At his blog, Joe G. has worked himself into a lather over the cardinality (or roughly speaking, the size) of infinite sets (h/t Neil Rickert):
Of Sets, Supersets and Subsets
Oleg the Asshole, Still Choking on Sets
Of Sets and EvoTARDS
Subsets and Supersets, Revisted [sic]
In particular, Joe is convinced that the sets {0,1,2,3,…} and {1,2,3,4,…} have different cardinalities:
…the first set has at least one element that the second set does not, ie they are not equal.
So if two sets are not equal, then they cannot have the same cardinality.
As a public service, let me see if I can explain Joe’s errors to him, step by step.
First of all, Joe, sets do not have to be equal in order to have the same cardinality, as Neil showed succinctly:
I give you two sets. The first is a knife and fork. The second is a cup and saucer. Those two sets are not equal. Yet both have cardinality two.
Two sets have the same cardinality if their elements can be placed into a one-to-one correspondence. That’s the only requirement. Here is a one-to-one correspondence (or “bijection”) for Neil’s sets:
knife <–> cup
fork <–> saucer
Note that there is nothing special about that one-to-one correspondence. This one works just as well:
fork <–> cup
knife <–> saucer
It makes no difference if the elements are numbers. The same rule applies: two sets have the same cardinality if their elements can be placed into a one-to-one correspondence. Consider the sets {0,1,2} and {4,5,6}. They can be placed into the following correspondence:
0 <–> 4
1 <–> 5
2 <–> 6
As before, there is nothing special about that correspondence. This one also works:
1 <–> 4
0 <–> 5
2 <–> 6
There are six distinct one-to-one correspondences for these sets, and any one of them, by itself, is enough to demonstrate that the sets have the same cardinality.
Now take the sets {1,2,3} and {2,4,6}. They can be placed into this one-to-one correspondence:
2 <–> 2
1 <–> 4
3 <–> 6
However, note that it’s not required that the 2 in the first set be mapped to the 2 in the second set. This one-to-one correspondence also works:
1 <–> 2
2 <–> 4
3 <–> 6
As in the previous example, there are six distinct one-to-one correspondences, and any one of them is sufficient to demonstrate that the cardinalities are the same.
Now consider the sets that are befuddling you: the infinite sets {0,1,2,3,…} and {1,2,3,4,…}. They can be placed into a one-to-one correspondence:
0 <–> 1
1 <–> 2
2 <–> 3
3 <–> 4
…and so on.
A one-to-one correspondence exists. Therefore the cardinalities are the same.
Here’s what I think is confusing you, Joe: there is a different mapping from the first set to the second that looks like this:
0 doesn’t map to anything
1 <–> 1
2 <–> 2
3 <–> 3
… and so on.
You observe that all of the elements of the second set are “used up”, while the first set still has an “unused” element — 0. You comment:
They both go to infinity but the first one starts one number before the second. That means that the first one will always have one element more than the second which means they are NOT the same size, by set standards.
But that’s silly, because you could just as easily choose a different mapping, such as:
0 doesn’t map to anything
1 doesn’t map to anything
2 doesn’t map to anything
3 <–> 1
4 <–> 2
5 <–> 3
… and so on.
Now there are three unused elements in the first set. We can also arrange for the “unused” elements to be in the second set:
nothing maps to 1
nothing maps to 2
nothing maps to 3
0 <–> 4
1 <–> 5
2 <–> 6
… and so on.
There are now three unused elements in the second set. By your logic, that would mean that the second set has a greater cardinality. It clearly doesn’t, so your logic is wrong.
You can even arrange for an infinite number of unused elements:
0 <–> 2
1 <–> 4
2 <–> 6
3 <–> 8
… and so on.
Now the odd integers are unused in the second set. Since there are now infinitely many “unused” elements in the second set, then by your logic the second set is infinitely larger than the first. Your logic is wrong, and the mathematicians are right.
Two sets have the same cardinality if they can be placed into a one-to-one correspondence. The sets {0,1,2,3,…} and {1,2,3,4,…} can be placed into such a correspondence, so they have the same cardinality.
END (Hi, KF!)
Some exercises for math genius Joe G:
1. Take any set A of integers. A has a cardinality. Leave A alone for five minutes without adding or removing any elements. Has A’s cardinality changed?
2. Take any set B of integers. B has a cardinality. Add 1 to each of the elements of B without adding or removing any elements. Has B’s cardinality changed?
3. Take any set C of integers. C has a cardinality. Multiply each element of C by 17 without adding or removing any elements. Has C’s cardinality changed?
4. Take the set {0,1,2,3,…}. It has a cardinality. Add 1 to each of its elements without adding or removing any elements. You’ll obtain {1,2,3,4,…}. Has the cardinality changed?
5. Take the set {0,1,2,3,…}. It has a cardinality. Multiply each element by 17 without adding or removing any elements. You’ll obtain {0,17,34,51,…}. Has the cardinality changed?
6. Flounder about uselessly, trying to explain why your method says that the cardinality changes in scenarios 5 and 6.
keiths,
I think a simpler way to describe the problem is to agree that Joe may add 1 (or 100 or infinity) to infinity as soon as he has arrived at that “number”. In other words, you can’t look back from infinity because there is no “there”, there.
Perhaps this will make Joe’s “arguments” easier.
Let S = 1 + 2 + 4 + 8 + 16 + …
Multiply this by 2; then 2S = 2 + 4 + 8 + 16 + …
But, as we can easily see, 2S = S – 1.
Subtract S from both sides and we get that S = -1.
Therefore “infinity” = -1, a finite number. So, anytime you encounter infinity, replace it with minus one.
In fact, take the series
F = 1 – ½ + 1/3 – ¼ + 1/5 – 1/6 + 1/7 – 1/8 + 1/9 – 1/10 + …
Add it up in the order given, and get ln2; but add it up in any order you wish and get any number you wish.
Poor, confused Joe:
Umm, Joe — you haven’t specified the speed of each train, or whether it’s constant, or the distance between “markers” on each number line, and whether it’s constant, so you can’t conclude that Einstein 1’s count is greater than Einstein 2’s at any particular point in time.
However, let’s give you the benefit of the doubt and assume that the trains always travel at the same finite speed and that the distance between markers on Einstein 2’s number line is constant and twice that of Einstein 1’s. Then Einstein 1 will accumulate marks twice as fast as Einstein 2.
Here’s where you screw up: you assume that infinity is just a really big finite number, and that if Einstein 1 has accumulated more marks at every finite point in time, then he will also have more marks “at infinity”. But infinity is not a point, Joe. The Einsteins, traveling at finite speed, will never reach infinity.
Get it through your head, Joe: infinity is not just a big finite number. It is fundamentally, qualitatively different. It violates some of our intuitions.
You are clinging to an intuition that has been shown to be wrong. It’s time to take the next step toward intellectual maturity and let go.
I missed this the first time:
Neither did Bill Pascal or Ron Descartes.
I for one can’t wait for Joe’s Principia. Once he has ironed out a few kinks, naturally.
Patrick:
Sure he’s arguing that! After all, you can verify it. Just subtract infty from both sides and you get 1 > 0, which is true.
So it is verified: this is another example where it fits the known data.
😉
Joe G,
If you ever have to build an infinitely long fence, don’t put the posts 6′ apart. Instead, put them 12′ apart so that you only have to dig half as many holes! 😛
Joe has been making progress. He figured out that sets {1,2,3,4,…} and {−1,−2,−3,−4,…} have the same cardinality.
It remains a mystery how he reasoned to that. The two sets have no members in common, so he couldn’t have used his original comparison method based on subsets. He couldn’t have used his “line method,” either: the two sets of numbers live on opposite sides of the number line.
So he must have used Cantor’s standard method of finding a bijective mapping between the two sets. In this case one can use f(n) = −n. Each member of the positive set is mapped onto one, and only one, member of the negative set, and vice versa. Thus the two sets have the same cardinality.
That’s how sets are compared, Joe.
Well, it might have helped that Winston Eward posed the question 🙂
Winston may succeed where even Sal failed.
I don’t know about that. Joe is reacting to Ewert’s probing with his usual hostility:
It is amusing to see Joe take on mathematics, the font of right reason, right under Barry’s nose.
Eventually Winston had to pull “Do you know who I am?” on Joe. That made an impact.
Allegedly infinite size? LOL indeed.
Winston just suggested that Joe google him – that might help 🙂
Not if you spell his name Eward.
Hey Joe, don’t forget to do your homework:
Winston Ewert points Joe to an essay by Stephan Kulla, a German student, proposing an alternate definition of cardinality. Under Kulla’s proposed definition, proper subsets of countably infinite sets are smaller than their supersets.
Joe is ecstatic:
Before you get too excited, Joe, let me draw your attention to a few things:
1. It’s a student essay on a personal website.
2. It’s unpublished and unreviewed.
3. It suffers from the same flaw as “Joe’s cardinality”.
In fact, any definition of cardinality is flawed if it holds that a countably infinite set is larger than its countably infinite proper subset. I’ve already explained why, but you’re repeating the error in your latest screed, so I’ll try again.
You and Kulla both believe (though for different reasons) that the set of positive even numbers {2,4,6,…} is smaller than the set of natural numbers {1,2,3,…}. In your case, that is because you can set up the following mapping:
1 maps to nothing
2 maps to 2
3 maps to nothing
4 maps to 4
…and so on.
When you remove the matching elements, you’re left with {1,3,5,…}. In set notation, that would be:
{1,2,3,…} – {2,4,6,…} = {1,3,5,…}.
{2,4,6,…} is big. So is {1,3,5,…}. If you can subtract something that is infinitely big from {1,2,3,…} and still have an infinitely big set left over, then {1,2,3,…} must be bigger than both of them, right?
Wrong. The fact is that {1,2,3,…} can be converted into {2,4,6,…} without adding or removing any elements. You simply multiply each element by 2.
If one set can be converted into another without adding or removing any elements, by merely operating on each existing element, then the sets have the same cardinality.
{1} can be converted into {3} by multiplying each element by 3. They have the same cardinality.
{5,9,11} can be converted into {11,19,23} by multiplying each element by 2 and adding one. They have the same cardinality.
{1,2,3,…} can be converted into {2,4,6,…} by multiplying each element by 2. They have the same cardinality.
In desperation, you are now making this argument:
No, Joe, you don’t have to remove the 0 in order to add 1 to it. And even if you did, it wouldn’t help your case at all. You could still convert {0,1,2,3,…} to {1,2,3,4,…} by removing one element at a time, operating on it, and adding it to a result set. It would work like this:
1. Start with A = {0,1,2,3,…} and B = {}.
2. Remove an element from A.
3. Add 1 to the element.
4. Add the element to B.
5. Repeat steps 2-4 for each remaining element of A.
Here’s how the sets would evolve, iteration by iteration:
A = {0,1,2,3,…} B = {}
A = {1,2,3,4,…} B = {1}
A = {2,3,4,5,…} B = {1,2}
A = {3,4,5,6,…} B = {1,2,3}
…and so on.
After you’ve done this for all the elements of set A, you’ll have:
A = {} B = {1,2,3,4,…}
At each iteration, you removed exactly one element from A and added exactly one element to B. Yet according to “Joe math”, the new set B is one element smaller than set A was.
Where did the missing element go, Joe? Did the Designer poof it out of existence? Can you identify the step where it went missing?
OK, Joe. Commence flailing.
How about F(n) = |n| for {…,-3,-2,-1,0,1,2,3,…} ?
RTH:
Not sure what you’re asking, cupcake. 🙂
Joe,
After you’ve told us where the missing element went, here’s another question for you.
You claim that {1,2,3,…} is twice as big as {2,4,6,…}. If that were true, then it would be impossible to map {1,2,3,…} to {2,4,6,…} using F(n) = 2n, because we would exhaust the elements of {2,4,6,…} before all of the elements in {1,2,3,…} had been mapped.
Which elements of {1,2,3,…} are left over when we attempt this “impossible” mapping, Joe?
Recommence flailing.
keiths,
F maps n to absolute (positive) value of n for all integers from negative infinity to positive infinity.
But it’s not a bijection. With the exception of 0, two elements of {…,-3,-2,-1,0,1,2,3,…} map to every one element of {0,1,2,3,…}. It’s not a one-to-one correspondence, so it doesn’t allow us to compare cardinalities.
This one does, however:
…
-3 <–> 6
-2 <–> 4
-1 <–> 2
0 <–> 0
1 <–> 1
2 <–> 3
3 <–> 5
…and so on.
That is a one-to-one correspondence,,so we can conclude that {…,-3,-2,-1,0,1,2,3,…} and {0,1,2,3,…} have the same cardinality. They are both countably infinite.
keiths,
Maping to square roots must have similar issues…
I think you mean “mapping to squares”. Mapping to square roots wouldn’t be possible unless you included both the reals and the imaginaries.
Yeah, it would be the same problem. Except for 0, two elements of {…,-3,-2,-1,0,1,2,3,…} would map to every element in {0,1,4,9,16,..}.
You can still form a bijection, though:
…
-3 <–> 36
-2 <–> 16
-1 <–> 4
0 <–> 0
1 <–> 1
2 <–> 9
3 <–> 25
… and so on.
Since there’s a bijection, the cardinalities are the same.
Sqaure roots would have both a positive and negative correspondant, was (I think) my point, for N>0. n<0 has other problems, as you noted.
Poor Joe flunks his homework assignment:
The odd numbers are still there, Joe. It’s just that they’ve been doubled. The original set was {1,2,3,4,5,…} and the new set is {1+1, 2+2, 3+3,4+4,5+5,…}, which of course is the same as {2,4,6,8,10,…}. I didn’t remove any elements; I doubled each existing element and left it in place. Yet according to Joe Math, the second set is now magically half the size of the first.
Which elements magically disappeared, Joe? Give me the values of n for which n+n does not appear in the second set.
No, I’m converting the first set into the second and saying that they have the same cardinality. That should be obvious even to you, because no elements are added or removed. Each element is simply doubled in its place.
And if you insist that I have to remove each element in order to operate on it, that’s fine. I already explained how that would work for {0,1,2,3,…} and {1,2,3,4,…}. Here’s how it would work for {1,2,3,4,…} and {2,4,6,8,…}:
1. Start with A = {1,2,3,4,…} and B = {}.
2. Remove an element from A.
3. Double the element.
4. Add the element to B.
5. Repeat steps 2-4 for each remaining element of A.
Here’s how the sets would evolve, iteration by iteration:
A = {1,2,3,4,…} B = {}
A = {2,3,4,5,…} B = {2}
A = {3,4,5,6,…} B = {2,4}
A = {4,5,6,7,…} B = {2,4,6}
…and so on.
After we’ve done this for all the elements of set A, we have:
A = {} B = {2,4,6,8,…}
At each iteration, we removed exactly one element from A and added exactly one element to B. Yet according to Joe Math, the new set B is only half as big as set A was. Where did the missing elements go, Joe? Did the Designer poof them out of existence? Identify the step in which they disappeared, and show me the values of n for which n+n does not appear in set B.
Resume flailing.
Many high school 9th grade students can get the concept of cardinality. In fact, the kids interested in math pick this stuff up in elementary and middle school.
I suspect it may blow Joe’s mind to be told that the set of rational numbers has the same cardinality as the set of integers.
Make an array of rational numbers on a grid labeled 1 to infinity along the top, and 1 to infinity down the left side. Each rational number can be placed in the box that is at the intersection of the number along the top with the number along the left side.
Then count by starting at 1/1, 2/1, 1/2, 1/3, 2/2, 3/1, 4/1, 3/2, 2/3, 1/4, 1/5, 2/4, etc. So the rational numbers can be put into one-to-one correspondence with the integers.
Luckily for us, Joe doesn’t get it, he doesn’t want to get it, and he probably never will get it. He is floundering in shallow mathematical waters that are nevertheless way over his head, much to the amusement of a large internet audience.
For that I am grateful, and so is Hermagoras of AtBC:
Indeed. I am finding this immensely entertaining.
This reminds me of all the discussions with the intelligent design creationists of UD about genetic algorithms. Many of them appeared unable to distinguish between the model of evolutionary mechanisms and the implementation of the model.
This same confusion of the map and the territory is demonstrated by Joe’s response. Leaving aside the fact that he has clearly never understood even high school level math, he seems incapable of distinguishing between the value of an element in a set and the index of that element.
It seems that the ability to accept a literal reading of scripture is linked to the inability to consider abstractions.
It just gets better and better.
Here is Joe’s latest:
LOL.
On Dembski’s blog. Priceless.
Hey Joe,
Let’s use your brilliant discovery to compare the cardinality of two sets A and B where A = {…,-3,-2,-1,0,1,2,3,…} and B = {…,-6,-4,-2,0,2,4,6,…}.
Let’s find a mapping function that gives us a bijection (a one-to-one correspondence).
F(n) = 2n works, so according to Joe Math, set A is twice as big as set B. Great! What was that twit Cantor thinking?
But wait a minute… F(n) = 2n + 10 also works. So according to Joe Math, set A has twice the cardinality of set B plus 10.
And F(n) = 2n – 10 also works, so set A has twice the cardinality of set B minus 10.
F(n) = -2n works, so set A has negative two times the cardinality of set B.
F(n) = -2n – 9,510,348,551,087,402 also works, so set A has negative two times the cardinality of set B minus 9,510,348,551,087,402.
What’s the right answer, Joe?
Screw this Joe Math. Let’s go back to Cantor.
Silly mathematicians thought that they were establishing a one-to-one correspondence, when in fact they were… well, establishing a one-to-one correspondence.
Joe, are you really incapable of plugging numbers into a formula?
F(-7) = 2(-7) + 10 = -4
F(-6) = 2(-6) + 10 = -2
F(-5) = 2(-5) + 10 = 0
F(-4) = 2(-4) + 10 = 2
F(-3) = 2(-3) + 10 = 4
…and so on.
F(n) = 2n + 10 maps every element of {…,-3,-2,-1,0,1,2,3,…} to exactly one element of {…,-6,-4,-2,0,2,4,6,…}, and every element of {…,-6,-4,-2,0,2,4,6,…} is mapped to by exactly one element of {…,-3,-2,-1,0,1,2,3,…}. It’s a one-to-one correspondence. (It’s a frickin’ linear function, Joe. Did you think it would loop back on itself?)
“However that ‘mapping’ just means they made each set equal to the other, equal in size . . .”
He looks like he’s starting to get it!
“. . . as well as equal in membership . . .”
But he spins out at the turn.
“. . . and the equation is the actual difference in cardinality between the two original sets.”
And crashes into the wall.
If the two sets have the same size, Joe, they have the same cardinality. That’s what cardinality means.
Another attempt to penetrate Joe’s infinitely thick skull.
Joe, I showed you how to convert {1,2,3,…} to {2,4,6,…} without adding or removing any elements.
You politely and plaintively ask:
This may help. Imagine you have a countably infinite collection of objects, to which you have applied sticky labels with the numbers “1”, “2”, “3”, etc., written on them. Each label is unique; no two labels contain the same number.
Now imagine that for each object, you peel off the existing label and replace it with a new label. On the new label you write a number that is twice the number on the old label.
The objects remain the same throughout. You haven’t added or removed any objects — you’ve just changed the labels. Because the objects remain the same, the cardinality of the set remains the same. As you like to say, it’s so obvious that even a fourth-grader could understand it.
Yet according to Joe Math, the set miraculously lost half of its objects when it was relabeled. Where did they go, Joe? Who removed them?
Joe asked:
I showed him, and now he is backpedaling. Seems the mapping function no longer matters — just part of the mapping function:
No, Joe, it doesn’t accomplish the same thing. When you change the mapping function from F(n) = 2n to F(n) = 2n + 10, every single element of the first set now maps to a different element of the second set.
But let’s suppose you’re actually right, and that tacking a constant onto the end of the mapping function “accomplishes the same thing.”
In that case, F(n) = 2n + 0 accomplishes the same thing as F(n) = 2n + 10 which accomplishes the same thing as F(n) = 2n – 666,666 . If they all accomplish the same thing, then any one of them can be used to establish the relative cardinality of the two sets. Yet you say that only F(n) = 2n + 0 is permissible. On what basis?
Imagine that.
It’s not not understanding the maths. It’s not understanding the maths AND insisting he knows the maths better than anyone else. My mum has probably never heard of cardinality, but I wouldn’t laugh at her for it.
Not that I want Joe to change.
Exactly. It’s Joe’s flaming Dunning-Krugerism that earns our mockery, not his mathematical illiteracy.
According to this, the set of negative integers {-1,-2,-3,…} has negative the cardinality of the set of positive integers {1,2,3,…} … whatever that means.
Also, the set {1,1/2,1/3,1/4,1/5,…} has half the cardinality of the set {1/2,1/4,1/6,1/8,…}. And the set of nonnegative powers of two {1,2,4,8,16,32,…} has twice the cardinality of the set {2,4,8,16,32,…}. And if you take the set of all integer powers of two {…,1/16,1/8,1/4,1/2,1,2,4,8,16,…} and double each of its members you get the same set back, proving that this set has twice its own cardinality (and also half).
Joe, is it really this hard to learn something about how Cantor’s theory of large cardinals works, instead of inventing your own theory by piling half-baked intuition on top of half-baked intuition? Yes, it’s complicated and unintuitive, but reality’s like that sometimes; in the long run, you’re better off learning to cope with this rather than insisting that reality bend to your will.
keiths,
Gordon Davisson,
Joe, going by your declaration that f(x)=2x doubles the cardinality, what would you do with the function:
f(x)=x-1?
Since the coefficient of x in that function is singular both sets must have the same cardinality. Let’s see…
x -> f(x)
1 -> 0
2 -> 1
3 -> 2
4 -> 3
5 -> 4
6 -> 5
But this is exactly the relation that started everything and you said didn’t have equal cardinalities. You have just created a contradiction. Isn’t that supposed to mean you are wrong?(Law of Non-Contradiction or some such rot.)
Edited to add:
JoeG asked for equations.
:::Joe G said…”So no equations- I asked for equations, Mr Math.”
:::9:46 AM
I assume f(x)=x-1 is an equation but I could always be wrong.
Joe,
I don’t think Joe can distinguish the difference between the Cardinals and the Ordinals.
Ok, so we need a pattern. Let’s try this. We can use three sets. Set A is just the non-negative integers: {1,2,3,4,5,6,7,8,9,10,11,12…}. Set B is all the non-negative even integers: {0,2,4,6,8,10,12,14,16,18,20,22…}.
For set C, instead of a function or math let’s perform marriage*. Joe you wanted a pattern. The pattern(although you could also think of it as a rule) is this; the ordinal position of the element in set B is chosen for marriage by the cardinal value(which shares the same ordinality) of the element in set A they are to be married to. The first element of set C will be the visual joining of the first elements of the other two sets and so on. So the first element of set C will be 10. The second will be 22. The third will be 34. ditto, ditto, etc, etc, and so forth. Set C is: {10,22,34,46,58,610,712,814,916,1018,1120,1222,…}.
The question is: which set, A, B, or C, has the largest cardinality. Which set has the smallest. Which set has the middle cardinality. (BTW, this is a trick question since everyone, but you, knows they all have the same cardinality.)
*Since right wingers are all infatuated with marriage, right?
Aardvark:
Hi Aardvark,
I thought about making that point yesterday, but decided against it for the following reasons.
Joe tells us that F(n) = 2n + 0 and F(n) = 2n + 10 “accomplish the same thing” with respect to {…,-3,-2,-1,0,1,2,3,…} and {…,-6,-4,-2,0,2,4,6,…}. What he may be fumbling toward, in his confused and inarticulate way, is that both functions establish a one-to-one correspondence between the two sets. The additive constant is arbitrary. As long as it is evenly divisible by 2, we have a one-to-one correspondence.
Similarly, F(n) = -2n works as well as F(n) = 2n. The sign is arbitrary. As long as the absolute value of the coefficient is 2, we achieve a one-to-one correspondence.
If we throw out negative 2 as the coefficient and retain positive 2, and throw out all additive constants except 0, then we are left with F(n) = 2n (though Joe still needs to justify his preference for positive 2 and 0 over the other possibilities).
In the case of {1,2,3,4,…} and {0,1,2,3,…}, the function F(n) = n – 1 produces a one-to-one correspondence, as you point out. However, nothing about that function is arbitrary. If you change the coefficient, the sign of the coefficient, or the additive constant, a one-to-one correspondence is no longer achieved.
Thus Joe could argue that the entire mapping function is relevant in this case, and that the cardinalities therefore differ by 1.
(I’m probably giving Joe WAY too much credit here.)
He would still be completely wrong, of course, and the relabeling argument shows why, but at least he wouldn’t be contradicting his earlier statement about the relative cardinalities of {0,1,2,3,…} and {1,2,3,4,…}.
Joe,
OK, Joe, here’s a pattern for you:
{1} has the same cardinality as {2}
{1,2} has the same cardinality as {2,4}
{1,2,3} has the same cardinality as {2,4,6}
{1,2,3,4} has the same cardinality as {2,4,6,8}
By using Joe’s Principle of Extension to Infinity, we can conclude that
{1,2,3,4,…} has the same cardinality as {2,4,6,8,…}.
We have thus contradicted Joe’s Earlier Principle of Cardinality by Mapping Function.
Good job, Joe.
After thinking all day and coming up empty, Joe wants to change the subject:
C’mon, Joe, don’t give up — an entire internet audience is awaiting your floundering, expletive-laced Joexplanation of how relabeling a set of objects makes half of them vanish into thin air. We also want to know what to do when Joe’s Principle of Extension to Infinity collides with Joe’s Earlier Principle of Cardinality by Mapping Function. Should we apply the theory of relativity? Should we “go fuck ourselves”?
Don’t disappoint us.
Of course we don’t need set theory to discuss nested hierarchies, but it is applicable and useful.
Joe’s main difficulty with the concept of infinity is a failure to realize that infinity is a journey, not a destination.
When Joe compares sets {1,2,3,…} and {2,4,6,…}, he argues that the first of these has twice as many numbers on any finite segment between 0 and some positive number.
This reasoning works for finite segments. However, infinity is not simply a Really Big Positive Number. It is a never-ending sequence of numbers that grow without limit. The even integers have no need to catch up to all integers. Both sequences keep going forever.
Joe’s latest post: Is Set Theory Misleading? where he starts with:
Here’s a clue, Joe. Mathematics is not about reality. Mathematics is important for its methods, which turn out to be very useful. We don’t expect mathematical truths to be directly applicable to reality. We expect them to be applied to the methods used to study reality.
From Joe’s latest.
keiths:
Joe:
All we did was change the labels, Joe. Nobody removed any elements, but you claim that half of them are now missing. Why?
keiths:
Joe:
What directions, Joe? Show us the directions, and then show us how you would deal with my example: