This ensures we can apply Zorn’s Lemma when considering chains in E, but is not as strong as the full Axiom of Choice. If the set E is finite or countable, for instance, then A2 applies automatically.
Definitions: We define a relation C such that x C y iff there are entities v, w such that v P x, y P w and v C w.
Note: This gives a broader causal relation which automatically satisfies “if x C y and x P z then z C y” as well as “if x C z and y P z then x C y”, loosely “anything which is caused by a part is caused by the whole” and “anything which causes the whole, causes the part”. So we don’t need to state those as extra premises.
We then define a further relation ⇐ such that x ⇐ y iff x = y, or there are finitely many entities x1, …, xn such that x1 = x, xn = y and xi C* xi+1 for i=1.. n-1.
Note: This construction ensures that ⇐ is a pre-order on E.
Say that a subset S of E is a “chain” iff for any x, y in S we have x ⇐ y or y ⇐ x. Say that S is an “endless chain” iff for any x in S there is some y in S distinct from x with y ⇐ x. We shall say that y is “uncaused” if and only if there is no z in E distinct from y with z C* y (this of course implies there is no z distinct from y with z C y, but it also implies that y isn’t part of anything which is caused by something distinct from y). Say that x is a proper part of y iff x is distinct from y but x P y.
A3. Let S be any endless chain in E; then there is some z in E such that every x in S is a proper part of z.
Lemma 1: For any chain S in E, there is an entity x in E such that x ⇐ y for every y in S.
Proof: Suppose S is not endless. Then there is some x in S such that for no other y in S is y ⇐ x. By the chain property we must have x ⇐ y for every member y of S. Alternatively, suppose that S is endless, then by A3, there is some z in E of which every x in S is a part. Now consider any y in S. There is some x not equal to y in S with x ⇐ y, so there are entities x = x1… xn = y with each xi C xi+1 for i=1..n-1. Further, as x P z we have z C x2 and hence z ⇐ y.
Lemma 2: For any x in E, there is some y in E such that y ⇐ x, and for every z ⇐ y, we must have y ⇐ z.
Theorem 3: For any x in E, there is some uncaused y in E such that y ⇐ x.
Proof: Take a y as given by Lemma 2 and consider the set S = {s: s ⇐ y}. By Lemma 2, y ⇐ s for every member of S, and if S has more than one element, then S is an endless chain. So by A3 there is some z of which every s in S is a proper part, which implies that z is not in S. But by the proof of Lemma 1, z ⇐ y, which implies z is in S: a contradiction. So it follows that S = {y}, which completes the proof.
We now partition E into three subsets. I are the “inert” entities, which do not cause anything and have no causes themselves. (Note that the new version allows there to be some of these, unlike the previous version; you can think of them as abstract entities like numbers, sets, propositions and so on, if you want to). Formally I = {x in E: there is no y distinct from x with x C y or y C x}. U are the “uncaused causes”—formally U = {x in E: there is no y distinct from x with y C x, but there is z distinct from x with x C z}. O are all the “other”, caused entities, so that formally O = {x in E: there is some y distinct from x with y C* x}.
B1. If S is any subset of U such that for any x, y in S we have x P y or y P x, (call such an S a “chain of parts”), then there is some entity z of which all members of S are parts.
B2. Suppose that y ⇐ x and z ⇐ x. Then there is some entity w such that: w ⇐ x; w ⇐ y or y P w; w ⇐ z or z P w.
(EDIT: Restated to ensure that Theorem 4 properly follows.) Informally, the idea is that y and z can’t independently cause x without any further causal explanation. So there must be some common cause, however each of them may be part of that common cause.
Definition: Say that entities x and y are causally-connected if and only if x=y or there are finitely-many entities x=x1,..,xn=y with each xi C xi+1 or xi+1 C xi for i=1..n-1.
B3. Any two entities x, y in O are causally-connected.
Informally, O doesn’t “come apart” into disconnected components, such as a bunch of isolated universes. Premises B1-B3 turn out to be necessary for Theorem 6 to hold, as well as sufficient (see below). So they can’t be made any weaker!
Theorem 4: For any x in O, there is a unique entity f(x) in U such that: f(x) ⇐ x, and any other y in U with y ⇐ x satisfies y P f(x).
Proof: For any x in O, define a subset U(x) = {y in U: y ⇐ x}; this is non-empty by Theorem 3. Consider any chain of parts S that is a subset of U(x). If it has at least two members, then by B1 there is some z in E of which all members of S are parts, and such a z must be in U. (If not, then note any w C z would also satisfy w C s for each member s of S, which would require them all to be equal to w). Also since y ⇐ x for any member of S and y P z we have z ⇐ x. So z is also a member of U(x). Or if S is a singleton—say {z} - then clearly all members of S are parts of z, and z is also in U(x). By application of Zorn’s Lemma to U(x), there is a P-maximal element f(x) in U(x) such that there is no other y in U(x) with f(x) P y. By B2, for any other y in U(x) there must be some z in U(x) with f(x) P z and y P z; given f(x) is maximal we have z = f(x) and so y P f(x). This makes f(x) the unique maximal element of U(x).
Theorem 5: For any x, y in O, f(x) = f(y) if and only if x and y are causally-connected.
Proof: It is clear that if f(x) = f(y) then x is causally-connected to y (just build a path backwards from x to f(x) and then forward again to y). Conversely, consider any two x, y in O:
a) If x C y, then f(x) is in U and satisfies f(x) ⇐ y so we have f(x) P f(y). Since x is not f(x), we have f(x) C x2 ⇐ x for some x2, and hence f(y) C* x2 ⇐ x i.e. f(y) ⇐ x which means f(y) P f(x), and so f(x) = f(y).
b) If z is in U with z C x and z C y, then z P f(x) so f(x) ⇐ y and f(x) P f(y); similarly, f(y) P f(x) so that f(x) = f(y).
The result now follows by induction on the length of the causal path connecting x to y.
Theorem 6: If O is non-empty, then there is a single entity g in U such that: f(x) = g for every x in O, and y P g for every y in U.
Proof: Assuming O is non-empty, take any element y in O, and set g = f(y); then the result that f(x) = g for any x in O follows from Theorem 5 and B3; further, for any y in U, there is some x in O with y C* x, so by Theorem 4, y P f(x). If there are no elements of O (meaning there are none in U either) then the Theorem is trivial.
Finally, note that B1, B2 and B3 are entailed by the statement of Theorem 6. For B1, we can just take g as the relevant z. For B2, we can take g as the relevant w. B3 follows using using the first part of Theorem 5 (just track from x back to g, then forward to y again).
I’m just about done now, so unless there are errors in the above proof will leave it. What are the residual weak points? Well, B2 and B3 have been weakened a bit, but are still basically unjustifiable (we can imagine them being false without absurdity) and the above re-work shows they are needed for the uniqueness conclusion (Theorem 6). Also, we have the weakness of not deriving anything else useful about g.
Note: This gives a broader causal relation which automatically satisfies “if x C y and x P z then z C y” as well as “if x C z and y P z then x C y”, loosely “anything which is caused by a part is caused by the whole” and “anything which causes the whole, causes the part”. So we don’t need to state those as extra premises.
This will lead to a problem.
Consider assumption A3:
A3. Let S be any endless chain in E; then there is some z in E such that every x in S is a proper part of z.
Consider any endless chain consisting of at minimum two elements. Consider two elements in that chain, x and y, such that x C y. x and y are both proper parts of z. Therefore, x C z, and z C y. But then we have a longer chain; using x C z C y in place of x C y. Each element of that longer chain must then, by A3, be a proper part of a larger entity, z2. But then, similarly, we can construct the chain using x C z C z2 C* y. There are therefore an infinite number of entities z, z2, z3, z4… and so on, each including the one before it as a proper part (and nothing that is not part of the one before it).
Furthermore, anything which is a proper part of anything else is a part of such an infinitely recursive loop by default.
Consider any endless chain consisting of at minimum two elements. Consider two elements in that chain, x and y, such that x C y. x and y are both proper parts of z. Therefore, x C z, and z C* y.
It follows that z C y but it does not follow that x C z or that y C z.. The “whole” z may be a cause of its parts, without in turn being caused by its parts. Note that by construction of C it is true that if x is a cause of y and x is a part of z, then z C y. However, it is not generally true that if x is a cause of y and z is a part of x then z C y.
As an example of the intuition behind this: suppose I have a thermostat box containing two circuit boards. Board 1 is connected into my home heating system; Board 2 is a spare not connected into anything. It is true that Board 1 causes my heating to come on. It is true that the thermostat (of which Board 1 is part) causes my heating to come on. It is false that Board 2 (which is part of the thermostat) causes my heating to come on.
But then we have a longer chain; using x C z C y in place of x C* y.
You are right that when adding z, we now get a longer chain {x, y, z}, but this won’t in general be an “endless chain” (the new z may well be an end).
Consider any endless chain consisting of at minimum two elements. Consider two elements in that chain, x and y, such that x C y. x and y are both proper parts of z. Therefore, x C z, and z C* y.
It follows that z C y but it does not follow that x C z or that y C* z.
No, you need x C y and z P y to get x C z (be careful about which way the P relation is going).
The intuition is “Anything which is a cause of the whole is a cause of the part”, not “Anything which is a cause of the part is a cause of the whole”. Again, there are intuitive examples here. (Compare me baking a cake for a child’s birthday party vs just buying the cake from a shop, and putting a few sprinkles and candles on the top. In the second case, I am a cause of some part of the cake as presented to the child, but not the whole cake, and if someone says “Wow that cake tasted delicious!” I’d have to admit I didn’t make it, only decorated it).
I make a cake. I am a cause of the cake. The cake contains eggs. I am not the cause of the eggs. I think “what causes the whole, causes each part” is a bad intuition to have.
In general, I think it is an error, and a source of confusion, to think of things rather than events having, or being, causes. I know people sometimes do, and I’ve gone along with it in #1 above, but I think it’s a mistake.
Why would anyone assume A3? It seems really arbitrary. Exception: you might believe A3 because you believe in an entity of which all others are parts. See below.
If E includes an entity V of which all others are parts (call it “the universe”) then, provided C is reflexive, V C* anything-you-like. And I think it’ll then turn out that the way all your theorems work is that V is the canonical uncaused cause of everything. Which is a bit dull and wouldn’t satisfy many theists. Perhaps something more interesting happens if you make C irreflexive instead, so that things don’t count as causes of themselves.
Fair points, though there is in fact a lot of disagreement about what are the basic relata of the causal relation: see the SEP entry for example. When we apply causation to entities (which we can sometimes do, as in your example) then “A causes B” probably means something like “at least one event in which A is involved is a cause of every event in which B is involved”.
On counterexamples to “what causes the whole, causes the part” : possibly an even stronger counterexample considers just one of the atoms in the cake. However, we must be careful here: it is only some temporal part of the egg (or of the atom) which is part of the cake; the eggs/atoms in their full temporal entirety are NOT parts of the cake in its full temporal entirety. We could perhaps treat the relevant temporal part (“egg mixed into cake” or “atom within cake”) as an “entity” in its own right, but then it does seem that by making the cake, I am a cause of all the events which involve that particular “entity” (since I put the egg/atom into the cake in the first place).
In any case, note that the most recent version of the argument doesn’t actually need to assume this “cause-whole ⇒ cause-part” applies to C, since it only ever uses the constructed relation C instead. The conclusion is still interesting, since if nothing Cs the entity g, then nothing Cs it either, and if g causes some whole of which each entity is a part, that is still an interesting property of g. The argument makes no assumptions on whether C is reflexive or not.
On A3, I’m not totally sure of the circumstances in which we can aggregate entities together and treat them as parts of a single entity, but if the entities are causally related (and particularly if they are causally-related in an odd way, like an endless chain), then it does make some sort of sense to do this aggregation. After all, we immediately want to ask the question “How could there be an endless chain?” a question which does treat the “chain” as some sort of an entity to be explained. If entities are not causally related (they are in different universes), lumping them together seems much less natural.
Finally, on the “maximal entity” approach, CCC I believe discussed this in the original thread before I lifted here, and he seems to find it theologically interesting.
As discussed, I have a new version which preserves the proof structure, but weakens the premises about as much as possible.
A1. The collection of all entities is a set E, with a causal relation C and a partial order P, such that x P y if and only if x is a part of y.
Note: This merges the assumption that P is a partial order into the overall set-up; that feature of P now gets used earlier in the argument.
A2. The set E can be well-ordered.
This ensures we can apply Zorn’s Lemma when considering chains in E, but is not as strong as the full Axiom of Choice. If the set E is finite or countable, for instance, then A2 applies automatically.
Definitions: We define a relation C such that x C y iff there are entities v, w such that v P x, y P w and v C w.
Note: This gives a broader causal relation which automatically satisfies “if x C y and x P z then z C y” as well as “if x C z and y P z then x C y”, loosely “anything which is caused by a part is caused by the whole” and “anything which causes the whole, causes the part”. So we don’t need to state those as extra premises.
We then define a further relation ⇐ such that x ⇐ y iff x = y, or there are finitely many entities x1, …, xn such that x1 = x, xn = y and xi C* xi+1 for i=1.. n-1.
Note: This construction ensures that ⇐ is a pre-order on E.
Say that a subset S of E is a “chain” iff for any x, y in S we have x ⇐ y or y ⇐ x. Say that S is an “endless chain” iff for any x in S there is some y in S distinct from x with y ⇐ x. We shall say that y is “uncaused” if and only if there is no z in E distinct from y with z C* y (this of course implies there is no z distinct from y with z C y, but it also implies that y isn’t part of anything which is caused by something distinct from y). Say that x is a proper part of y iff x is distinct from y but x P y.
A3. Let S be any endless chain in E; then there is some z in E such that every x in S is a proper part of z.
Lemma 1: For any chain S in E, there is an entity x in E such that x ⇐ y for every y in S.
Proof: Suppose S is not endless. Then there is some x in S such that for no other y in S is y ⇐ x. By the chain property we must have x ⇐ y for every member y of S. Alternatively, suppose that S is endless, then by A3, there is some z in E of which every x in S is a part. Now consider any y in S. There is some x not equal to y in S with x ⇐ y, so there are entities x = x1… xn = y with each xi C xi+1 for i=1..n-1. Further, as x P z we have z C x2 and hence z ⇐ y.
Lemma 2: For any x in E, there is some y in E such that y ⇐ x, and for every z ⇐ y, we must have y ⇐ z.
Proof: This follows from Zorn’s Lemma applied to pre-orders.
Theorem 3: For any x in E, there is some uncaused y in E such that y ⇐ x.
Proof: Take a y as given by Lemma 2 and consider the set S = {s: s ⇐ y}. By Lemma 2, y ⇐ s for every member of S, and if S has more than one element, then S is an endless chain. So by A3 there is some z of which every s in S is a proper part, which implies that z is not in S. But by the proof of Lemma 1, z ⇐ y, which implies z is in S: a contradiction. So it follows that S = {y}, which completes the proof.
We now partition E into three subsets. I are the “inert” entities, which do not cause anything and have no causes themselves. (Note that the new version allows there to be some of these, unlike the previous version; you can think of them as abstract entities like numbers, sets, propositions and so on, if you want to). Formally I = {x in E: there is no y distinct from x with x C y or y C x}. U are the “uncaused causes”—formally U = {x in E: there is no y distinct from x with y C x, but there is z distinct from x with x C z}. O are all the “other”, caused entities, so that formally O = {x in E: there is some y distinct from x with y C* x}.
B1. If S is any subset of U such that for any x, y in S we have x P y or y P x, (call such an S a “chain of parts”), then there is some entity z of which all members of S are parts.
B2. Suppose that y ⇐ x and z ⇐ x. Then there is some entity w such that: w ⇐ x; w ⇐ y or y P w; w ⇐ z or z P w.
(EDIT: Restated to ensure that Theorem 4 properly follows.) Informally, the idea is that y and z can’t independently cause x without any further causal explanation. So there must be some common cause, however each of them may be part of that common cause.
Definition: Say that entities x and y are causally-connected if and only if x=y or there are finitely-many entities x=x1,..,xn=y with each xi C xi+1 or xi+1 C xi for i=1..n-1.
B3. Any two entities x, y in O are causally-connected.
Informally, O doesn’t “come apart” into disconnected components, such as a bunch of isolated universes. Premises B1-B3 turn out to be necessary for Theorem 6 to hold, as well as sufficient (see below). So they can’t be made any weaker!
Theorem 4: For any x in O, there is a unique entity f(x) in U such that: f(x) ⇐ x, and any other y in U with y ⇐ x satisfies y P f(x).
Proof: For any x in O, define a subset U(x) = {y in U: y ⇐ x}; this is non-empty by Theorem 3. Consider any chain of parts S that is a subset of U(x). If it has at least two members, then by B1 there is some z in E of which all members of S are parts, and such a z must be in U. (If not, then note any w C z would also satisfy w C s for each member s of S, which would require them all to be equal to w). Also since y ⇐ x for any member of S and y P z we have z ⇐ x. So z is also a member of U(x). Or if S is a singleton—say {z} - then clearly all members of S are parts of z, and z is also in U(x). By application of Zorn’s Lemma to U(x), there is a P-maximal element f(x) in U(x) such that there is no other y in U(x) with f(x) P y. By B2, for any other y in U(x) there must be some z in U(x) with f(x) P z and y P z; given f(x) is maximal we have z = f(x) and so y P f(x). This makes f(x) the unique maximal element of U(x).
Theorem 5: For any x, y in O, f(x) = f(y) if and only if x and y are causally-connected.
Proof: It is clear that if f(x) = f(y) then x is causally-connected to y (just build a path backwards from x to f(x) and then forward again to y). Conversely, consider any two x, y in O:
a) If x C y, then f(x) is in U and satisfies f(x) ⇐ y so we have f(x) P f(y). Since x is not f(x), we have f(x) C x2 ⇐ x for some x2, and hence f(y) C* x2 ⇐ x i.e. f(y) ⇐ x which means f(y) P f(x), and so f(x) = f(y).
b) If z is in U with z C x and z C y, then z P f(x) so f(x) ⇐ y and f(x) P f(y); similarly, f(y) P f(x) so that f(x) = f(y).
The result now follows by induction on the length of the causal path connecting x to y.
Theorem 6: If O is non-empty, then there is a single entity g in U such that: f(x) = g for every x in O, and y P g for every y in U.
Proof: Assuming O is non-empty, take any element y in O, and set g = f(y); then the result that f(x) = g for any x in O follows from Theorem 5 and B3; further, for any y in U, there is some x in O with y C* x, so by Theorem 4, y P f(x). If there are no elements of O (meaning there are none in U either) then the Theorem is trivial.
Finally, note that B1, B2 and B3 are entailed by the statement of Theorem 6. For B1, we can just take g as the relevant z. For B2, we can take g as the relevant w. B3 follows using using the first part of Theorem 5 (just track from x back to g, then forward to y again).
I’m just about done now, so unless there are errors in the above proof will leave it. What are the residual weak points? Well, B2 and B3 have been weakened a bit, but are still basically unjustifiable (we can imagine them being false without absurdity) and the above re-work shows they are needed for the uniqueness conclusion (Theorem 6). Also, we have the weakness of not deriving anything else useful about g.
This will lead to a problem.
Consider assumption A3:
Consider any endless chain consisting of at minimum two elements. Consider two elements in that chain, x and y, such that x C y. x and y are both proper parts of z. Therefore, x C z, and z C y. But then we have a longer chain; using x C z C y in place of x C y. Each element of that longer chain must then, by A3, be a proper part of a larger entity, z2. But then, similarly, we can construct the chain using x C z C z2 C* y. There are therefore an infinite number of entities z, z2, z3, z4… and so on, each including the one before it as a proper part (and nothing that is not part of the one before it).
Furthermore, anything which is a proper part of anything else is a part of such an infinitely recursive loop by default.
This leads to trouble in the proof of theorem 3.
It follows that z C y but it does not follow that x C z or that y C z.. The “whole” z may be a cause of its parts, without in turn being caused by its parts. Note that by construction of C it is true that if x is a cause of y and x is a part of z, then z C y. However, it is not generally true that if x is a cause of y and z is a part of x then z C y.
As an example of the intuition behind this: suppose I have a thermostat box containing two circuit boards. Board 1 is connected into my home heating system; Board 2 is a spare not connected into anything. It is true that Board 1 causes my heating to come on. It is true that the thermostat (of which Board 1 is part) causes my heating to come on. It is false that Board 2 (which is part of the thermostat) causes my heating to come on.
You are right that when adding z, we now get a longer chain {x, y, z}, but this won’t in general be an “endless chain” (the new z may well be an end).
It does, because y P z.
x C y, and y P z. Therefore, x C z.
No, you need x C y and z P y to get x C z (be careful about which way the P relation is going).
The intuition is “Anything which is a cause of the whole is a cause of the part”, not “Anything which is a cause of the part is a cause of the whole”. Again, there are intuitive examples here. (Compare me baking a cake for a child’s birthday party vs just buying the cake from a shop, and putting a few sprinkles and candles on the top. In the second case, I am a cause of some part of the cake as presented to the child, but not the whole cake, and if someone says “Wow that cake tasted delicious!” I’d have to admit I didn’t make it, only decorated it).
I make a cake. I am a cause of the cake. The cake contains eggs. I am not the cause of the eggs. I think “what causes the whole, causes each part” is a bad intuition to have.
In general, I think it is an error, and a source of confusion, to think of things rather than events having, or being, causes. I know people sometimes do, and I’ve gone along with it in #1 above, but I think it’s a mistake.
Why would anyone assume A3? It seems really arbitrary. Exception: you might believe A3 because you believe in an entity of which all others are parts. See below.
If E includes an entity V of which all others are parts (call it “the universe”) then, provided C is reflexive, V C* anything-you-like. And I think it’ll then turn out that the way all your theorems work is that V is the canonical uncaused cause of everything. Which is a bit dull and wouldn’t satisfy many theists. Perhaps something more interesting happens if you make C irreflexive instead, so that things don’t count as causes of themselves.
Fair points, though there is in fact a lot of disagreement about what are the basic relata of the causal relation: see the SEP entry for example. When we apply causation to entities (which we can sometimes do, as in your example) then “A causes B” probably means something like “at least one event in which A is involved is a cause of every event in which B is involved”.
On counterexamples to “what causes the whole, causes the part” : possibly an even stronger counterexample considers just one of the atoms in the cake. However, we must be careful here: it is only some temporal part of the egg (or of the atom) which is part of the cake; the eggs/atoms in their full temporal entirety are NOT parts of the cake in its full temporal entirety. We could perhaps treat the relevant temporal part (“egg mixed into cake” or “atom within cake”) as an “entity” in its own right, but then it does seem that by making the cake, I am a cause of all the events which involve that particular “entity” (since I put the egg/atom into the cake in the first place).
In any case, note that the most recent version of the argument doesn’t actually need to assume this “cause-whole ⇒ cause-part” applies to C, since it only ever uses the constructed relation C instead. The conclusion is still interesting, since if nothing Cs the entity g, then nothing Cs it either, and if g causes some whole of which each entity is a part, that is still an interesting property of g. The argument makes no assumptions on whether C is reflexive or not.
On A3, I’m not totally sure of the circumstances in which we can aggregate entities together and treat them as parts of a single entity, but if the entities are causally related (and particularly if they are causally-related in an odd way, like an endless chain), then it does make some sort of sense to do this aggregation. After all, we immediately want to ask the question “How could there be an endless chain?” a question which does treat the “chain” as some sort of an entity to be explained. If entities are not causally related (they are in different universes), lumping them together seems much less natural.
Finally, on the “maximal entity” approach, CCC I believe discussed this in the original thread before I lifted here, and he seems to find it theologically interesting.
I agree that “x C y, and y P z. Therefore, x C z” is wildly unintuitive, causes problems, and is just plainly wrong. But...
...
...actually, looking back, you’re right. I apologise; I misread the definition of C* (I read w P y instead of y P w).
I’m going to have to look through it again before I can comment further.