To infinity and beyond.

Ricky · 2009-07-09 10:28:48

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?

George,Y · 2009-07-09 17:47:43

"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

Ricky · 2009-07-09 18:08:05

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.

George,Y · 2009-07-17 20:33:37

"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.

George,Y · 2009-07-17 22:04:40

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, ...X-1}
Now, we have {1, 2, 3, ...X-1} U {-} in A U {-}
and {1,2,3,...X-1, 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, ...X-1} 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 X-1 and the largest one in C is X
Also X and X-1 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{...,X-3,X-2,X-1,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!"

Ricky · 2009-07-18 03:56:24

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.

George,Y · 2009-07-28 23:14:32

The element at the right end in C
(The largest infinity in C)

in this special case, it is also Amount(C)

soroban · 2009-07-29 08:44:38

.

. .

.

Ricky · 2009-07-29 16:57:06

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".

George,Y · 2009-08-22 01:39:20

Please reread post 30, thanks.

btw: I know the Hilbert Hotel tale, and post 30 is just a refutal on it.

George,Y · 2009-08-22 01:46:41

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.

George,Y · 2009-08-22 02:29:58

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 guest-moving technique is employed:
1-2 2-3 3-4 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 process-not 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(quasi-number?) 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.

Ricky · 2009-08-23 04:16:17

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."

Math Is Fun Forum

#26 2009-07-09 10:28:48

Re: To infinity and beyond.

#27 2009-07-09 17:47:43

Re: To infinity and beyond.

#28 2009-07-09 18:08:05

Re: To infinity and beyond.

#29 2009-07-17 20:33:37

Re: To infinity and beyond.

#30 2009-07-17 22:04:40

Re: To infinity and beyond.

#31 2009-07-18 03:56:24

Re: To infinity and beyond.

#32 2009-07-28 23:14:32

Re: To infinity and beyond.

#33 2009-07-29 08:44:38

Re: To infinity and beyond.

#34 2009-07-29 16:57:06

Re: To infinity and beyond.

#35 2009-08-22 01:39:20

Re: To infinity and beyond.

#36 2009-08-22 01:46:41

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#37 2009-08-22 02:29:58

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#38 2009-08-23 04:16:17

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