-4
$\begingroup$

I apologise if this has been asked before or is otherwise an ill-formed question.

Consider the following predicates:

  • $B(x)$: "$x$ began to exist".

  • $C(x)$: "$x$ has a cause".

Let $U$ be the observable universe.

Consider the following . . .

$$\begin{align}(\forall u\in U)&(B(u)\implies C(u))\tag{1}\\ &B(U).\tag{2} \end{align}$$

That is to say that, for all $u$ in the observable universe $U$, if $u$ began to exist, then $u$ has a cause; the observable universe $U$ began to exist.

What must one assume to deduce that the universe has a cause, then?

As an analogy, consider the following predicates:

  • $B'(x)$: "$x$ has finite cardinality".

  • $C'(x)$: "$x$ is a natural number".

Consider the counter model of $S=\{1,42\}$.

Then we have that $(\forall s\in S)(B'(s)\implies C'(s))$ and $B'(S)$, but not $C'(S)$.

$\endgroup$

closed as unclear what you're asking by Alfred Centauri, Emilio Pisanty, G. Smith, Jon Custer, ZeroTheHero Aug 16 at 0:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ The most common (and in my mind, correct) answer from physicists is: what do you even mean? What does it mean for something to “have a cause”, mathematically? This is an incredibly vague notion that historically has not led to progress in cosmology. You know this, which is why you are trying to dress up this notion in unnecessary notation, but that doesn’t make it any less vague. $\endgroup$ – knzhou Aug 15 at 18:18
  • 1
    $\begingroup$ Welcome to PhysicsSE. I think this is a question for mathsSE. I can't see any physics in it. Love the 42, tho. ;) $\endgroup$ – StudyStudy Aug 15 at 18:18
  • 3
    $\begingroup$ It’s like those misguided philosophy papers that start with “definitions” like “let $C$ equal consciousness”, as if you could convert something you don’t understand to something you do by assigning it a letter. $\endgroup$ – knzhou Aug 15 at 18:19
  • $\begingroup$ My question is sincere, @knzhou. $\endgroup$ – Shaun Aug 15 at 18:44
  • 2
    $\begingroup$ Well, you asked “how do we know the universe has a cause?” The simple answer is that we do not know that at all. And we certainly do not use formal logic to think about questions like that. There is no physics in formal logic. $\endgroup$ – G. Smith Aug 15 at 19:27
1
$\begingroup$

There seems to be a problem with this notation: the first line only applies to members of the set $U$, while the second line claims the predicate $C$ is also true and can be applied to the set beside (some) elements. It is not defined whether $C(U)$ is even defined.

You can try this by making $U$ the natural numbers, and making "having a cause" correspond to being a successor of a natural number. All numbers except the first have a cause... but that does not indicate anything about the set having a cause. Which, were it true, would have the semantic interpretation that "the set of natural numbers are a successor of some natural number" - plain nonsense.

Notation cannot fix this. You need to define the semantics of causes to make this work, either as philosophy, math or physics.

$\endgroup$
  • $\begingroup$ Thank you. I provided a counter model of my own in the original edition of the question. $\endgroup$ – Shaun Aug 15 at 19:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.