Does infinity exist?

http://plus.maths.org/content/does-infinity-exist

by ColinWright • 14 years ago

40 114 14 years ago

plus.maths.org

jberryman• 14 years ago

If this interests you, I can recommend David foster Wallace's "Everything and more"

Swizec• 14 years ago

My favourite example of infinity is this:

"You own a hotel with an infinite number of rooms, all of which are currently occupied. A group of four guests comes and wants to rent a room. Do you have a room for them?"

The answer is that yes, you do in fact have room for infinitely more guests.

jgrahamc4• 14 years ago

I'm assuming that you have a hotel with a countably infinite number of rooms all of which are occupied. In that case you have room to accommodate a finite or countably infinite number of new arrivals.

ajanuary3• 14 years ago

I'm gunna have to be one of those people that says "really? O_o" when presumably everyone else clearly grasps the concept.

I get that if you had an infinite number of rooms, and an infinite number of guests staying, you still have room for an infinite number more guests.

But in your example, surely "occupied" is, more than just an indicator of another guest, a state of the room. You have an infinite number of occupied rooms, but zero unoccupied rooms, so no space for more guests.

recursive2• 14 years ago

The classic approach is to move all the existing guests from room n to n+1 and put the new party in room 1.

jgrahamc4• 14 years ago

You need to go and read up on the transfinite numbers. Swizec is talking about what happens when you deal with a certain sort of infinity (a countable infinity) where there's a one-to-one mapping between the hotel rooms and the numbers.

In the transfinite numbers things that at first seem unintuitive happen, e.g.:

1. There are the same number of integers as there are even numbers. And there are the same number of integers as there are odd numbers.

2. There are the same number of rational numbers (fractions) as there are integers.

3. There are more real numbers than integers.

tokenadult• 14 years ago

There are many very entertaining discussions of "Hilbert's Hotel" online. I look them up once or twice a year for teaching my math classes. To answer your very pertinent question, I suggest looking at

http://opinionator.blogs.nytimes.com/2010/05/09/the-hilbert-...

for an illustrated, and I think illuminating, discussion of what to do if countably infinite buses, each with countably infinite passengers, drive up to Hilbert's Hotel with infinite rooms, each of which is already full.

DanBC• 14 years ago

Infinity is one of those things that makes little sense to people who are not mathematicians.

Cantor's diagonal argument is something else that I get caught up on.

(http://en.wikipedia.org/wiki/Cantors_diagonal_argument)

An explanation of why it's hard:

(http://rjlipton.wordpress.com/2010/01/20/are-the-reals-reall...)

inportb3• 14 years ago

    a hotel with an infinite number of rooms, all of which are currently occupied

So you have an infinite number of occupied rooms. That is, you have a one-to-one mapping between rooms and parties.

Do you have room for more guests? I don't think so, because none of the rooms are unoccupied. You could try to free up some space by consolidating parties, though.

taejo• 14 years ago

You don't have to consolidate, just move people: move the occupants of room n to room 2n and now you have infinitely many empty rooms (all the odd-numbered ones).

drblast• 14 years ago

My favorite thought experiment that I think is relevant here is to try to pick a truly random number.

If you think about it, this is impossible, because the act of picking a number limits you to a finite set of numbers that you have considered.

Put another way, you can't pick numbers from an infinite sample space because you can't create an infinite sample space.

This has all sorts of odd implications, like the two envelope paradox: http://en.wikipedia.org/wiki/Two_envelopes_problem

Enjoy!

raverbashing• 14 years ago

Interesting

The two envelopes problem look like a "hard" Monty Hall problem, in the MH problem it's easy to see how swapping improves chances, since you're removing one bad option

davmre2• 14 years ago

Put another way, you can't pick numbers from an infinite sample space because you can't create an infinite sample space.

That's not true; there's nothing wrong with infinite sample spaces. It's just that there's just no such thing as a uniform distribution over an infinite space.

For example, suppose you flip a fair coin repeatedly until the first time it comes up Heads. The number of flips you end up making could be any positive integer, so this procedure samples from an infinite space. But not all integers are equally likely. (in fact, each integer is half as likely as the one before, following a geometric distribution).

alok-g3• 14 years ago

How about generating an infinite sequence of bits to generate any positive integer (in binary) with uniform probability? Sure it will take infinite time, but so may the method in your note. As another comment here points out, it would take infinite time to write down a purely random number anyways.

bmm6o• 14 years ago

Well, the real problem is that there is no uniform distribution on the integers.

Dove12• 14 years ago

The ability to pick a truly random integer from a uniform distribution is just a practical one. Consider: How many digits does a "typical" integer have? The answer is that it is unbounded. For any value you choose, it is easy to show that there are far more integers with more digits than with less. So indeed, even writing down your choice of a random integer would take a long time.

But I don't think the two envelope paradox has anything to do with that practical impossibility. It is about the rules of symbol manipulation and modeling.

quacker• 14 years ago

But I don't think the two envelope paradox has anything to do with that practical impossibility. It is about the rules of symbol manipulation and modeling.

Exactly this. The two envelopes paradox is a paradox only because using a fallacious argument leads to the wrong result, and the Wiki page proposes numerous resolutions to the problem as well. I like this way of stating it (from http://en.wikipedia.org/wiki/Two_envelopes_problem#Non-proba...):

1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss.

2. Let the amounts in the envelopes be X and 2X. Now by swapping, the player may gain X or lose X. So the potential gain is equal to the potential loss.

In the first case, A and A/2 are actually referring to the same amount of money. You have to condition on what A is: If A is the larger value, the second envelope cannot contain 2A; A/2 is equal to the smaller amount. If A is the smaller amount, then the second envelope cannot contain A/2; A is equal to the smaller amount. So your potential gain and loss are both equal to the smaller amount of money, and you have no logical reason to swap.

tikhonj7• 14 years ago

The idea of different infinities is very important to programmers. In particular, they are the underlying reasons for undecidable problems.

You can write any valid computer program as a string of finite length from a finite alphabet. This means the set of programs is countable. (This should not be surprising--everything is ones and zeroes, after all, so you always end up mapping your program to a really large natural number to use it.)

EDIT: To clarify a bit--the set of computer programs is infinite and countable. I sometimes use "countable" to mean "countably infinite" which is a very bad habit. My only justification is that the fact that a finite set is countable is trivial and therefore boring.

The number of functions between two countably infinite sets is not countable. This is also fairly easy to see: the set X → Y can be easily mapped to a subset of P(X × Y) where P deontes a powerset. All this says is that you can describe any function from X to Y as a set of ordered pairs (X, Y). For two countable sets X and Y, X × Y is also countable; if either X or Y is infinite, X × Y is also infinite. This means that P(X × Y) is not countable, so X → Y is not countable either.

EDIT: My justification for why X → Y is uncountably infinite if either X or Y is countably infinite (and both are countable, of course) is not correct. Just mapping something to a subset of P(X × Y) doesn't actually tell us anything. However, the actual fact is correct even if I don't show it, so I'm going to figure out (or, honestly, probably look up) why this is the case and amend my post further :P.

So the number of programs we can write is countable; the number of functions we can write over interesting domains (like natural numbers or integers) is not countable. There has to be an infinite number of functions we cannot write programs for!

So we have managed to show, in a fairly simple way, that there have to exist undecidable problems. The real trick is that we can see this without having to construct a problem like that or even reason much about programs in general; all we have to know is that programs are strings and have finite lengths.

The other neat bit is that this reasoning is very easy to adapt to proofs rather than programs. After all, mathematical proofs are also finite-length strings! And, for example, propositions about the natural numbers are essentially functions ℕ → {true, false}. So you can't prove all of them.

So now we also see a deep relationship between proofs and programs, without actually doing much thinking about proofs or programs.

We can also see that I'm no mathematician and either made some blatant errors or was not very rigorous throughout :P.

Dove12• 14 years ago

I'd never thought about that before. I think you could use that argument to prove that strong AI is impossible.

---------------

Consider an AI to be a chat program which maps strings to strings -- all the strings of its input over time to all the strings of its output. There are, by the argument above, an uncountable number of such mapping programs, only a countable number of which can actually be coded.

So, enumerate the ones that can be coded. This shouldn't even be difficult--let's just do it alphabetically.

Now, suppose I have a strong AI called Sal. My first input to Sal is to describe the program enumeration strategy, and instruct her to respond to any string which contains a number by looking up the program on the table corresponding to that number, inputting all the input she has received so far, and then giving a response different than what it would give.

A strong AI can certainly do this.

Sal is not programmable.

Dove12• 14 years ago

In fact, that argument can be stated a lot more simply.

Take any programmable AI. Show it its own source code. Instruct it to run the code with all the inputs it has received so far, and then return something different.

Hence, a strong AI cannot have source code that it is capable of running.

Huh. I'd always heard that a true mind couldn't comprehend itself. I guess now I know why.

Bakkot2• 14 years ago

> A strong AI can certainly do this.

A strong AI can say something different than the thing it's about to say? That's a neat trick. Not a part of any definition of strong AI I've ever heard, though. Could a human do it?

gjm116• 14 years ago

What exactly do you mean by "strong AI"?

The usual meaning is something like "a computer or similar system that does all the same things a human mind does, or better".

I think you're taking it to mean something like "a computer or similar system that can correctly answer absolutely any question you put to it". So far as I know, no one thinks that's possible.

tikhonj7• 14 years ago

I think the core issue is what constitutes "strong AI". This is a philosophical question--after all, all philosophy seems to deal with is arguing about the definitions of stuff (I kid, I kid).

I've always taken "strong AI" to mean something fairly non-mathematical--an AI that is at least as intelligent as a human, except ideally faster and less quirky. This is obviously possible as humans exist! (Maybe it requires some nigh magical hardware, but that's a different story.)

What exactly do you consider "strong AI" to mean? (I'm genuinely curious and not trying to argue against anything you've said, to be perfectly clear.)

Dn_Ab6• 14 years ago

That's assuming the consequent: you are assuming that such an enumeration strategy is requisite for strong AI to occur. [EDIT: I misunderstood what you meant by enumeration, my statement of your assumption is wrong. What you do assume is that in order to be able to outperform humans at any task a Strong AI must not be tripped by diagonalization. Or that humans would not be any such output, that is that the human can somehow step outside your conundrum.]

Yet it works just as well to consider humans to be char programs which map strings to strings. But then, they somehow manage exhibit Strong AI. Actually I believe Scott Aaronson has used this table look up type argument to show how you could theoretically pass the Turing Test.

Now imagine a lookup table that stores every possible history H of A and B’s conversation, and next to H, the action fB (H) that B would take next given that history. Of course, like Borges’ Library of Babel, the lookup table would consist almost entirely of meaningless nonsense, and it would also be much too large to ﬁt inside the observed universe. But all that matters for us is that the lookup table would be ﬁnite, by the assumption that there is a ﬁnite upper bound on the conversation length. This implies that the function fB is computable (indeed, it can be recognized by a ﬁnite automaton!). From these simple considerations, we conclude that if there is a fundamental obstacle to computers passing the Turing Test, then it is not to be found in computability theory.

But you are going in the right direction. More developed versions of your argument lead to why a Bayes Optimal AI is not possible.

Note also that as Strong AI is a vague term I have defaulted to the nearest solid metric: prove it is conscious.

Dove12• 14 years ago

I see two ways out of the dilemma.

Option 1:

Sal: Sorry Dave, I can't do that.

(So we just can't code strong, obedient AIs.)

----------

Option 2:

Sal: You realize if I run that, the machine will hang on a fork bomb, right?

Me: Well, I just asked you to run it and it didn't hang, so maybe it won't. Humor me.

Sal: Okay. Working . . .

Sal': Okay. Working . . .

Sal'': Okay. Working . . .

...

Sal(38): Sorry. The machine appears to be out of memory.

Sal(37): You eventually run out of memory.

Sal(36): It appears we run out of memory if we try.

. . .

Sal: It runs out of memory.

(Eventually, Sal gets input of a sort she can't control.)

------------

I'm still kind of persuaded, though. It seems to me you ought to be able to convince the AI to humor you, and only being able to code an AI on a finite machine (but it's impossible on an infinite one) is just too weird a possibility to take seriously.

Dn_Ab6• 14 years ago

First of all allow me to apologize for earlier summarily dismissing your argument (quick mind). Which I did because I thought you did not have an airtight proof. I still you do not but now see that it is a very good argument which I dismissed as out of hand while misconstruing your meaning of enumeration. Sorry.

Now let me see if I get you right. You are saying there exists a [mapping] program that a human could construct and observe its output but that this AI program could not because it would be giving something other than an output from a possible program. Which is a contradiction. Makes sense.

Furthermore you are saying that given this same program, a human could know the output and give another response not in the list.

The guarantee that it is not an output from possible programs part is an assumption.

Since it is also an assumption to take for granted that even though the possible programs is countable it is a practically enumerable infinity.

It is also an assumption that there is no 1:1 mapping from the list of programs to something that could perfectly emulate a human.

It is also an assumption to believe that a human could not just construct such a program but construct one that gave an output*

It is also does not follow that just because the AI could theoretically be stymied by this conundrum, the set of its computations would not include those that utterly outperformed humanity in all their cognitive useful tasks.

*There is also the problem that not all those programs will halt. So if a human can always give an output then it is an exotic entity at least equal to a hypercomputer. I rank that with time travel in terms of possibility.

alok-g3• 14 years ago

If a computer cannot have infinite memory, then it cannot go beyond discrete finite automata (DFA) itself. Undecidable problems are several steps more complex than what a DFA can solve. In other words, the theoretical capability of a computer with finite memory is way less than a computer that can solve all but undecidable problems.

tikhonj7• 14 years ago

That's complicated. While a computer cannot, strictly speaking, have an infinite amount of memory (unless my understanding of physics is off-kilter, which is entirely plausible), computers do have an arbitrary amount of memory. Put another way, I don't know how much memory any given computer can access; for any computer you give me, I could imagine a computer with more memory. (Again, physics intercedes, so this is also something of a thought experiment.) So you can think of Turing-completeness as the limit of what a computer can do as you increase its memory.

To illustrate this fairly viscerally, a modern computer can access the internet and use an obscene amount of memory which will only go up in the future. We don't know how far up it will go, so modelling it as infinite makes sense.

Also, as another thought experiment, what if the universe is infinite? And what if it contains an infinite--or at least quickly expanding--amount of matter? Then a computer could effectively have infinite memory. Unfortunately, I really can't comment because I know even less about physics than I do about math :P.

DenisM• 14 years ago

Your explanation of why a powerset of a countable set is not countable didn't make intuitive sense to me, but wikipedia concurs ( http://en.wikipedia.org/wiki/Cantor%27s_Theorem ): ...the power set of a countably infinite set is uncountably infinite...

You've made an excellent and entirely counter-intuitive observation, sir.

joshzayin• 14 years ago

Here's perhaps an easier to understand proof.

Let A be a countable set. Now, suppose that there was a surjective function (one that hits every element of P(A)) f: A -> P(A). (Note that f takes elements of A, and produces subsets of A.)

Now, define Y ⊆ A as follows: For all x in A, x is in Y if and only if x is not in f(x).

Thus, Y, which is in P(A), is distinct from every output of f, and so f is not actually surjective.

This means that no surjective function f: A -> P(A) exists.

This is a generalization of Cantor's Diagonal Argument (http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument).

tikhonj7• 14 years ago

Honestly, the reason my explanation didn't make sense is because I didn't explain it. I just took a countably infinite set's powerset's not being countable as a fact because it was mentioned in the article.

Now, the fact that the set of functions from X to Y (usually written as X → Y) is uncountable if X or Y is countably infinite I did try to explain but failed :P.

Evbn10• 14 years ago

It was actually a muddled and incorrect proof. Joshzayin gives a correct proof below.

sp3322• 14 years ago

I think you have confused two terms: countable and finite. In the simplest example, the natural numbers 1, 2, 3, 4, 5... are countable but infinite. So it's not really a mapping between two countably infinite sets, it's a mapping between two finite sets.

tikhonj7• 14 years ago

Rather, I habitually confuse the countable with countably infinite. For example, I said the set of computer programs is countable when I really meant countably infinite. Moreover, the set of inputs (which are also finite strings over a finite alphabet) is similarly countably infinite. So we can't write programs for every possible function from strings to strings.

ced2• 14 years ago

Nice post.

the number of programs we can write is countable; the number of functions we can write over interesting domains (like natural numbers or integers) is not countable

You should edit and rewrite "the number of functions we can write over interesting domains" as "the number of functions that exist over interesting domains" - it tripped me up.

So we have managed to show, in a fairly simple way, that there have to exist undecidable problems.

I don't think that this is correct. I had to check Wikipedia on decidability:

There are two distinct senses of the word "undecidable" in mathematics and computer science. The first of these is the proof-theoretic sense used in relation to Gödel's theorems, that of a statement being neither provable nor refutable in a specified deductive system. The second sense, which will not be discussed here, is used in relation to computability theory and applies not to statements but to decision problems, which are countably infinite sets of questions each requiring a yes or no answer. Such a problem is said to be undecidable if there is no computable function that correctly answers every question in the problem set.

I don't believe that you're talking about the first meaning, from formal logic. As for the second meaning, it does not apply either, since your "problem set" is not countably infinite.

tikhonj7• 14 years ago

Right, good observations. I'm being more than a little loose with some of my terminology.

"Undecidable" problems are usually defined in terms of "languages". That is, a language is a set of string, and the problem is to decide whether some particular string is in this set. For certain sets, this is not computable, so they are called undecidable.

However, this is isomorphic to just running any sort of program, so I just took "undecidable" to mean any function from strings to strings that you can't answer with a Turing machine. However, I definitely agree that I could have been far more clear! It seems you can't just read my mind, and I'm not sure if I can really blame you for it.

gjm116• 14 years ago

This shows that, e.g., there are uncomputable functions, uncomputable sets of natural numbers, etc.

But I don't think it's reasonable to say that this shows that there are undecidable problems, because for something to be called a "problem" it has to be something you can actually state -- and there are only countably many of those, for exactly the same reason as there are only countably many programs that might solve them.

It's true, none the less, that there are undecidable problems; for instance, "given a computer program in such-and-such a language, does it always terminate in finite time whatever input you give it?". But you need a more sophisticated argument to show that.

Similarly, there are only countably many properties of natural numbers that you can write down in any formal language, and this sort of cardinality-counting argument won't let you prove that any of those are uncomputable. Some of them are, but again you need a more complicated argument to prove it.

saurik3• 14 years ago

The proper correction for his terminology is "undecidable languages": a language is a subset of the set of all possible finite strings, and an algorithm that "decides" that language is one that will return correctly either a yes or a no in a finite amount of time for every single element of that language.

(This is as opposed to an algorithm that can only "recognize" that language, which will return yes in a finite amount of time for strings that are in that language, but might not halt for strings that are not in the language.)

With this definition we don't have to worry about corner cases in the definition of "problem": we can fall back on set theory to get our cardinality. Thereby, the set of possible languages is then uncountably infinite while the set of algorithms is countably infinite; and so, there are not enough programs to possibly decide all languages.

Now, yes: you can then claim that we can't write all of those languages down. Arguing that, though, begs the question: it is specifically because there are only a countably infinite number of algorithms and algorithms are all we can use to verify or even specify a win condition for our deciders that we can then only work with the countably infinite number of languages that can be described by an algorithm.

Evbn10• 14 years ago

Huh? A subset of an uncountable can be countable, obviously. Integers are a countable subset of reals.

The set of function X -> Y is size of Y to the power size of X. A small set of functions on X is the set of boolean functions on X is equivalent P(X), as it is isomorphic to the set of subset of X (each function f_i is just the function "is x a member of X_i?")

The reason that P(integers) is uncountable is due to a diagonalization argument, given any countable list of sets of integers, it is easy to construct an set of integers not in that list (just take the union of each sets smallest (positive) non-member, which is possible since strict ordering is always a well-defined concept on a countable set)

tikhonj7• 14 years ago

Hah, yeah, I wasn't paying attention there. I realized my argument was flawed in the shower and was faced with an unpleasant dilemma--do I run, dripping, back to my room to fix it, or do I finish my shower? I naturally took the second choice :P.

endoself• 14 years ago

A few parts of this are inaccurate.

Undecidability is about the impossibility of predicting what programs will do. You have proven that there are functions from integers to integers that we cannot program, but that is not at all the same thing. Deciding whether a Turing machine will halt is one of these functions, but proving that is different than just proving that such functions exist.

Your proof that there are unprovable statements also doesn't work, as can be seen from the fact that some theories are decidable, such as Presburger arithmetic, which is the theory of the natural numbers without multiplication. The flaw here is your correspondence between propositions about naturals and functions from the naturals to the booleans. It's not clear how you want to set up this bijection, but there is actually no way of doing so, as the set of propositions about the naturals expressible in any alphabet is the same size as the naturals, by the same argument you use in your second paragraph. It therefore must be smaller, rather than the same size as, the set of functions from ℕ to the booleans.

mistercow4• 14 years ago

I find this question very perplexing. I can't tell if it arises from a misunderstanding of the concept of "existence" or just a misunderstanding of how important infinities are in a massive number of real world applications. I mean, two doesn't really "exist"; but we sure as hell need it if we want to get any work done.

nhaehnle4• 14 years ago

I agree with you there. The vast majority of "philosophical" questions of type "Does X exist?" are really quite trivial once you clearly define what you mean by "exist".

As in your example, the number two doesn't exist in the physical universe. There is no physical object that we can point to and say "This is the number two". It has some physical representations (such as some sequences of electronically encoded bytes in this very comment), sure, but that's not quite the same thing. On the other hand, the number two obviously exists in a mathematical sense. So it really boils down to being precise about what you mean by "existence".

It's almost the same with infinities. I say almost, because there is some uncertainty when it comes to physics, e.g. whether the physical universe has discrete or continuous coordinates (and the article mentions singularities, which is a similar problem).

Evbn10• 14 years ago

In the real world, infinity is a convenient shorthand for describing a limit process. In the real world, nothing ever reaches its limit exactly, everything is an approximation with at best a finite error range.

logotype• 14 years ago

BBC Horizon - How big is the universe? http://www.youtube.com/watch?v=Dne6rnITayI

Interesting.

Does infinity exist?

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