Mathematicians Measure Infinities, Find They’re Equal
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Mathematicians Measure Infinities, Find They’re Equal
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Re: Mathematicians Measure Infinities, Find They’re Equal
#2[deleted]
Re: Mathematicians Measure Infinities, Find They’re Equal
#3I find Cantor's diagonal argument unconvincing.
The claim is that there are more real numbers in the range from zero to one than there are natural numbers. To see that this is false simply realize that you don't actually have to write a decimal point to specify the real numbers in this range. Without the decimal point these real numbers just become natural numbers. Can a rational person believe that there are infinite sequences of digits in the form of real numbers but not infinite sequences of digits in the form of natural numbers? The natural numbers are just an infinite sequence of finite numbers. If you believe n is a natural number then you must also believe that n*10 is a natural number. One more digit! There is always one more digit (that is what infinity implies). If there really are an infinite number of natural numbers then some of them must be of a transfinite number of digits or else you would be including numbers in the list more than once.
The problem with Cantor's argument comes down to the fact that the procedure he uses to find a number not in the set is essentially the same as the procedure he uses for creating the infinite set in the first place. The only difference is our understanding of randomness. His procedure for finding a number not in the set may not seem very random but it might be as random as any other. A truly random coin could theoretically come up heads every time. The important part of his argument is that the infinite list of real numbers has no repeats. The diagonalization procedure similarly ensures that there are no repeats. On the one hand he claims the infinite set of real numbers exists. On the other hand he argues that the diagonalization that yields a number not in the set has not already been done. He takes away infinity and then gives it back!
There is only one infinity. It means "repeat". It is simply the interplay of finite state with process. You can think of it as an "infinite loop" in programming. To say that one infinity is smaller than another is to deny that the smaller is infinite. Infinite means without bound.
Re: Mathematicians Measure Infinities, Find They’re Equal
#4I find Cantor's diagonal argument unconvincing. The claim is that there are more real numbers in the range from zero to one than there are natural numbers. To see that this is false simply realize that you don't actually have to write a decimal point to specify the real numbers in this range. Without the decimal point these real numbers just become natural numbers. Can a rational person believe that there are infinit…
So if I understand you correctly, you find it unconvincing, and therefore generations of mathematicians who study these things must all be wrong.
Perhaps you simply don't understand the argument in detail, and are relying on your intuition. And perhaps your intuition is faulty.
Which seems more likely?
So let me try to provide a better insight for you. Consider the collection of natural numbers, including 0. Call it N. We all agree that N, also described as the set of non-negative integers, is infinite.
Now imagine flipping a coin at time t_0, t_1, t_2, etc. If you're worried that this will take infinite amounts of time, we can suppose that each flip - because we are practised - takes half the time of the previous flip, so all the flips can be done in finite time. However, we're in the realm of Pure Mathematics now, chasing the puzzle for its own sake, and not worrying about practicalities.
So what might the result be? Well, you might get all heads, you might get all tails, you might get alternating heads and tail, in practice, of course, you'll get something that looks random.
Let's think about all the possible results of flipping the coin. All possible results. Let's let F be the set of all possible results obtained from flipping the coin are each of t_0, t_1, t_2, and so on. We can think of F as functions from N to {H,T}.
So we have F and N. Let's wonder if it's possible to have a function from N to F that hits every element of F. Suppose we can.
So we have m:N -> F, and for every f in F, there is an n in N such that m(n)=f.
Do you think that's possible? Because hundreds of thousands of mathematicians say that it's not possible.
Re: Mathematicians Measure Infinities, Find They’re Equal
#5I find Cantor's diagonal argument unconvincing. The claim is that there are more real numbers in the range from zero to one than there are natural numbers. To see that this is false simply realize that you don't actually have to write a decimal point to specify the real numbers in this range. Without the decimal point these real numbers just become natural numbers. Can a rational person believe that there are infinit…
So if I understand you correctly, you find it unconvincing, and therefore generations of mathematicians who study these things must all be wrong. Perhaps you simply don't understand the argument in detail, and are relying on your intuition. And perhaps your intuition is faulty. Which seems more likely? So let me try to provide a better insight for you. Consider the collection of natural numbers, including 0. Call it…
It is possible. The inverse of m is called q. The function q takes an infinite sequence of coin flips and one by one changes every heads to 1 and every tails to 0. The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation. This will not take forever because each change will only take half as long as the previous one. A transfinite natural number can exist since there are an infinite number of natural numbers. If we stop at 1-bit numbers then the natural numbers are not an infinite set. Likewise, if we stop at 2-bit numbers then the natural numbers are not an infinite set. Our only option is to concede that transfinite natural numbers do actually exist and that they can be put into correspondence with all sequences of coin flips.
Hundreds of thousands of mathematicians are wrong.
Re: Mathematicians Measure Infinities, Find They’re Equal
#6I find Cantor's diagonal argument unconvincing. The claim is that there are more real numbers in the range from zero to one than there are natural numbers. To see that this is false simply realize that you don't actually have to write a decimal point to specify the real numbers in this range. Without the decimal point these real numbers just become natural numbers. Can a rational person believe that there are infinit…
[I'll try a non technical argument to convince you. It's also not a complete argument, so you must think about this for a while.]
> If you believe n is a natural number then you must also believe that n x 10 is a natural number. One more digit!
If you interpret the natural number in this way, the important property is that they have only a finite amount of "interesting" digits. Almost all their digits are zero.
You can define the set of number that have a finite amount of non zero digits. Let's call them the "Very Boring" numbers.
The set of the "Very Boring" numbers is infinite, but it's the same infinite that the Natural numbers.
You can extend this set to include the periodic numbers, for example 46.2222222222222... and 462.2222222222222... and 4622.2222222222222... ... Let's call them the "Boring" numbers. You still get the same infinity.
You can also 71.3535353535... and 713.5353535353... and 7135.3535353535... and all the periodic numbers. Now you have the "rational" numbers. You still get the same infinity.
The Cantor's diagonal argument fails with Very Boring, Boring and Rational numbers. Because the number you get after taking the diagonal digits and changing them may not be Very Boring, Boring or Rational.
--
A somewhat unrelated technical detail that may be useful:
Most of the times you don't prove that the cardinal of real numbers between 0 and 1 is a bigger infinity than the cardinal of the natural numbers.
I's much easier to consider the infinite strings of digits like "0.765653625367523765..." or "0.5265362556..." or "0.000073468763478..." and also the one with repetitions like "0.0006767000000..." or "0.0072257822222222...". This is essentially a copy of the real number, but in this copy "0.2999999999999..." is different from "0.300000000000000..."
This trick makes much easier to prove that the diagonal ´+1 in each one digit is not in the list. Then it's possible to fix the details and use the real numbers instead of the infinite strings of digits.
Re: Mathematicians Measure Infinities, Find They’re Equal
#7Earlier quoted context omitted.
So if I understand you correctly, you find it unconvincing, and therefore generations of mathematicians who study these things must all be wrong. Perhaps you simply don't understand the argument in detail, and are relying on your intuition. And perhaps your intuition is faulty. Which seems more likely? So let me try to provide a better insight for you. Consider the collection of natural numbers, including 0. Call it…
It is possible. The inverse of m is called q. The function q takes an infinite sequence of coin flips and one by one changes every heads to 1 and every tails to 0. The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation. This will not take forever because each change will only take half as long as the previous one. A transfinite natural numbe…
So I asked: is it possible to have m:N -> F such that for every f in F, there is an n in N such that m(n)=f?
Your reply says yes, but then your construction does not do it. In particular you said:
> The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation.
But the set N does not have transfinite natural numbers, so q does not map F to N, it maps F to something else.
So I ask again, is it possible to have m:N -> F, such that for every f in F, there is an n in N such that m(n)=f?
Re: Mathematicians Measure Infinities, Find They’re Equal
#8I find Cantor's diagonal argument unconvincing. The claim is that there are more real numbers in the range from zero to one than there are natural numbers. To see that this is false simply realize that you don't actually have to write a decimal point to specify the real numbers in this range. Without the decimal point these real numbers just become natural numbers. Can a rational person believe that there are infinit…
[I'll try a non technical argument to convince you. It's also not a complete argument, so you must think about this for a while.] > If you believe n is a natural number then you must also believe that n x 10 is a natural number. One more digit! If you interpret the natural number in this way, the important property is that they have only a finite amount of "interesting" digits. Almost all their digits are zero. You c…
>I's much easier to consider the infinite strings of digits like "0.765653625367523765..." or "0.5265362556..." or "0.000073468763478..." and also the one with repetitions like "0.0006767000000..." or "0.0072257822222222...". This is essentially a copy of the real number, but in this copy "0.2999999999999..." is different from "0.300000000000000..."
This trick makes much easier to prove that the diagonal ´+1 in each one digit is not in the list. Then it's possible to fix the details and use the real numbers instead of the infinite strings of digits.
What am I missing?
"0.765653625367523765..." could be assigned the transfinite natural number beginning "1765653625367523765..."
"0.000073468763478..." could be assigned the transfinite natural number beginning "1000073468763478..."
Transfinite natural numbers must exist otherwise you do not have an infinite set.
Re: Mathematicians Measure Infinities, Find They’re Equal
#9Earlier quoted context omitted.
It is possible. The inverse of m is called q. The function q takes an infinite sequence of coin flips and one by one changes every heads to 1 and every tails to 0. The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation. This will not take forever because each change will only take half as long as the previous one. A transfinite natural numbe…
So I asked: is it possible to have m:N -> F such that for every f in F, there is an n in N such that m(n)=f? Your reply says yes, but then your construction does not do it. In particular you said: > The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation. But the set N does not have transfinite natural numbers, so q does not map F to N, it ma…
>But the set N does not have transfinite natural numbers, so q does not map F to N, it maps F to something else.
How many natural numbers are there? How many bits does it take to represent the average natural number? If you believe the natural numbers do not include transfinite numbers then how do you pick a successor when counting? There are infinite picks to be made so some of the picks must be transfinite. What I am calling a transfinite natural number must exist in N because N is an infinite set.
Assume that N has only finite numbers in it but is itself an infinite set. Would you care to tell me which number (or numbers) are listed twice? But then it is not really a set!
Re: Mathematicians Measure Infinities, Find They’re Equal
#10Earlier quoted context omitted.
So I asked: is it possible to have m:N -> F such that for every f in F, there is an n in N such that m(n)=f? Your reply says yes, but then your construction does not do it. In particular you said: > The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation. But the set N does not have transfinite natural numbers, so q does not map F to N, it ma…
>But the set N does not have transfinite natural numbers, so q does not map F to N, it maps F to something else. How many natural numbers are there? How many bits does it take to represent the average natural number? If you believe the natural numbers do not include transfinite numbers then how do you pick a successor when counting? There are infinite picks to be made so some of the picks must be transfinite. What I…
I think you are confusing "arbitrary long" with "infinite".
The leading/trailing zeros are confusing. So many digits are confusing. Let's pick an easy model: The "naïve roman numbers" (This is a made up name, not a technical name.)
Let's consider the set that has
I
II
III
IIII
IIIII
IIIIII
...
IIIIIIIIIIIIIIIIIIIIIIII
IIIIIIIIIIIIIIIIIIIIIIIII
IIIIIIIIIIIIIIIIIIIIIIIIII
...
...
i.e. all the strings that have a bunch of "I" characters.
Each element has a finite amount of "I". If you delete one of them you get a shorter string/number.
The set has strings that are as long as you want. If you print them in a 8pt times new roman font, you can pick one of them that will be long enough to wrap around the Earth, one of the is long enough to reach the Moon, ...
This is an infinite set, where each element has a finite amount of "I".