Does a Line Have Twice as Many Numbers?

[Thudlow_Boink]

@Exapno_Mapcase

There are not twice as many integers as there are even integers. They are in a one-to-one correspondence.

This is true if you’re talking about cardinality, and is an example of what’s sometimes known as Galileo’s paradox.

But, as Chronos noted upthread, cardinality isn’t the only notion of “size” in mathematics. There are (roughly) twice as many integers as there are even integers in the first n integers, and this remains true as n approaches infinity. This sort of size comparison is simple and trivial when you talk about even integers but becomes far more interesting when you talk about something like how many prime numbers there are.

[Topologist]

@Exapno_Mapcase

You can’t have infinity plus one as an ordinal.

I just want to address this point: Yes, ω+1 is a perfectly valid ordinal. An ordinal is (an equivalence class) of a well-ordered set. One such is ω, which is the set of natural numbers, {1, 2, 3, …}. You get ω+1 by adding another element greater than any element in ω, call it 1’, say. So ω+1 = {1, 2, 3, …, 1’}, where 1’ is the largest element and there are countably infinitely many elements in the set less than 1’.

(Somewhat confusingly, but, it turns out, conveniently, a mathematician would usually call that largest element ω, not 1’, so ω+1 = {1, 2, 3, …, ω}.)

ω+2 is then {1, 2, 3, …, 1’, 2’}, and so on. 2ω = ω+ω = {1, 2, 3, …, 1’, 2’, 3’, …}, one countably infinite set on top of another.

[septimus]

@Exapno_Mapcase

… There are not twice as many integers as there are even integers. They are in a one-to-one correspondence…

The point is that there is also a one-to-two correspondence. (And also a seven-to-three correspondence, and so on.) Just because cardinal number equality is defined using one-to-one correspondences does NOT mean you are prohibited from considering other correspondences!

[Exapno_Mapcase]

@Topologist

I just want to address this point: Yes, ω+1 is a perfectly valid ordinal. An ordinal is (an equivalence class) of a well-ordered set. One such is ω, which is the set of natural numbers, {1, 2, 3, …}. You get ω+1 by adding another element greater than any element in ω, call it 1’, say. So ω+1 = {1, 2, 3, …, 1’}, where 1’ is the largest element and there are countably infinitely many elements in the set less than 1’.

(Somewhat confusingly, but, it turns out, conveniently, a mathematician would usually call that largest element ω, not 1’, so ω+1 = {1, 2, 3, …, ω}.)

ω+2 is then {1, 2, 3, …, 1’, 2’}, and so on. 2ω = ω+ω = {1, 2, 3, …, 1’, 2’, 3’, …}, one countably infinite set on top of another.

Here I yield, as my memory is obviously faulty.

@septimus

The point is that there is also a one-to-two correspondence. (And also a seven-to-three correspondence, and so on.) Just because cardinal number equality is defined using one-to-one correspondences does NOT mean you are prohibited from considering other correspondences!

Not here, though. :mad:

[Thudlow_Boink]

@septimus

The point is that there is also a one-to-two correspondence. (And also a seven-to-three correspondence, and so on.) Just because cardinal number equality is defined using one-to-one correspondences does NOT mean you are prohibited from considering other correspondences!

Okay, but in that sense, you can say with just as much justification that there are twice as many positive even integers as positive integers…

2 and 4 correspond with 1, 6 and 8 correspond with 2, 10 and 12 correspond with 3, etc.

[septimus]

@Thudlow_Boink

Okay, but in that sense, you can say with just as much justification that there are twice as many positive even integers as positive integers…

Sure. I implied this with my “And also a seven-to-three correspondence, and so on.”

I don’t think I’m being dogmatic — to the contrary I’m countering the dogmatic claim that there aren’t twice as many integers as even integers. There ARE twice as many; that’s seen very clearly by displaying a 2-to-1 mapping. The wrong viewpoint almost seems to argue that since infinity equals infinity it must not also equal two times infinity.

Exercise: Display a bijection between the set of real numbers and the power set of the integers. (This may be slightly harder than it looks.)

[Thudlow_Boink]

@septimus

I don’t think I’m being dogmatic — to the contrary I’m countering the dogmatic claim that there aren’t twice as many integers as even integers. There ARE twice as many; that’s seen very clearly by displaying a 2-to-1 mapping. The wrong viewpoint almost seems to argue that since infinity equals infinity it must not also equal two times infinity.

It’s true that Aleph[sub]0[/sub] = 2 × Aleph[sub]0[/sub] (or, in fact, Aleph[sub]0[/sub] = n × Aleph[sub]0[/sub] for any nonzero n).

What seems to be bothering some people (e.g. Exapno) is that terminology like “twice as many” is not standard with regard to infinite sets (AFAIK), and it makes it sound as though you’re asserting that the cardinalities are not the same, since that would be what it means in the case of finite sets.

[Chronos]

It’s slightly harder because of the 0.999… problem (or more precisely, the 0.111… problem, in binary). But it’s easy to define a 1:1 correspondence from the reals to a subset of the power set of the integers, or from the power set of the integers to a subset of the reals.

EDIT:

It’s true that Aleph0 = 2 × Aleph0 (or, in fact, Aleph0 = n × Aleph0 for any nonzero n).

Nitpick: For any nonzero finite n. Or at least countable.

[Quartz]

@Jim_B

When I first heard this joke, I thought it was ridiculous. Infinity plus one, is still infinity, I would tell myself. But now I am not sure.

Read up on Hilbert’s Hotel.

Does a line have twice as many numbers as a ray?

Think about it. A line goes on to infinity in both directions. A ray just one.

A line has two sets of coordinates, being a start point and an end point; a ray has one set, the start point, but also a velocity and a direction which can be combined into a vector for two sets of numbers.

[Thudlow_Boink]

@Chronos

Nitpick: For any nonzero finite n. Or at least countable.

Right. That’s what I meant, but I should have specified.

[Francis_Vaughan]

@Chronos

Nitpick: For any nonzero finite n. Or at least countable.

Actually no. Aleph[sub]0[/sub] * Aleph[sub]0[/sub] is still Aleph[sub]0[/sub]
Moreover Aleph[sub]0[/sub][sup]n[/sup] is still Aleph[sub]0[/sub] for all countable n. You just use the diagonalisation argument.

When you get to 2[sup]Aleph[sub]0[/sub][/sup] you get a bigger number. The continuum hypothesis says that this number (Beth[sub]0[/sub]) is the next countable number, which is Aleph[sub]1[/sub].

[Wendell_Wagner]

Jim B., does anything we’re saying make sense to you? You haven’t returned to the thread since the OP. All we’re doing now is talking among ourselves. We should be explaining this to you. Please tell us what you’ve learned in this thread and we’ll explain further.

[Chronos]

Re-read what I said, Francis Vaughan. We’re in agreement. Except for this:

Moreover Aleph[sub]0[/sub][sup]n[/sup] is still Aleph[sub]0[/sub] for all countable n. You just use the diagonalisation argument.

There, you did mean to say “for all finite n”. Aleph[sub]0[/sub] ^ Aleph[sub]0[/sub] is no less than 2 ^ Aleph[sub]0[/sub].

[Francis_Vaughan]

@Chronos

Re-read what I said, Francis Vaughan. We’re in agreement. Except for this:

There, you did mean to say “for all finite n”. Aleph[sub]0[/sub] ^ Aleph[sub]0[/sub] is no less than 2 ^ Aleph[sub]0[/sub].

Oops, yes, for finite n.

You had only written n x Aleph[sub]0[/sub] = Aleph[sub]0[/sub], when n is finite, or at least countable - which are of course different. n countable is bigger than finite, and is Aleph[sub]0[/sub][sup]2[/sup], and this can be extended for all finite n powers, not just 2. Aleph[sub]0[/sub][sup]n[/sup] = Aleph[sub]0[/sub].

The point being that Beth[sub]0[/sub] = 2^Aleph[sub]0[/sub] is bigger than Aleph[sub]0[/sub]. We define Aleph[sub]1[/sub] to be the next number bigger than Aleph[sub]0[/sub], and the continuum hypothesis says that Aleph[sub]1[/sub] = 2^Aleph[sub]0[/sub] = the cardinality of the powerset of the integers.
We can continue to take powersets, and thus create bigger and bigger infinite numbers. And do this in ever more infinite ways.

[Jim_B]

@Wendell_Wagner

Jim B., does anything we’re saying make sense to you? You haven’t returned to the thread since the OP. All we’re doing now is talking among ourselves. We should be explaining this to you. Please tell us what you’ve learned in this thread and we’ll explain further.

I find it all very fascinating. I don’t understand it all. But I assumed that would happen when I posted my question;).

[jz78817]

Population of the universe: None. Although you might see people from time to time, they are most likely products of your imagination. Simple mathematics tells us that the population of the Universe must be zero. Why? Well given that the volume of the universe is infinite there must be an infinite number of worlds. But not all of them are populated; therefore only a finite number are. Any finite number divided by infinity is as close to zero as makes no odds, therefore we can round the average population of the Universe to zero, and so the total population must be zero.*

• Hitchhiker’s Guide to the Galaxy

[Hari_Seldon]

@Exapno_Mapcase

This whole post confuses me to all heck. The cardinal numbers are the ordinary counting numbers, but the ordinals show place in a count. Among the numerals 1, 2, 3, 4, 5, 6, 7, 8, three is the third number. Cardinal is the ordinal number. I don’t get what you say about “Both are infinite ordinal numbers.” You can’t have infinity plus one as an ordinal.

Why can’t you have infinity plus one as an ordinal. It is an ordinal (order type) and differs from omega (sorry, I don’t know how to do these things) in having a greatest element. And all the other order types mentioned are also ordinals in good standing. And all the ones named are countable ordinals, meaning their cardinality is the same as that of the integers.

Each of those ordinals can be modeled as the set P + {P} where P consists of all the preceding ordinals together with one more element, namely P itself. So, for example, 0 is the empty set (no elements), 1 has one element, namely 0, 2 has two elements, 0 and 1, omega consists of all the finite integers and the elements of omega+1 are all the finite integers together with that as an element. So omega + 1 = omega + {omega}, the latter being the one element set whose element is omega.

The set of all the countable ordinals is denoted Omega and is the first uncountable ordinal. It is formally undecidable if its cardinality is the same as that of the line. The assumption that it is is called the continuum hypothesis, continuum being one name for the line.