Tuesday, June 7, 2022

The incoherence of Spinoza's mode ontology

According to Spinoza, I am a mode and God is the only substance. But I am not directly a mode of God. I am a mode of a mode of a mode of … a mode of God, with infinitely many “a mode of” links in between.

This is incoherent. It is an infinite chain with two ends, one being me and the other being God. But any infinite chain made of direct links has at most one end: it would have to be of the form 1:2:3:4:…, with one endpoint, namely zero. We can stick on another chain running in the other direction, like …:iv:iii:ii:i, and get the two ended sequence 1:2:3:4:…:iv:iii:ii:i. But this two-ended sequence is not a chain, because there is no connection between any of the arabic numbered nodes and any of the roman numbered nodes.

20 comments:

Daryl said...

I'm not super familiar with Spinoza, but why would he think that we're not direct modes of God?

Alexander R Pruss said...

Spinoza thinks the finite follows directly only from the finite. Why? I think he thinks the infinite just can't directly determine the finite... This is all deeply mysterious to me. Here's a piece on it: https://www.jstor.org/stable/20116348

Alexander R Pruss said...

In this case, my "intuition" comes from the following:

Theorem: There is no connected infinite graph with the properties (i) every node lies on at most two edges and (ii) there are two nodes ("endpoints") that each lie on only one edge.

Alexander R Pruss said...

Nope, that's not a theorem.

For the record, here's a proof of my theorem.

Suppose there is a graph G with (i) and (ii). Let A and B be distinct endpoints. Let a_0 = A. Now inductively define a sequence (a_n) of nodes for n>0 each of which lies on exactly two edges, one of these connecting a_n to a_(n-1). Since A is an endpoint, it is connected to exactly one node. Call that node a_1. Then a_1 lies on at most two edges, on at least one. If a_1 lies on exactly one edge, then the nodes A and a_1 are connected to each other and to nothing else, and hence {a_0, a_1} is a finite subset of the nodes of G that are not connected to anything outside the subset, contradicting the claim that G is a connected infinite graph. So a_1 lies on exactly two edges. Suppose now that a_1,...,a_n have been defined with the property that each lies on exactly two edges, one of them connected a_k to a_(k-1) for each k<=n. Since a_n lies on two edges, and is connected to a_(n-1), let a_(n+1) be the node that a_n is connected to via the other edge it's on. The remaining thing to prove is that a_(n+1) is on two edges. It is on at least one, the one that connects it to a_n. Suppose that's the only edge is on. Then a_0 is on exactly one edge, connecting it to a_1, a_(n+1) is on exactly one edge, connected it to a_n, and a_k for k strictly between 0 and n+1 is on exactly two edges, connecting it to a_(k-1) and a_(k+1). Thus {a_0,...,a_(n+1)} is a finite subset of the nodes of G that are not connected to anything outside the subset, contradicting the claim that G is a connected infinite graph.

So, we have now defined the sequence a_0,a_1,.... Of these, a_0 is connected to a_1, and to nothing else, every note a_k for k other than zero is connected to a_(k-1) and a_(k+1) and to nothing else (since each node is on at most two edges). Thus, {a_0,a_1,...} is a non-empty set of nodes of G no member of which is connected to anything outside the set. Since G is a connected graph, that set of nodes must contain ALL the nodes of G. Hence it must contain two endpoints, since G does. But it only contains one endpoint, namely a_0, since a_k for k non-zero lies on TWO edges. Contradiction!

Zsolt Nagy said...

Contradiction?!?
Contradiction to what exactly?!?
What is the contradiction there in your proof?
I guess, that's supposed to be your proofless theorem as an oxymoron, since the given and provided proof doesn't proof the made claim and theorem. Also that theorem doesn't have a proof, the negation have already been proven to be true.
So all that is there to it, I guess.

Alexander R Pruss said...

The derived claim that there is only one endpoint contradicts the assumption at the beginning of the proof that there are two endpoints.

(It's an annoyingly tricky proof.)

Zsolt Nagy said...

Hm, but the proof starts with a finite graph with two endpoints and it is supposedly concluding with some kind of an infinite graph with only one endpoint and not with an arbitrary one. Do I understand your given proof correctly?
I guess so and if so, then why not start with an infinite graph with two endpoints from the get go?!?
After all you want to show, that any arbitrary infinite graph has no two endpoints and not only some?!?

Alexander R Pruss said...

I meant: "Suppose there is an INFINITE CONNECTED graph G" at the beginning. The theorem only applies to infinite connected graphs, so there would be no point thinking about a finite or disconnected graph.

Zsolt Nagy said...

What?!? Right at the beginning an INFINITE CONNECTED graph G have been supposed?!?
What?!?
And then how does follow from that INFINITE CONNECTED graph G, that "hence {a_0, a_1} is a finite subset of the nodes of INFINITE CONNECTED G that are not connected to anything outside the subset"?
What?!?
I'm so confused by your "proof" and I don't think, that's because of the language barrier between us - English being not my native language.
Well if so, and That supposed to be the problem here, then why not give a formal proof of your theorem?
That might also help to understand this "mess" of a proof.

Zsolt Nagy said...
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Zsolt Nagy said...

I still don't understand, how does follow from that INFINITE CONNECTED graph G, that "hence {a_0, a_1} is a finite subset of the nodes of INFINITE CONNECTED G that are not connected to anything outside the subset"?
How is {a_0, a_1} a finite subset of the nodes of INFINITE CONNECTED G not connected to anything outside the subset, when G is an INFINITE CONNECTED graph?!?
Certainly the finite subset {a_0, a_1} is connected t other nodes of the INFINITE CONNECTED G. Or not?!?
I'm still very much confused here.

Alexander R Pruss said...

Pretty much nobody in mathematics, outside of work on foundations or computer verifications, gives formal proofs, because doing so is way too time-consuming. Mathematics papers almost always have proofs written in prose.

As for your question why {a_0,a_1} is not connected to anything else, remember that the proof is a reductio ad absurdum. I started by assuming that A was an endpoint. I labeled it a_0. Because a_0 is an endpoint, it is connected to exactly one other node, which I called a_1. Now, we prove that a_1 does NOT lie only on one edge. For if a_1 were to lie only on one edge, then it would be connected to only one thing. It IS connected to a_0, we already assumed. The assumptions we have so far would then ensure that (a) a_0 is connected to a_1 and only to a_1 and (b) a_1 is connected to a_0 and only to a_0. It follows from this that {a_0,a_1} is not connected to anything on its outside. But that contradicts the assumption that G is an infinite connected graph. So, indeed, it is false that a_1 lies on only one edge. Hence, it must lie on two.

Zsolt Nagy said...

So the answer to my question is, that the sub set finite sub set {a_0,a_1} of nodes of graph G is actually connected to something outside of the sub set, namely to other nodes of the infinitely connected graph G.
If so, then my question kind of remains still unanswered: Why write in your proof that the sub set {a_0,a_1} wouldn't be connected to anything else outside of the sub set, when that's not actually the case here?
I know, that your proof supposed to be a reduction ad absurdum. But the only absurd thing, I'm seeing here, is your "mess", that you are calling a "proof".

Zsolt Nagy said...

Particularly I don't understand this:
"If a_1 lies on exactly one edge, then the nodes A and a_1 are connected to each other and to nothing else, and hence {a_0, a_1} is a finite subset of the nodes of G that are not connected to anything outside the subset,..."
Why assuming a_1 lieing only on exactly one edge?!?
If then that assumption leads to a contradiction in conjunction with other assumptions like the considered graph G being supposedly infinite, then wait follows from that contradiction exactly?!?
It might be that the considered graph is actually only finite and not infinite!!!
This is all so confusing.

Zsolt Nagy said...

If a_1 lies on exactly one edge, then the nodes A and a_1 are not connected to each other and to nothing else, and hence {a_0, a_1} is a finite subset of the nodes of G that are connected to something outside the subset.

Zsolt Nagy said...

Like wise for any finite subset {a_0,...,a_(n+1)}:
{a_0,...,a_(n+1)} is a finite sub set of nodes of G, which are connected to something outside of the sub set, namely to other nodes of the infinite graph G.

Yeah, it's like a big giant "red herring" not contributing anything substantial to the "proof" of your "theorem" Why have it that even in there in that "proof" of yours then?!?

Zsolt Nagy said...

Also let me define now the following sequence:
a_0=A (one endpoint of the infinitely connected graph G)
a_1=B (the other endpoint of the infinitely connected graph G)
and the nodes are connected in the following way:
a_0,edge,a_2,edge,a_4,edge,...,edge,a_5,edge,a_3,edge,a_1
Then this is an infinitely connected graph with two endpoints. Therefore, there is such an infinitely connected graph with two endpoints.

I can shred your "mess" of a "proof" for your "theorem" to pieces.
You can not do the same with mine.

Zsolt Nagy said...

So then by reductio ad absurdum (or obviously and trivially saying) any {a_0,...,a_(n+1)} is a finite subset of nodes of G, which are connected to something outside of the sub set, namely to other nodes of the infinite graph G.
Hence, {a_0,a_1,...} is a non-empty SUBset of nodes of G with some members of which being connected to something outside the set. Since G is a connected graph, that set of nodes must contain SOME of the nodes of G, but sadly not all even if that SUBset is already containing an infinite amount of nodes (just like the set of infinitely many odd numbers is just a subset of all natural numbers).
I guess, this is the result of your "proof"/"reductio ad absurdum" here.
Or how am I supposed to understand this "proof"/"reductio ad absurdum" of yours correctly here?!?

Zsolt Nagy said...

So your intuitions, Alexander, on infinity are relying on a false "theorem" here. Then why have those false intuitions then?!?
Because based upon those you can "justify" your scepticism and objections towards other positions and claims?
But such scepticism and objections based upon false intuitions and false "theorems" are worth nothing. In my opinion such faulty or fallacious justifications might even be harmful by giving the false impression of there to be a proper warrant for a believe, when there is actually no such proper warrant for it or just only fallacious ones.
I rather don't believe in a true proposition, than believe in a false proposition based upon fallacious or even false reasoning and deduction.
And your so called "theorem" is not just lacking any proper proof or deduction here, but it is also false because of its negation being simply provable and deductable by giving a proper example, which has been provided before multiple times here already.

Maxim said...
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