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Does infinity exist?

plus.maths.org

21–30 of 127 posts

Re: Does infinity exist?

#21
post #6

Earlier quoted context omitted.

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 occupie…

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

I can see how that works if you have an infinite number of rooms and an infinite number of guests, because you still have an unoccupied room to move them up into. But how does that work when you've stated that all rooms are occupied?

I'm really struggling to square up my understanding of how the former problem works with my intuition that if every room is in an occupied state, you can't magic up some in an unoccupied state.

Re: Does infinity exist?

#22
post #16
post #12

Earlier quoted context omitted.

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

... which would consolidate the occupants of rooms n and 2n (and there are people in room 2n to begin with, if all of the rooms are indeed currently occupied). edit: I see that the folks in 2n would now be in 4n , and so on down the chain. You'd always have some guests that are in the process of being evicted (i.e. between rooms); then again, that's negligible because we're not talking about real life (™). Clever!

If you move every guest from n to 2n, all of the odd rooms will open up, and you have room for a countably infinite number of new guests.

I believe David Hilbert was the first to describe this paradox. Source: http://en.m.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Gr...

Re: Does infinity exist?

#23

Earlier quoted context omitted.

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

Aren't there already guests in room n+1?

Yes. The statement as given doesn't work out. If you say "an infinite number of guests and an infinite number of rooms" it works. But when every room is endowed with the property of being occupied there is no room for more guests.

Re: Does infinity exist?

#24
post #13

Earlier quoted context omitted.

No, those guests now stay in room n+2.

I don't get it. Can't I just as easily say that n+2 is also occupied? After all, they're all occupied by definition, aren't they? For example. All blocks are either red or blue. You have an infinite number of blocks that are all red. Do you have any blue blocks? No. Where is the mistake in my logic?

n+2 was occupied, but it's not anymore because you moved them to n+3.

Re: Does infinity exist?

#26
post #13

Earlier quoted context omitted.

No, those guests now stay in room n+2.

I don't get it. Can't I just as easily say that n+2 is also occupied? After all, they're all occupied by definition, aren't they? For example. All blocks are either red or blue. You have an infinite number of blocks that are all red. Do you have any blue blocks? No. Where is the mistake in my logic?

>Can't I just as easily say that n+2 is also occupied?

No, because the former occupants of n+2 are now in n+3.

Re: Does infinity exist?

#27
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.

Re: Does infinity exist?

#28
post #5

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 e…

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.

Re: Does infinity exist?

#29

Earlier quoted context omitted.

Aren't there already guests in room n+1?

Yes. The statement as given doesn't work out. If you say "an infinite number of guests and an infinite number of rooms" it works. But when every room is endowed with the property of being occupied there is no room for more guests.

That's the sort of thing that happens when semantics meets mathematics; Swizec screwed up his statement of the problem. Swizec and the people agreeing that the answer is yes are confusing the problem's literal statement with the common statement of similar problems.

Re: Does infinity exist?

#30

Earlier quoted context omitted.

I don't get it. Can't I just as easily say that n+2 is also occupied? After all, they're all occupied by definition, aren't they? For example. All blocks are either red or blue. You have an infinite number of blocks that are all red. Do you have any blue blocks? No. Where is the mistake in my logic?

n+2 was occupied, but it's not anymore because you moved them to n+3.

no, n+3 is also full. See? Why do you get to win that back-and-forth and not me? After all, the problem stated that all rooms are full, even N+3, 4, 5, etc. Why do you get to construct a new room without also constructing a new pre-existing occupant?
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