
 Ricky
 Moderator
Re: To infinity and beyond.
And if mathematicians force people to accept their own idea in instituations like university, in the name of abstractness, in the name of beauty or in the name of intellect.
Do you see engineers learning about Cantor sequences? Biologists learning about Dedekind cuts? Psychologists learning about Hilbert's space filling curve?
No. Your premise is false.
Regardless of what particular infinite set you define Try to shift all elements one position rightwards by some pratical algorithm (in this case +1) Then add in the starting element lost in the shifting process (in this case 1) And then it is undeniable logic truth that I have just created an infinite set with one element more than the old one And now it is time to reveal what the old set lacks  the right end And the right end is the infiniteth set revealed sometimes infinity, sometimes infinitesimal so the paradox is discovered.
For the statement in bold, how can you say the new set has "one more"? What does "one more" mean with infinite sets?
For the statement in italics, What do you mean "what the old set lacks"? What does it mean that the right end is revealed? I see no right end...
Feel free to talk more on this, but I would also like to try again to introduce you to the concept of cardinality. I failed miserably before, not because I did anything wrong, but because it was overly complicated. I got a new book on Set theory and logic, which was an absolutely marvelous introduction to it.
Let's start our discussion on finite sets and try to compare their size. Take two sets:
{1, 2, 3} and {a, b, c}
Now each has a size of 3, and so they have the same size. But to say this, one needs to understand what integers are. While seemingly simple, integers can be rather complicated. So I would like to introduce a way to say whether two sets are the same size or not that does not involve any counting.
We say that two sets have the same size if the elements of one set can be paired with the elements of the other set, with no repeats and no leftovers. So my two sets from above have the pairs:
(1, a) (2, b) (3, c)
And since we've used all elements from both sets, I conclude that both sets have the same size. Notice here that we did absolutely nothing with counting. It does not matter how many pairs there are, just that I can pair with no repeats and no leftovers. The integers were not used at all for this pairing.
Is this definition ok with you?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: To infinity and beyond.
"We say that two sets have the same size if the elements of one set can be paired with the elements of the other set, with no repeats and no leftovers. "
Fine, I really like this definition, Ricky.
Let us pair the set {1,2,3,...} Set A to {2,3,4,...} Set B by the rule of adding 1 to each of the element of Set A perfect pairs, right? They have the same amount of elements, as definition.
Now let me show you something you won't find in a set theory book Define Set C = {1} U Set B Without counting Set C over, we know Set C has more element than Set B. Right? Since Set B is included in Set C, but no the other way around. It has one more, since the complimentary of {1} in Set C is Set B, and {1} has one element, do you agree?
Now we are facing a tough problem Set A and Set C appear all the same {1,2,3,...} except Set C has one more element than Set A or Set A has one less element than Set C
by transivity
X'(yXβ)=0
 Ricky
 Moderator
Re: To infinity and beyond.
Without counting Set C over, we know Set C has more element than Set B. Right? Since Set B is included in Set C, but no the other way around.
This is false, according to the definition that you "really like".
It has one more, since the complimentary of {1} in Set C is Set B, and {1} has one element, do you agree?
Again, no I do not. As in the definition, you can pair the elements in set C with those of set B with no left overs and no repeats. Therefore, they are the same size.
Now let me show you something you won't find in a set theory book
You're wrong if you think mathematicians don't talk about and in fact praise Hilbert's Grand Hotel, precisely the "problem" which you are stating.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: To infinity and beyond.
"You're wrong if you think mathematicians don't talk about and in fact praise Hilbert's Grand Hotel, precisely the "problem" which you are stating."
Very good you mentioned Hilbert's Hotel. Because it is the first false proof that paved the way for false set theory.
I am thinking of a way to better present how it went wrong to you, since it has alrealdy taken so deep root.
X'(yXβ)=0
 George,Y
 Super Member
Re: To infinity and beyond.
Okay, here it is how it went wrong:
By traditional textbook pairing method, it seems possible to pair each and every element of Set B to corresponding one of Set C
{1, 2, 3, 4, ...}     ... {1, 2, 3, 4, ...}
But the question is have the pairing exhausted Set B or Set C?
The answer is not. When one tries to do the traditional pairing, they actually are doing inductive counting: 1 to 1  1 pairs 1 to 1 & 2 to 2  2 pairs 1 to 1 & 2 to 2 & 3 to 3  3 pairs Good?
1 to 1 & 2 to 2 & 3 to 3 & ... & N to N  N pairs So that the same to above & N+1 to N+1  N+1 pairs
Such inductive method seems powerful, but it underestimates the scale of the infinite sets. Let us admit bluntly that either Set B or Set C has an amount of elements that is larger than any N you name, N as an integer
How can you finish pairing all elements of them with only N, O, P, Q, ... pairs? I mean, no matter how far your counting and pairing go, they only appear to pair the elements over, but actually they fail to cover the overwhelming amount of elements in either Set B or Set C
Take it in another way, when you draw the graph of 1/x^2, you draw it uprising against y axis, and you don't draw it intersecting y axis, because you know getting larger and larger doesn't mean reaching an infinite amount that exceeds any number, don't you? Same applies here.
The pairing is not complete. And this is why the proof is false. (this false proof fooled numerous folks to accept one set and its subset can have the same amount of elements just because they cannot prove it wrong)
Now let me show you some real proof. Just changing the method a little, we strictly define the pairing as complete. By complete pairing, I define: If one set A pairs to another set B completely, and it pairs to a third set C completely, which doesn't equal to B, set A cannot pair to set B U set C.
Take it in simpler form: A set cannot completely pair to its real subset.
This is not hard at all, anyone would think it makes sense unless they are fooled by Hilbert's Hotel. After all, it took a long time for people to accept Hilbert's Hotel after they failed to find an angle to prove it wrong.
But there is an angle, now. Let us move on, By simple, straightforward complete pairing definition, we know/assume (whatever you call it) Set A is different than Set C And here is something interesting about it we cannot find which one element is different by traditional pairing, 1 to 1, 2 to 2, 3 to 3, ... all fine?
But wait, we know there is difference already, shall we give up searching just because we cannot go to the end of the cave? Come on, let's try a different approach:
We know there is at least one element that Set C has and Set A doesn't that makes Set C one element larger than Set A And we pair them all up to find out who For convenience, we name the missing element X, and it pairs to "" to Set A U {} Suppose both A and C are ordered set from small to large. We know X is somewhere in C, and we would like to put {} in A U {} to in a place corrsponding X that all the elements before X in C and all the elements before  in A U {} are the same
So all the elements before are {1, 2, 3, ...X1} Now, we have {1, 2, 3, ...X1} U {} in A U {} and {1,2,3,...X1, X} in C Here we go Can there be an element X+1 in both A and C? if so , we will have {1, 2, 3, ...X1} U {} U {X+1} in A remember, A does not include X and it is absurd and contradictory So there is no X+1
Hence we have proved the missing element is at the right end, the largest one in A is X1 and the largest one in C is X Also X and X1 is the total amount of elements in Set C and Set A X=∞ where ∞ > any N, N is any number one can name
And here we are not over yet if X = ∞ X 1 = ∞ X 2 = ∞ X/2 = ∞ X/N=∞
from the other direction
1<∞ 2<∞ 100000000000000000000000000000000<∞ 2^100000000000000000000000000000000<∞
There is no way for ∞ to transit to some finite integer Yeah C={1,2,3,...}U{...,X3,X2,X1,X}
(My Paradox)
And do you still now believe it is a set of "all integers"
The only possible logic conclusion is there is no such "all integers" thing The notion of {1,2,3,...} or {1}U{1,2}U{1,2,3}U... is a self contradictory definition it is like a definition of relationship < and > and = or a definition of A & B, where A>B as well as B>A
Astonished? You may be suprised to find there is actually no empirical evidence of "all integers"
When a poet looked upon the night sky and counts: "one star, two stars,...countless stars!" we call it romantic
But when a mathematician counts "one, two, three, ... and theeeeeeeeeeeeeeeen, infinite of them!"
X'(yXβ)=0
 Ricky
 Moderator
Re: To infinity and beyond.
The pairing is not complete. And this is why the proof is false. (this false proof fooled numerous folks to accept one set and its subset can have the same amount of elements just because they cannot prove it wrong)
Name an element from set B or C that is not paired.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: To infinity and beyond.
The element at the right end in C (The largest infinity in C)
in this special case, it is also Amount(C)
X'(yXβ)=0
Re: To infinity and beyond.
 Ricky
 Moderator
Re: To infinity and beyond.
George,Y wrote:The element at the right end in C (The largest infinity in C)
in this special case, it is also Amount(C)
Please prove that there exists a "right end in C".
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 George,Y
 Super Member
Re: To infinity and beyond.
Please reread post 30, thanks.
btw: I know the Hilbert Hotel tale, and post 30 is just a refutal on it.
X'(yXβ)=0
 George,Y
 Super Member
Re: To infinity and beyond.
It reminds me of a bilogical experiment.
The existence of bacteria
At first people do not know bacteria.
People simply didn't see it  like you don't see or induce the last element
Then Paster evacuated a glass bottel containing soup.
His soup went bad much slower than usual.
He then concluded no matter anyone could see it or not, the logic is there must be something in the soup that lives with the air and make the soup bad.
And the same applies to my proof in Post 30. The last element exists by logic induction, no matter the mathematical induction can find it or not. If it cannot find it, it is the method's fault, not that the element does not exist.
X'(yXβ)=0
 George,Y
 Super Member
Re: To infinity and beyond.
Hilbert Hotel is a typical false proof by ambiguilty,
let me simplify it a little, I think this is the original version
The hotel rooms are filled up with guests, 1, 2, 3, 4, ... the hotel room numbers corresponds to "all" of natural numbers.
And then a new guest comes in.
No problem, there is a new room for you. guest in room 1 goes to room 2, ... now the room 1 is available.
Neat? Wait
Back to the beginning, what does it mean by they are all full? That means there was not a single room empty The amount of guest groups=the amount of hotel rooms To simplify the terminology, assume one room accomodates at most 1 guest and now
the amount of guest=the amount of hotel rooms (1)
How can a manipulation makes a room empty and the equation
the amount of guest +1 = the amount of hotel rooms? (2)
This is self contradictory, at least.
To justify it, the ambiguous guestmoving technique is employed: 12 23 34 wait
In this process, the amount of guest still equals to the amount of rooms The only improvement is that some one room has squeezed in 2 people
If you are telling me, Ricky, you can make up an empty room by putting a guest into another guest's room, I absolutely agree. But may I ask: when can you even out the guests to make sure each room has only one guest and still a room is emptied?
I bet you couldn't do that through out the processnot during the moving process as above reason, not after the moving process either. Because for all the guests as a whole, they need exactly the same amount of rooms as before if they don't share. The identity (1) must hold, the law of conservation of guests and the law of conservation of hotel rooms ensures you cannot arrive at identity 2, forever.
If anyway, the law of conservation of hotel rooms is allowed to break by this condition: another room can be added if there is every room has changed its host (although they don't guarantee the room's condition  it might be the janitory's room)
Definitely this room is added at last
Finally, eventually, there is a new room for the new Sir/Madam But could you tell me
What is the number of the new room? Now I reveal a secret room in this case And this room is Room Infinity, The Room at last, The Room with the largest number(quasinumber?) on the door Trinity!!!
Don't wanna believe it? But it is true. Admitting set {all the natural numbers} has a secret member Infinity isn't the end of the world, Ricky. Or perhaps it is yours, but it is not mine. And I sincerely wish you could walk out of this psychological obstacle.
X'(yXβ)=0
 Ricky
 Moderator
Re: To infinity and beyond.
George, you keep posting about this "element at infinity". Prove it exists, or you have no claim to its existence.
"That which can be asserted without proof, can be dismissed without proof."
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
