keiths' infinity "proof"

More like poof
N= {0,1,2,3,4,...,Ln}
SLn finite subset of N that is also an improper subset.
keiths flails away:
Hey asswipe, you cannot use that which you are trying to prove as proof. No one can prove N is infinite. If N is the set of natural NUMBERS and infinity is NOT a number, then it doesn't belong in the set. However the largest number does belong in the set largest number Ln, ie greatest element.
And keiths, if Ln = largest number, then Ln + 1 is still the largest number.
SLn is a finite subset of N. Ln= largest number {0,1,2,3,4,...,Ln} the Ln is INSIDE of the brackets keiths. It is the largest NUMBER. So SLn would be the name of the finite set that starts at 0 and goes to the largest number. And the set of natural numbers is also the set that starts at 0 and goes to the largest number.
More like poof
1.Definition: Set A is infinite if every finite subset of A is also a proper subset of A.That just moves the problem back to defining "finite". But anyway...
2. Proof that infinite sets exist:
a. Assume that infinite sets do not exist. By the definition above, that means that every set has at least one finite subset that is not a proper subset.
OK.
b. Consider all finite subsets S1, S2, etc. of the set N of natural numbers. Every such subset contains a greatest element Ln (if there were any elements greater than Ln, then one of them would be the greatest element instead of Ln).
OK
c. Ln is a finite natural number. Therefore Ln+1 exists and is also a finite natural number.
Ln + 1 would be the new Ln. You just covered that in "b".
d. Ln + 1 is not a member of Sn, but it is a member of N.
No, it would be a member of Sn. Ln+1=Ln, just as YOU said in "b"
Therefore every Sn is a proper subset of N, thus contradicting a).
Not if one follows YOUR directions.
OK, do it for SLn.
SLn= {0,1,2,3,4,...,Ln}
SLn + 1 = {0,1,2,3,4,...,Ln}
OK, do it for SLn.
SLn= {0,1,2,3,4,...,Ln}
SLn + 1 = {0,1,2,3,4,...,Ln}
N= {0,1,2,3,4,...,Ln}
SLn finite subset of N that is also an improper subset.
keiths flails away:
The only improper subset of N is N itself, and N is infinite, not finite.
Hey asswipe, you cannot use that which you are trying to prove as proof. No one can prove N is infinite. If N is the set of natural NUMBERS and infinity is NOT a number, then it doesn't belong in the set. However the largest number does belong in the set largest number Ln, ie greatest element.
And keiths, if Ln = largest number, then Ln + 1 is still the largest number.
SLn is a finite subset of N. Ln= largest number {0,1,2,3,4,...,Ln} the Ln is INSIDE of the brackets keiths. It is the largest NUMBER. So SLn would be the name of the finite set that starts at 0 and goes to the largest number. And the set of natural numbers is also the set that starts at 0 and goes to the largest number.
24 Comments:
At 1:15 PM, socle said…
Joe,
Suppose S7 = {1, 3, 6, 11}. Then L7 = 11, the largest element of S7. But L7 + 1 = 11 + 1 = 12 is a natural number which is not a element of S7.
The conclusion is that S7 is a proper subset of the natural numbers, N.
You can do this for any Sn. Therefore any finite subset of N is a proper subset of N.
By keiths' definition of infinite sets, this means N is infinite.
At 2:49 PM, Joe G said…
OK, do it for SLn.
SLn= {0,1,2,3,4,...,Ln}
SLn + 1 = {0,1,2,3,4,...,Ln}
That's a finite subset of N that is also an improper subset.
At 5:01 PM, socle said…
Joe,
No one can prove N is infinite.
vs.
We count and count and count FOREVER.
XD
At 6:11 PM, Joe G said…
Unfortunately no one can count forever.
At 8:25 PM, socle said…
And the set of natural numbers is also the set that starts at 0 and goes to the largest number.
What's this 'largest number'? Each natural number has a successor. As you said the other day, given any n in N, you can just add 1 to it to get a larger number.
At 9:18 PM, Joe G said…
What's this 'largest number'?
The top number in the count. No one told me I was in charge of keeping track of it. As I said I was under the impression there was a computer that was counting away.
As you said the other day, given any n in N, you can just add 1 to it to get a larger number.
Yeah, it's called "counting by ones." Sooner or later you get to unchartered territory where no one has counted before. A number so ridiculously large that just to properly pronounce it would take a year. So large that it might as well be infinity as it satisfies all equations in which infinity is used and all the paper in the world could not hold the number.
In a "finite" set with that as the Ln, the cardinality of the set of nonnegative integers is twice that of the cardinality of positive even integers.
At 12:36 AM, Rich Hughes said…
"In a "finite" set with that as the Ln, the cardinality of the set of nonnegative integers is twice that of the cardinality of positive even integers"
Uhhuh.
At 1:19 AM, socle said…
Yeah, it's called "counting by ones." Sooner or later you get to unchartered territory where no one has counted before. A number so ridiculously large that just to properly pronounce it would take a year. So large that it might as well be infinity as it satisfies all equations in which infinity is used and all the paper in the world could not hold the number.
I'm not sure what difference it makes what numbers have been counted before, though. Let's just say for the sake of argument that there is a supercomputer deep inside a mountain in Colorado counting by ones, and meanwhile I am doing math here in my living room. If I'm doing a calculation that results in a number that this computer has not counted yet, is that a problem?
It's very easy to concoct very large natural numbers that no computer could ever count to by ones, even if they ran from now until the heat death of the universe. If I make up such a number, the size of the set of natural numbers doesn't increase, does it? We already knew that that number is somewhere in the list (1, 2, 3, 4, ...).
At 7:26 AM, Joe G said…
I'm not sure what difference it makes what numbers have been counted before, though.
I never said it made a difference. I was just setting up a scenario. A scenario that should tell you that Cantor's concept is nonsense.
At 7:42 AM, Joe G said…
Richie doesn't seem to grasp the fact that what is good enough for my Ln is also good enough for infinity...
At 9:10 AM, socle said…
I never said it made a difference. I was just setting up a scenario. A scenario that should tell you that Cantor's concept is nonsense.
I'm still not convinced of that. Here's the thing I believe you are missing: The natural numbers is not just a very large set that is somehow growing as the "LKN" marches forward. All the positive integers are already in the set.
Consider this related puzzle/experiment: You roll a single fair 6sided die until the first "4" comes up. Then you write down the number of rolls it took. How many distinct outcomes are there for this experiment? If you prefer, you can assume that in each run of the experiment, it will only take a finite number of rolls to get a "4".
The set of outcomes is infinite, no?
At 9:17 AM, Joe G said…
The natural numbers is not just a very large set that is somehow growing as the "LKN" marches forward. All the positive integers are already in the set.
Then what is the highest integer in that set? Infinity is not a number.
The set of outcomes is infinite, no?
I'm not convinced of that. :)
At 9:27 AM, socle said…
Then what is the highest integer in that set? Infinity is not a number.
There is no highest integer in the set. If someone claims than n is the highest integer in N, we can refute that by producing n + 1, which is also in N.
I'm not convinced of that. :)
Heh. If it's not infinite, then it's finite, by definition. That would mean that the set of outcomes has a largest element, say K (of course we know the set of outcomes is nonempty).
Do you know of a fixed number K, such that if you roll a die K times, you always will get at least one "4"?
At 10:07 AM, Joe G said…
I know that no one can roll any die an infinite number of times.
I know that no one can count forever.
Do you know of a fixed number K, such that if you roll a die K times, you always will get at least one "4"?
100. I have always had a 4 well before reaching 100, so 100 is a definite lock.
At 10:44 AM, socle said…
For the purposes if this experiment, why don't we assume that we're in heaven, or maybe that we are riding on Einstein's Train, which in principle can run forever.
100. I have always had a 4 well before reaching 100, so 100 is a definite lock.
Here's what I found: If we give 300 million people a fair die (roughly the population of the US), and ask them to roll their die 100 times, the probability that at least one person will not get a "4" is about 0.97, or 97%. It's certainly possible I have made an error, so please correct me if necessary.
At 11:11 AM, Joe G said…
Do it. You haven't found anything until you actually do it.
Maybe Vegas could have a contest come to Vegas, roll a die 100 times and if you do NOT roll a "4", you will win millions of dollars.
At 1:31 PM, socle said…
Do it. You haven't found anything until you actually do it.
Does this rule also apply to ID experts such as KF, who routinely carry out armchair probability calculations such as this to support their claims?
The point is that there is no number which is an absolute "lock", so the outcomes for this experiment comprise an infinite set.
Maybe Vegas could have a contest come to Vegas, roll a die 100 times and if you do NOT roll a "4", you will win millions of dollars.
Sure, why not? Your chances of winning this dice game are a bit better than twice your chances of winning the powerball lottery jackpot on a single ticket, if my numbers are right.
At 1:44 PM, Joe G said…
LoL! That KF et al., even grant your position a place at the probability table, is more than it deserves.
And perhaps 100 is too few. But I would love to see it played out.
And if there is anyone left after 100 let them keep going until they hit a 4. I bet they won't get close to infinity...
At 2:15 PM, socle said…
And perhaps 100 is too few. But I would love to see it played out.
If 100 is too few, meaning that there is a nonzero chance of rolling 100 times with no 4's, then 101 is also too few, since the probability just decreases by a factor of 5/6 for each additional roll. Then 102 is too few, and so on.
You will never find a number which is truly a "lock".
If you do want to see it played out, perhaps Gary Gaulin could set up a computer simulation, Inshallah.
At 3:42 PM, Joe G said…
I have yet to get beyond 10 rolls without hitting a 4.
And computer simulations are a different ballgame. When you are rolling a die you cannot exactly duplicate each and every physical movement there are intangibles involved.
At 5:14 PM, socle said…
And computer simulations are a different ballgame. When you are rolling a die you cannot exactly duplicate each and every physical movement there are intangibles involved.
That sounds much more ambitious than what I had in mind. The experiment assumes a fair die, which means each of the 6 numbers has an equal chance (1/6) of coming up. You could model that very well with a pseudorandom generator without getting into the physics of rolling dice.
At 7:28 AM, Joe G said…
Again there are physical parameters that cannot be matched by a computer simulation.
At 3:20 PM, socle said…
Again there are physical parameters that cannot be matched by a computer simulation.
I suppose, but we don't need to keep track of all that. Just like with Einstein's Train, we didn't need to know all the physical parameters such as the number of cars or the width of the track.
At 6:55 AM, Joe G said…
I never said anyone has to keep track of all that. It's just that the physical parameters are part of rolling a die.
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