Regular readers of TSZ will remember the hilarity that ensued when former commenter JoeG grappled unsuccessfully with the cardinality (loosely, the size) of various infinite sets. In honor of that amusing episode, I’m posing a new problem involving an infinite set.

Here’s the problem:

Consider the set containing every real number that can be described using a finite number of English words. For example, “thirty-three” and “two point eight” obviously qualify as members of the set, but also “pi minus six”, “the cube root of e”, and “Zero Mostel’s age in years on July seventh, nineteen sixty-three”, all of which designate specific real numbers. The set is infinite, of course.

Prove that the set of all such numbers takes up exactly zero percent of the real number line.

Of course Joe G, in his current incarnation as “ET” at Uncommon Descent, is welcome to participate remotely. For laughs.

Does reaching zero percent involve dividing by infinity?

I’m not sure that there’s a point to this thread.

“Zero percent” means nothing without a suitable technical definition.

Using traditional mathematics, the real numbers you describe would constitute a set of measure zero, with the standard Lebesgue measure. However, somebody using intuitionist mathematics would likely have a different answer.

Yes, JoeG was confused about mathematics. But why try to pick a fight with him, when he is unable to post here?

Not a formal proof, but I’ll give it a go

The cardinality of the proposed set is N0 (the cardinality of the set of natural numbers)

The cardinality of the set of all real numbers ‘c’ is larger, so it’s an infinitely larger set than the other one

And incapable of making sense anywhere.

Glen Davidson

petrushka:

No, the proof doesn’t require division by infinity.

Neil:

That’s okay. Leave it to the folks who get it.

dazz,

That’s correct, but the challenge is in proving it.

Neil:

I explained that already:

The laughs have already begun:

“But Einstein still holds.” LMAO.

I dont think JoeG can be faulted for not knowing the answer to this. He can only be faulted if he thinks his intuition on the matter trumps that of an actual mathematician.

How did you even get started on this topic? Did it begin with ‘the set of all possible 300 amino acid proteins’ or something of that sort? …or maybe it had to do with ‘cosmic fine-tuning’?

RodW,

I certainly wouldn’t expect Joe G to be able to solve the problem. The point of including him is to elicit gems like “But Einstein still holds.”

The problem isn’t original with me. I’ll reveal the source later.

Did JoeTard just say that R is countable? lulz

dazz,

Nah, he’s still struggling with a problem from the earlier thread:

I think comments on this thread ought to be closed.

As a pragmatist, I will leave them open. Alan is free to change that if he wants.

Here’s my reasoning:

If I leave comments open, there will actually be very few such comments anyway.

If I close comments then the moderation thread will be flooded with complaints.

Make them thoughts into prayers perhaps?

Mung:

Neil:

Your reasoning is bogus, but at least you managed to make the right decision this time, unlike in the Paley thread.

As dazz correctly intuited, but didn’t prove, the set described in the OP has the same cardinality as the set of natural numbers. By mathematical convention, that cardinality is referred to as ℵ0 (“aleph-null”).

The overall proof can therefore be decomposed into two subproofs:

1) Prove that the cardinalities are the same; and

2) Prove that any set of real numbers with cardinality ℵ0 will occupy zero percent of the real number line.

While people ponder the possibilities for a proof, let me address Joe G’s “Einstein” argument:

Besides the hilarious conflation of math with physics, Joe is making another dumb error here. He’s correct that we can map elements from the second set onto elements of the first set in a way that leaves unmapped elements in the first set. What he doesn’t seem to realize is that we can also do it the other way around.

For example, suppose we map each number n in the first set to 6n in the second set:

1 <-> 6

2 <-> 12

3 <-> 18

4 <-> 24

…and so on.

Now it’s elements in the

secondset that are left unmapped. Cantor was smart enough to see the problem and avoid it. Joe isn’t. By Joe’s reasoning, the first set is both larger and smaller than the second.Joe G responds:

Okay. Then let’s use JoeMath’s sophisticated technique of SET SUBTRACTION to determine the relative sizes of these two sets:

A = {1,2,3,4,5…}

B = {1.1, 2.1, 3.1, 4.1, 5.1…}

Subtract B from A, and all the elements of A are left unmatched. Subtract A from B, and all of the elements of B are left unmatched. Therefore A is larger than B, and B is larger than A.

Again:

Joe’s response:

Okay, let’s apply JoeMath™ SET SUBTRACTION to the following two sets, which contain matching numbers (the first two elements of each set):

A = {1,2,3,4,5…}

B = {1,2, 3.1, 4.1, 5.1…}

Subtract B from A, and infinitely many elements of A are left unmatched. Subtract A from B, and infinitely many elements of B are left unmatched. Therefore A is larger than B, and B is larger than A, according to JoeMath™. Another fail.

How do you know? You can’t use JoeMath™ SET SUBTRACTION, because there are no matching elements. Hmmm. Wait — I have an idea! Let’s use Cantor’s method, which actually works.

Yes, indeed. You

dounderstand that.Joe G:

Duh, indeed. There is no zero “left of the decimal point”, unless you’re talking about the implicit leading zeros, and those won’t help you.

Again:

Moved comment to guano

Moved comments to guano. Please discuss moderation issues in the appropriate thread.

In response to my latest demonstration of the failure of JoeMath™ SET SUBTRACTION, Joe offers this incisive rebuttal:

Poor Joe:

keiths:

Joe G insists that the leading zeros help him…

…and then describes a procedure in which the leading zeros

help him in the slightest:don’tSo besides the leading zeros problem, the elements don’t

actuallyhave to match, according to Joe. They merely have to Joematch, meaning thatsomethingin one matchessomethingin the other.But even the revised Joematching “procedure” fails, by leading to a contradiction. Consider the following four sets:

A = {1,2,3,4,5…}

B = {2,4,6,8,10…}

C = {3.1, 3.2, 3.3, 3.4, 3.5…}

D = {3.2, 3.4, 3.6, 3.8, 3.10…}

1. First consider the original sets A and B:

A = {1,2,3,4,5…}

B = {2,4,6,8,10…}

Since only half of the elements in A are Joematched by elements of B, the two sets have different cardinalities according to JoeMath™.

2. Now consider sets C and D. According to Joe’s revised “procedure”, the “3” in each element of C matches the “3” in the corresponding element of D. Every element has a Joematch, so JoeMath™ SET SUBTRACTION leaves no elements when you subtract C from D. Therefore C and D have the same cardinality, according to JoeMath™.

3. Now repeat the procedure for B and D:

B = {2,4,6,8,10…}

D = {3.2, 3.4, 3.6, 3.8, 3.10…}

The “2” in element B0 matches the “2” in element D0, and

the “4” in element B1 matches the “4” in element D1,

the “6” in element B2 matches the “6” in element D2,

and so on.

JoeMath™ SET SUBTRACTION leaves no unJoematched elements in set B. Therefore, according to JoeMath™, sets B and D have the same cardinality.

4. And finally, do it for sets A and C:

A = {1,2,3,4,5…}

C = {3.1, 3.2, 3.3, 3.4, 3.5…}

The “1” in element A0 matches the “1” in element C0, and

the “2” in element A1 matches the “2” in element C1,

the “3” in element A2 matches the “3” in element C2,

and so on.

JoeMath™ SET SUBTRACTION leaves no unJoematched elements in set A. Therefore, according to JoeMath™, sets A and C have the same cardinality.

5. Notice:

A has the same cardinality as C, by step #4.

C has the same cardinality as D, by step #2.

D has the same cardinality as B, by step #3.

Therefore, by transitivity, A has the same cardinality as B, according to JoeMath™.

6. But in step #1, JoeMath™ told us that A and B have

differentcardinalities.7. It’s a blatant contradiction, and JoeMath™ fails again.

Prepare for another goalpost move.

Imagine that.

Meanwhile, Joe

resurrects choo-choo math:Joe:

No, Joe. You told us that the trains are moving at 1 mile per minute. After 10 hours, the trains will have gone 600 miles. Train A will have hooked 600 rings, and Train B will have hooked 300. It’s simple arithmetic.

And no, “your detractors” don’t predict that the counts will be the same.

You are talking about the equivalent of two finite sets:

A1 = {1,2,3…599, 600}

B1 = {2,4,6…598, 600}

A1 has twice as many elements as B1.

But we don’t care about A1 and B1. We care about A and B, whereObviously.A = {1,2,3…}

B = {2,4,6…}

Those are infinite sets, not finite ones.

And as I just demonstrated, JoeMath™ says that A and B have the same cardinality, and it also says that they have different cardinalities. A blatant contradiction.

JoeMath™, and its “choo-choo math” variant, are useless.

Heh. Joe went and quietly altered his OP to fix the arithmetic error I pointed out. He doesn’t mention the correction, of course.

keiths:

Joe G:

There will never be a point in time in which the two sets are infinite, either. Your choo-choo math is therefore irrelevant to the problem, which asks about the cardinality of two infinite sets.

Damn, Joe. This really isn’t that difficult.

Joe G has a new OP, and of course it’s a mess.

Joe, instead of making these endless ad hoc patches to your “procedure”, why not just sit down and figure out, once and for all, how it’s supposed to work?

Given two sets A and B, how does one determine — using JoeMath™ — whether the cardinalities are the same?

Lay out the complete procedure.Poor Joe. In the comment above I asked him to lay out the complete JoeMath™ procedure for comparing the cardinalities of two sets. My request apparently spooked him, because instead of responding, he is now arguing that the very concept of an infinite set is problematic:

This is the same Joe who confidently told us that

So Joe was jabbering about infinite sets then even though he now tells us that infinity and sets “were never supposed to go together.” Nice foot shot, Ace.

Joe G:

The set of positive real numbers

a well-defined collection of them, Ace. Give me a real number, and I’ll tell you whether it belongs to the set.isOn the choo-choo math front, Joe writes:

At every point along the journey, the sets are finite, Joe. This is obvious.

You’re talking about finite sets, not infinite ones.

Last call for solutions to the problem posed in the OP. I’ll reveal the answer tomorrow.

As promised, here’s how to do the proof stipulated by the OP.

Step 1 is to show that the cardinality of the set in question — that is, the set containing every real number that can be described using a finite number of English words — is the same as the cardinality of the natural numbers. This can be achieved by setting up a one-to-one correspondence between the two sets. But how?

The key is to deal not with the numbers themselves, but with their verbal descriptions. Every description is a sequence of English words, and all of the descriptions are finite. That means they can be placed in alphabetical order, forming an infinitely long alphabetized list.

Note:Many numbers will have more than one description. For example, 2 can be described as “three minus one”, “the cube root of eight”, “the number of legs Barack Obama had in 2012”, and many more. In these cases, we can keep the first description of a given number (going by alphabetical order) and throw out the rest.The first description in the alphabetized list can be paired with the natural number 1, the second can be paired with 2, and so forth. It’s a one-to-one correspondence, so the cardinalities of the two sets are the same.

That completes step 1.

Step 2 is to show that any set with the same cardinality (ℵ0) as the natural numbers will occupy zero percent of the real number line.

Consider the infinite series

For values of x less than 1, the series converges to the value x/(1-x).

Start out with x = 1/2. The series is then

…which converges to a value of 1.

Now associate the corresponding natural number with each of these terms, so that 1 corresponds to 1/2, 2 corresponds to 1/4, 3 corresponds to 1/8, and so on. At the position of each of the natural numbers, we’ll shade in a portion of the real number line corresponding to the size of the corresponding term.

Since the first term is 1/2, we’ll shade in a segment of length 1/2 starting at the number 1. That means that half of the segment from 1 to 2 is shaded in.

Since the second term is 1/4, we’ll shade in 1/4 of the segment from 2 to 3.

Since the third term is 1/8, we’ll shade in 1/8 of the segment from 3 to 4.

…and so on.

We already know that the infinite series sums to 1, so that the total length of all the shaded segments is 1. Yet each of those segments includes the corresponding natural number, so we know that the total space on the number line occupied by the natural numbers must be less than 1, which is the total length of the shaded segments.

We can repeat the entire process for a smaller value of x, say x = 1/4. In this case, the shaded lengths will sum to 1/3, so we know that the total space on the number line occupied by the natural numbers must be less than 1/3.

In fact, we can decrease x as much as we like, as long as it remains positive. Each time we decrease x, the infinite sum also decreases. In fact, the infinite sum approaches 0 as x approaches 0. But we know that the total space occupied by the natural numbers must be less than the infinite sum, no matter how small x gets. That means that the space occupied by the natural numbers can only be 0, which of course corresponds to zero percent of the number line.

Step 2 is then complete, and that completes the proof.

H/T to James Tanton, Mathematician in Residence at the Mathematical Association of America, who presented this problem in his Teaching Company course

The Power of Mathematical Visualization.Joe G, a week ago:

Joe G, today:

So now we find that JoeMath™, which was supposedly superior to Cantor in its handling of infinite sets,

can’t even cope with the concept at all.Poor Joe. Cantor must be laughing from the grave.

While Joe regresses, let’s have some more fun with choo-choo math.

Recall that Joe was trying to demonstrate that the set of natural numbers {1,2,3…} has a greater cardinality than the set of even natural numbers {2,4,6…}. He asked us to imagine two choo-choo trains moving in parallel down two infinite tracks. Choo-choo Train A picked up a brass ring (not iron, mind you, but

brass) every mile. This was supposed to represent the set {1,2,3…}. Choo-choo Train B picked up a brass ring (not iron, mind you, butbrass) every two miles. This was supposed to represent the set {2,4,6…}.After the first mile, Choo-choo Train A would have picked up more brass rings (not iron, mind you, but

brass) than Choo-choo Train B, and this would remain true thereafter. (“Always and forever”, as Joe likes to say.) Therefore, according to Joe, the set {1,2,3…} must have a greater cardinality than the set {2,4,6…}.The error is obvious. The sets Joe is actually comparing are finite, not infinite. He is confusing two finite but growing sets with two infinite ones, and reaching an erroneous conclusion regarding the infinite sets.

Here’s a choo-choo math scenario that illustrates the problem. Let’s compare the cardinality of the odd natural numbers {1,3,5…} with that of the even natural numbers {2,4,6…}. Again, we’ll have two choo-choo trains moving in parallel down a pair of infinite tracks, but this time, instead of picking up brass rings (not iron rings, mind you, but

brass) they will instead pick upnumbers.Choo-choo Train A will pick up the odd natural numbers, and Choo-choo Train B will pick up the evens. The numbers are arranged in order, one per mile.

After the first mile, Choo-choo Train A has picked up the “1”, and so its set is {1}.

After two miles, Choo-choo Train B has picked up the “2”, so its set is {2}.

After three miles, Choo-choo Train A has picked up the “3”, so its set is {1,3}.

After four miles, Choo-choo Train B has picked up the “4”, so its set is {2,4}.

…and so forth.

If we check after every two miles, we find that the two sets are always the same size. “Always and forever”, as Joe likes to say. We therefore conclude, using choo-choo math, that the two sets have the same cardinality.

And whadda you know? Cantor gives the same answer! Has choo-choo math been vindicated?

Um, no.

Look what happens when you make a simple change to the starting conditions. Instead of arranging the natural numbers in order along the tracks, we arrange them in this order: {1,3,5,2,7,9,11,4,13,15,17,6,19,21,23,8…}.

All the natural numbers are still there. They’re just presented in a different order.

Now we start the trains. After every four miles, we compare the sets and find that Choo-choo Train A’s set always contains three times as many numbers as Choo-choo Train B’s set — always and forever. Therefore we conclude, using choo-choo math, that the set of odd natural numbers has a greater cardinality than the set of even natural numbers.

Oops.

Now that he’s given up on applying JoeMath™ to the cardinalities of infinite sets, I became curious about where Joetelligence would find its next outlet.

Turns out it’s the greenhouse effect. Check out

this gem:H/T Occam’s Aftershave at AtBC.