# Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the same?

by tahayassen
Tags: numbers
P: 271
 Quote by leroyjenkens I used to have a problem with infinity. I kept using it like it was a number. For example, I couldn't understand that the amount of numbers between both 0 and 1 and 0 and 2 were both the same.
 Quote by tahayassen Aren't some infinities larger than other infinities?
 Quote by Jack21222 In a way, yes, that's what Cantor demonstrated in the late 1800s. If you took the number of numbers in between 0-1 and divided it by the number of numbers between 0-2, you should get 1/2. Let x be the number of numbers between 0-1. There are an equal number of numbers between 0-1 and between 1-2, so the number of numbers between 0-2 is x + x, or 2x. So you have x/2x, and even if x is infinity, they cancel (they're the same infinity). I'm sure mathematicians will murder me for doing it that way, since I probably did all kinds of things wrong, but I think that's the general idea.
 Quote by leroyjenkens The problem is you're using infinity as if it's a number. You added infinity with infinity. That makes no sense if infinity isn't a number.
 Quote by Jack21222 It makes plenty of sense. For every number in the 0-1 set, there is a corresponding number in the 1-2 set. In my example, x is not necessarily infinity, it's the number of numbers in between 0-1. The concept of infinities cancelling out, and one infinity being "bigger" than the other, is used ALL THE TIME in calculus when dealing with limits. For example, consider (2^x)/(x!) As x goes to infinity, the top and bottom are both infinity. However, the bottom infinity is "larger" so the limit as it goes to infinity is zero.
 Quote by EricVT For x equal to infinity, both the numerator and denominator are infinitely large, but their ratio is not zero. For x approaching infinity -- but still finite -- the numerator and denominator also have finite values and their ratio is close to zero, but not zero. Taking the limits of functions like this is not the same as dividing infinity by infinity.
This is a discussion we had in another part of the forum and I'm wondering who is correct. The discussion is becoming increasingly confusing and annoyingly (regardless of the posts in between the discussion), no one with a "Science Advisor", "Homework Helper", or "PF Mentor" title is stepping in to end the argument, so I would appreciate it if you would end the argument.
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The cardinalities of the intervals [0 1] and [0 2] are the same. In fact, there are as many numbers in the interval [0 1] as there are real numbers.

 For example, consider (2^x)/(x!) As x goes to infinity, the top and bottom are both infinity. However, the bottom infinity is "larger" so the limit as it goes to infinity is zero.
This is a pretty poor explanation of what's happening. Taking limits in calculus is a process; we examine what happens as a quantity increases without bound. The limit of that function as x goes to infinity is zero because x! increases faster as x increases, not because it is a "larger quantity". "Infinity" in this context has a completely different meaning than the "infinity" used to describe the size of sets.
 P: 2,237 there is an infinite, but countably infinite amount of rational numbers between 0 and 1. this is also the case for the interval from 0 to 2. but there is an uncountably infinite amount of real numbers between 0 and 1. this is also the case for the interval from 0 to 2.
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## Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the same?

Jack212222 is wrong. Cantor's theory is nothing like what he describes. He seems to be confusing things with limits.

And yes, the cardinality of [0,1] is exactly the same as the cardinality of [0,2]. So both sets have the same size.

EricVT is wrong too, since he assumes that $\frac{+\infty}{+\infty}$ is defined when it is not. So the ratio doesn't even make sense.
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 Aren't some infinities larger than other infinities?
Yes, this is true. But the example of [0,1] and [0,2] is not a good example since both infinities are the same here.

However, we can look at the sets $\mathbb{N}$ and $\mathbb{R}$. Those are infinite sets, but the latter set is much larger than the former.

See the following FAQ post: http://www.physicsforums.com/showthread.php?t=507003 (also check out the sequels whose link is at the bottom of the thread).
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 Aren't some infinities larger than other infinities?
You also have to be careful how you define "larger". Micromass is completely correct about cardinality and that will be appropriate if 'larger' means "has more numbers in the set" as was originally asked. But one can also argue that the interval [0, 2] is twice as long, and so twice as large, as the interval [0, 1]. It depends upon what you are comparing.
 P: 1 Two sets have the same cardinaility if there exists a 1-1 correspondence between them. The function f(x) = 2x establishes such a 1-1 correspondence between the rationals in [0,1] and [0,2]. It also establishes a 1-1 correspondence between the reals in [0,1] and [0,2].
 P: 1,058 Some infinities are larger than others. Namely, uncountable infinities are larger than countable infinities and that is all there is to it. The number of real numbers between 0 and 1 is uncountably infinite. The number of real numbers between 0 and 2 is uncountably infinite. The number of real numbers period is uncountably infinite. These all describe the same cardinality. The natural numbers are countably infinite, the real numbers are uncountably infinite. Therefore the cardinality of natural numbers is smaller than that of the real numbers. There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.
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 Quote by 1MileCrash There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.
Sure there is. There are many types of uncountably infinite sets. For example, the set $\mathcal{P}(\mathbb{R})$ (power set of the reals) has a strictly larger cardinality than $\mathbb{R}$. and $\mathcal{P}(\mathcal{P}(\mathbb{R}))$ is even larger!! This process continues indefinitely.
P: 271
 Quote by 1MileCrash There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.
Ah, thanks for pointing that out.

Edit: I take my thank you back.
 P: 810 [quote]amount of numbers[quote] ^ This is the crux of the problem. Talking about the "amount of numbers" in a set or which set is "bigger" (or "smaller" or "equal in size" or the notion of "size" at all) is an imprecise claim. When you're comparing two sets, there are many equally legitimate ways to do it. You can for example... Designate a count to each set, where the count is either the number of elements (if that number is finite) or infinity. Under this definition, the reals have the same count as the natural numbers because both are infinite. Designate a cardinality to each set. We say two sets have the same cardinality if there is a a bijection between them. We order the cardinalities with a relation, saying that A <= B when there is a bijection between A and some subset of B. In this case, card N < card R. However, card [0, 1] = card [0, 2] The notion of subset, too, gives us a way to compare sets. People take offense to the idea that [0, 1] is in some sense "the same" as [0, 2] because they are conflating two different senses of "same". One sense where [0, 1] and [0, 2] are clearly different is via a subset relation. Since [0, 1] is a subset of [0, 2], we can claim that [0, 1] is "less than" [0, 2] in this sense. However, this notion is different than the notions of cardinality or count above because it is a partial order (the other two are total orders). This means that you can't compare certain pairs of sets this way: [0, 1] has no relation to [2, 3]. There's one last common one that only works for special kinds of classes (including the real numbers), and that is measure. The sets [0, 1] has a certain property that, when written down on a number line, it gives us a certain length: 1 - 0 = 1. Similarly, the set [2, 5] also has a length: 5 - 2 = 3. This leads us to the notion of a measure. However, this is a very restricted notion, as it applies only to sets endowed with a measure space structure. And some sets, such as {0} in R or Q in R, have a measure of 0, despite having an infinite count. I post this because I feel too many people quickly jump on the idea that cardinality is the most important way to define a "size" of a set. I don't believe this is the case. In every day problem solving, count and the subset relation are just as important (if not more).
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 Quote by micromass Sure there is. There are many types of uncountably infinite sets. For example, the set $\mathcal{P}(\mathbb{R})$ (power set of the reals) has a strictly larger cardinality than $\mathbb{R}$. and $\mathcal{P}(\mathcal{P}(\mathbb{R}))$ is even larger!! This process continues indefinitely.
Just as a curiosity, what's the connection between the discrete/countable infinite sets (sequences) $\mathcal{P}^{n}(\mathbb{R})$ and $\aleph_{n}$ ?
P: 1,058
 Quote by micromass Sure there is. There are many types of uncountably infinite sets. For example, the set $\mathcal{P}(\mathbb{R})$ (power set of the reals) has a strictly larger cardinality than $\mathbb{R}$. and $\mathcal{P}(\mathcal{P}(\mathbb{R}))$ is even larger!! This process continues indefinitely.
Ah.. I failed to think of deriving a set from an uncountable set in such a way that the cardinality must be greater. It doesn't make much sense at all to say that a set and it's power set have the same cardinality. That is easy to see.

Thank you.

Is this limited to the idea of power sets? It feels like this could only occur when you "build" a "higher order" uncountable set from a previously uncountable set.

Also- the power set of the naturals would certainly be of higher cardinality than the naturals and thus not at a one-to-one correspondence with the naturals and therefore uncountable, right?

Does that make P(N) have the same cardinality as R, and P(P(N)) have the same cardinality as P(R)?

Or am I oversimplifying the idea?
 HW Helper Sci Advisor P: 11,717 P^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.
P: 1,058
 Quote by dextercioby P^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.
I don't understand how that could be the case.
 HW Helper Sci Advisor P: 11,717 Who's P(N) ? Write an explicit formula using the 3 symbols: ..., { and }. Then calculate its cardinality.
P: 1,058
 Quote by dextercioby Who's P(N) ? Write an explicit formula using the 3 symbols: ..., { and }. Then calculate its cardinality.
Forgive me if I'm not understanding but,

The number of elements of P(N) is the number of possible subsets of natural numbers that can be formed.

That is surely uncountable, as I can combine any number of any natural numbers I want to form some subset. If I call one subset, the non-proper subset, the entire set of natural numbers, one subset, what's another subset? The subset of 1 million of any natural numbers? 2 million any natural numbers? Considering all singleton sets of each natural number is already a countably infinite number of subsets, and there are uncountably many subsets besides those, it doesn't work in my brain to call P(N) a set of countable cardinality.

EDIT:

I found this

http://www.earlham.edu/~peters/writing/infapp.htm#thm3

Which I guess is a much better and more formal version of what I'm thinking.

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