Since the collection is not identical to any of the coins, it is a different object than any of the coins. — litewave
It's not identical to any one of the coins but it is identical to both of the coins. So you're duplicating entities when you count both coins individually in addition to the collection as a whole. This post really makes this point clear. — Michael
Well, physical properties like weight reflect the subsuming nature of a collection: a collection doesn't add weight additional to the weights of its parts; it subsumes their weights. — litewave
And the same when it comes to counting the things that exist. The existence of the collection subsumes the existence of its parts. Either you count the collection and say that 1 thing exists, and weighs 3g, or you count its parts and say that 2 things exist, and collectively weigh 3g. You can't count both the collection and its parts and say that 3 things exist, else you then have to say that they collectively weigh 6g. — Michael
But I'll try to be more precise in future if this is a misuse of the term. — Michael
The point I'm making is that if we have a red ball and a green ball and a blue ball, then even though we can consider them in various configurations, e.g. (1) a red ball and a green ball, (2) a red ball and a blue ball, (3) a green ball and a blue ball, etc., it's not the case that there are multiple balls of each colour, and it's not the case that each configuration is a distinct entity in its own right, additional to the red ball, the green ball, and the blue ball. That realist interpretation of sets (what I think of as reification) is, I believe, mistaken. — Michael
I think you don't even need the set of everything to generate the problem. You just need any set that includes its own cardinality and it will blow up incoherently to a meaningless version of infinity. — Cuthbert
Either way:
1. There is a set of "all that exists"
2. There is a powerset for "all that exists" — ThinkOfOne
Is my wastebasket a set ? — magritte
This is still hardly a problem though, namely because of Leibniz's Law: there are predicates true of a set that are not true of its members. For instance, consider cardinality. The set {a, b, c} would be truly predicated of having the cardinality of 3, though none of its members have a cardinality of 3 — Kuro
I have a piece of metal that weighs 1g and a piece of metal that weighs 2g. So the collection of metal weighs 3g. This is the only metal that exists.
What is the total weight of all the metal that exists? 3g or 6g? Obviously 3g. You don't add the weight of the collection to the weight of its parts. So you can't say that the collection exists in addition to each of its parts. Unless you want to be a Platonist and say that the collection exists as some abstract, weightless object, which I think is absurd. — Michael
This is an inaccurate understanding of sets. Recall the axiom of extensionality. {a, b, c} and {c, b, a}, as well as {b, c, a} are all just the same set, because they have the exact same members and thus satisfy coextension. Sets, plainly as sets, are therefore invariant with respect to these configurations you use in your example, which are otherwise too fine-grained of a notion. There's a grain of truth here in that a realist interpretation of sets would indeed count {b} and b as separate, distinct objects and thus count two things, but this is unrelated to your configuration problem. — Kuro
In terms of function or use or conception, sure. But it terms of counting the number of things that exist, no. — Michael
Under the logic you are suggesting, there could be no valid numerical sets such as the set of prime numbers as you would suggest but they are all just multiples of 1. So, 1 is the only true member of the set of primes, or integers etc? Is that a consequence of the logic you are applying? — universeness
I have a piece of metal that weighs 1g and a piece of metal that weighs 2g. So the collection of metal weighs 3g. This is the only metal that exists.
What is the total weight of all the metal that exists? 3g or 6g? — Michael
You imply the weights are real for your 1g, 2g and 3g posit and then notional for your 6g step. — universeness
Either way:
1. There is a set of "all that exists"
2. There is a powerset for "all that exists"
— ThinkOfOne
The powerset will always strictly be cardinally larger than the set, and as you yourself understand, those subsets are not actually part of the original set (so there will exist members of the powerset not in the set, making the set not itself hence why the set doesn't exist) You've articulated precisely what I said in my post, so maybe this is a misreading? I'm not sure where exactly you're disagreeing with me or objecting — Kuro
Obviously it's wrong to say that 6g of metal exists, but this is what follows if you say that the collection exists as its own entity, distinct from the existence of the two individual pieces. — Michael
If the parts exist, their collection necessarily exists too.........Collections in a spacetime can have causal relations between them — litewave
I don't see much difference between a galaxy posited as an abstract entity and me as an actual entity — universeness
You cannot demonstrate all three physical quantities of weight at the same instant of time — universeness
What is so hard to understand about this? — Michael
“The whole is greater than the sum of the parts.” — universeness
For however many things have a plurality of parts and are not merely a complete aggregate but instead some kind of a whole beyond its parts...
In the case of the set {apple, pear} we just have an aggregate. — Michael
Combining an apple and a pear will have a quite distinct taste, compared to tasting an apple or tasting a pear. So, the combination produces a new entity of taste. — universeness
Your last argument was based on metal weights in the real world, I see no difference between separation based on physical weight and separation based on physical taste. Your claim that because I don't agree with you, it then follows that I just don't understand your logic is a matter for your own measure of your own arrogance.Right, so this shows that you clearly misunderstand what is being talked about. — Michael
Your claim that because I don't agree with you, it then follows that I just don't understand your logic is a matter for your own measure of your own arrogance. — universeness
What does this have to do with philosophy? It's pure Math. — Alkis Piskas
If you think that the set {apple, pear} means that we've combined an apple and pair into some new hybrid fruit then you don't understand what sets are. — Michael
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