4

Suppose I make the argument:

It is very unlikely that in a naturalistic universe, the constants have life sustaining values, since the majority of metaphysically possible universes do not have such constants.

I'm confused as to how to make sense of this argument. To begin with, there are infinitely many (possibly uncountably many, if constants are on a continum) possible universes, with infinitely many of them life sustaining and infinitely many non-life sustaining.

How are we to assess the ratio |S|/|T| where both S and T are infinite (uncountable) sets, in an argument like the one presented.

3
  • In the context of fine-tuning, "possible universes" are typically described by finitely many parameters that take values in bounded domains. In that case, there is Lebesgue measure on the relevant set of parameter values and "majority" means a subset of it with more than half of the set's measure. If the set is the segment [0,1], for example, [0,3/4] will reference a majority.
    – Conifold
    Oct 30 at 8:22
  • This assumes that the universal constants are assigned as some kind of random variable in some distribution. But, we don't know that the universal constants are assigned in such a way. It could be that they aren't arbitrary at all, and what we think of as "other possibilties" have probability 0, i.e. are impossisible. The premise of the whole question requires a very wide leap. How do we know if a parameter is metaphysically possible? Is anyone aware of some kind of law of metaphysics that tells us when a universal constant is permissible? Oct 30 at 18:02
  • This is not a question about philosophy of maths, but about measure theory; it should be migrated to Maths SE.
    – Corbin
    Oct 30 at 18:15

4 Answers 4

6

This is already a problem even for simple probability distributions in the real numbers, such as the normal (Gaussian) distribution. The reals are infinitely dense, so we cannot assign individual probabilities to each number one at a time (unless we're content to only have countably many possible outcomes, which is not the case for continuous distributions such as the normal). Instead, we assign a density to each number. The probability of a given interval being selected is equal to the integral of the density over that interval.

Normally, this definition is given in terms of distributions supported on the real numbers, because those distributions are some of the commonest to encounter. But there is nothing restricting us to the reals. As long as you can take integrals over a given sample space, you can extend this definition to that space, and now you have a perfectly good way of defining continuous probability distributions over it. However, for particularly esoteric sample spaces, you will likely need to figure out how to take integrals over it, and (probably) also verify that your notion of probability will comply with the Kolmogorov axioms. If you can't do those things, then you can't really have a meaningful notion of probability for that sample space.

1
  • Given density, can we alter the fine-tuning claim to the following: we say $|A|$ happen a majority of the time in the space A\cupB, iff int_{-inf}^{inf} a(da) > int_{-inf}^{inf} b(db)
    – Mani
    Oct 30 at 18:31
0

Assume the axiom of choice.

Then any two cardinals x, y are comparable, i.e. there holds exactly one of the three alternatives

x < y, x = y, y < x

If x < y then the division y : x is defined. If in addition y infinite then y : x = y. The proof is given by cardinal number theory, e.g. see "Jean Rubin: Set Theory for the mathematician".

Aside: Which constants do you mean in your question, the natural constants from the physics of the specific universe under consideration?

2
  • The constants of physics are generally considered to possibly take on a continuum of values, so this is no help; any interval for a continuous constant has the same cardinality as any other interval.
    – causative
    Oct 29 at 23:12
  • Correct, that is what I meant, the constants of physics. I am also curious how you would remedy the issue raised by @causative
    – Mani
    Oct 30 at 18:34
0

For reasons of transfinite cardinal addition and multiplication, the ratio of an infinite cardinal C to itself is much akin to the ratio 0/0. E.g., since one can take all the naturals N and subtract all the even numbers from N so as to represent a case of N - N, but one still has N-many numbers left over (the odds), then N/N = 1 is possible, as is such as = 2, and = 3, and so on. And by multiplicative inversion can this be checked: if N/N = 2, then 2 times N = N, which is also true of the cardinality, here. This is because |N| + |N| = |N|, then.

Moreover, since N times N = N (assuming the axiom of choice, at least, as far as I know...), then there is a way for N/N = N too.

The ratio of N to R would be like an infinitesimal, since N is infinitely smaller than R (much smaller, we often imagine).

The ratio N/n would evaluate to N. Suppose N/2 = N. This means that subtracting 2 from N that many times could yet mean N-many more left over (due to above kinds of reasons). Or, again, N by 2 = N, "and we are almost done."

"Basically," in the sphere of infinite sets, the theory of probability has to be adjusted accordingly. You might very well find out that, other things being equal, the number of worlds with life-supporting constants was equivalent in size to the number of worlds with life-blocking constants. (This perhaps only insofar as talk of the constants varying "just like that" is realizable in the first place: there could be something about all possible worlds such that their physical constants all have to be a very specific way, even if we thought there was some meta-possibility of the constants differing.)

ADDENDUM: Cofinitude

There is, however, a relation of "almost all" or "all but finitely many," i.e. the cofinitude relation, that, if it held in this kind of case, would open the door to the number of life-supporting worlds to be finite in number, even out of infinitely many possible worlds. I.e. we would have to claim that all but finitely many worlds are lifeless, which would be a great deal to insist upon with almost entirely a priori reasoning as one's only serious justification for making such a claim.

0

Let's make a simple example with a toy universe in which sustainability of life only depends on two parameters.

Let's say that the parameters are c, the speed of light, and G, the gravitational constants.

Too little gravity G and nothing sticks together so life doesn't end up forming. Too large the speed of light and other physics breaks down in mind bending ways to prevent life.

Now, let's imagine we can control c and G by rotating a knob bringing them from 0 to 100 percent which is the maximum possible value these can potentially get.

You can now represent all possible universes as a rectangle. At the bottom left corner you have universes with c=G=0 and at the top right you have c=G=100.

Every point within the rectangle is a universe and there are uncountably many of them (if you believe that space is infinitely divisible..).

Now, if we knew that only universes where G is between 10 and 20 percent are capable of hosting life then we can shade the corresponding area to see it represents only a minority of the possible universes.

How to mathematically make sense of the fact that the point, an object with no area, can be added up to other points to give you an area, is the job of measure theory, an extremely challenging and beautiful branch of mathematics.

Also be wary, it is highly speculative that we can say what the parameters of life supporting universe might be, a little less if we stick to life as we know it. Also, obviously there might be tons of these parameters rather than two, though the argument here extends immediately to volumes with three parameters and by analogy to many more, and with a bit of functional analysis (which studies vector spaces with infinitely many degrees of freedom) to infinitely many. Finally, that we can imagine this mathematically doesn't mean that what reality actually is, our model is just a model, the map is not the territory.

4
  • The issue is, here you assumed that our parameter space has finite measure. Now if we had a set of parameters (a_{0},..., a_{n}) (for the moment I'm assuming this set is finite), each of which can take on a finite continuum of values, then your analogy works. But if we had either (a) that each parameter itself extends infinitely far, or (b) the set of parameters itself is infinite, or even uncountable. I guess there might be some Hilbert space stuff going on here. Let me know if you can clarify how to address these issues.
    – Mani
    Oct 30 at 18:38
  • There is no such issue, everything generalises and all these universes lie in the same class of infinities. It is just mind boggling and not really worth the hassle, especially from a philosophically pov.
    – Three Diag
    Oct 30 at 21:56
  • So you might need some functional analysis and Hilbert or Banach spaces to be able to define integration, but that is only a technical complication on how to correctly define or compute such quantities, but adds very little to the essential intuition behind the problem
    – Three Diag
    Oct 30 at 23:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .