’t make reference to particular brute distinguishables—but then you’re abandoning the assumption that they are brute distinguishables, because now given two brute distinguishables x and y, they have exactly the same properties. And if it was just undefinability you were after, why did you introduce this assumption of brute distinguishables in the first place? Undefinable sets exist anyway, if you want to make that the basis of your argument.
Edit Jan 1 2013: Oops! There’s a mistake in the above paragraph—I said any two such infinite sets would be equally undefinable, but (provided we accept the definabillity of finite ones, which as I detailed I don’t think we should), this is incorrect, because they could be cofinite. Thus, the above paragraph should really talk about sets of brute distinguishables that are both infinite and co-infinite. This makes essentially no difference to the point.
Even if we accept your argument this far, despite all the mistakes I’ve pointed out, all it shows is that there can’t be an infinite set of brute distinguishables, that there can be at most finitely many. (Except really it shows that there can’t even be 4, by the argument above, or even 2, by the version with singletons. At which point you no longer really have brute distinguishability at all.) It’s not an argument against infinite sets, just that a particular class of things must be finite.
Even if we accept your conclusion that there can’t be infinite sets, this still doesn’t do what you want. You aren’t just trying to argue against infinite sets, you’re trying to argue against “infinite quantities”. What the hell are “infinite quantities”? You seem to be under the mistaken impression that “infinite quantities” is a sensible unified notion, when in fact there are many different systems of infinities we use depending on what the situation calls for and that don’t fit together. Here, why don’t I just point you to the discussion article I wrote on the matter some time ago.
And really, you should know better than to base an argument on the existence of something just because it’s “conceptually possible”, i.e. you personally find it possible to conceive of. That really has no bearing on whether it’s actually possible. That is as silly as making arguments based on P-zombies being conceptually possible.
I’m not even touching the rest.