Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know ultrafinitists want to require not only that mathematical objects be constructible, but be constructible given finite resources (such as time).

So I wonder about something like the famous "Graham's number"

Graham's number.

It's used as (non-binding) upper constraint so—are ultrafinitists OK with it? If I said some combinatorial number is less than $\infty$ they would maybe agree with this also? Unless $\infty$ "doesn't make sense" because it would "take too long" to "make" the upper bound one is talking about.

A related question which is probably better attached here: What about the operator $\forall$? $\forall x$ can be translated into English in two ways:

  • "for all $x$", which paints a picture of "going around and getting each and every $x$",
  • or "for any $x$", which paints the picture of a "gate" and if something comes up to your gate, and it is an $x$, then let it through—no "gathering" or "rounding up" is necessary. Just "leave the instructions with the guard" and the problem is solved passively.

(The second interpretation also seems more consistent with requiring $\exists x$ in some "resource-constrained" sense. If $\not \exists x$, the gate guard would stand there his whole life and never let anyone through, whereas in the "for all" sense of $\forall$, we would wander around the universe forever, check every object, and never find an $x$.)

share|cite|improve this question
I predict that either B.M.S. or Asaf will answer this question and have it accepted. – Git Gud Mar 14 '13 at 18:56
I don't know much about ultrafinitism, but the constructivist understanding of $\forall x.\phi(x)$ is very consiously like your second suggestion: To prove $\forall x.\phi(x)$, we must give some explicit method which, given $x$, produces a proof that $\phi(x)$ holds. Or put another way, we must explain how that guy at the gate will be sure to let through every $x$ that comes his way. – MJD Mar 14 '13 at 19:17
The first thing you said, "going around and getting all of the $x$", sounds to me much more like set comprehension than like quantification. Set comprehension has historically been a thorny issue, for just these sorts of reasons: perhaps $\phi$ is a perfectly good predicate, but does that really mean you can go around and get all the $x$ such that $\phi(x)$ holds? Standard set theory says no, not in general. – MJD Mar 14 '13 at 19:19
@MJD Thanks for the interesting responses. Are you suggesting AoC type problems occur with set comprehensions? I didn't quite follow that part. (I guess there must be a reason why $\forall x, \phi (x)$ isn't the same as $\{x | \forall x, \phi(x) \}$?) – isomorphismes Mar 14 '13 at 23:18
I'm not sure what you mean by AoC problems; I didn't say anything about the axiom of choice. The kind of set comprehension problem I had in mind is that even if $\phi(x)$ is a straightforward predicate like $x\notin x$, one must be very cautious about claiming that there is a set $\{ y\mid \phi(y)\}$, because sometimes there isn't. – MJD Apr 1 '13 at 4:06
up vote 1 down vote accepted

Ultrafinitists do not even believe in unlimited exponentiation ( so, much less in hyperoperators!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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