Different Infinities-Series on Infinity Part 6

[tomcuchta]

The continuum does not have cardinality aleph-one unless you admit the continuum hypothesis.

[iateyourgranny]

Correction: the first number is assigned to 1, the second number is assigned to 2, etc. There is no "diagonalization" done here. What exactly is your point of assigning "1" to 1, "00" to 2, "011" to 3, etc? Get your method right before you set out to prove something.

[Thetarget1]

I've got a question: Does this mean that there is an infinite number of infinities? And are there then different kinds of infinite series of infinities, like there are different kinds of infinite series of numbers?

[ForeverWiked]

So why are one of them 'countable' but the other is not?

The argument of one was that you can always make another number by changing a digit. Well you can always change a real number by adding 1 to it, so in what way is it countable

[srfriggen]

So in the end you created a transfinite number, is that right? Is there no way to set up a scenario where this type of argument was repeated and repeated essentially creating (and counting) every transfinite number?

[tommyrjensen]

The explanation of the continuum hypothesis is not correct. What CH actually states is the equality of c and aleph1. Where c can be defined as 2^aleph0 and aleph1 can be defined as the smallest cardinal which is bigger than aleph0 (once one has a proof that such exists). Cantor may have driven himself crazy by trying to prove c=aleph1, sometimes thinking that he had proved it, sometimes that he had disproved it.

[rezkogitans]

@Davidson1956: You write "The cardinality of the real numbers, aleph one, is probably the next largest infinity".

No. Aleph one, by definition, is necessarily the next large infinity.

[StitchedMusik]

I don't understand. Isn't this the same as aleph-null?

[fonsidream]

It is a partial hypothesis to my senses.

Why not consider all possibilities like 00, 01, 10, 11 and then prove the hypothesis?

[balachandra108]

All I got to say is: thank you! You made this very complex idea simple to understand.

[alwaystheantagonist]

I'm familiar with Cantor's diagonal argument (though I am definitely not claiming to know more than you) and I think I understood it differently. Not meaning to antagonize you or anything but aren't you supposed to treat that first string of 0s and 1s as just one element of a Set (as that first string, correct me if i'm wrong, is just one number)? That said, applying the 1 to 1 correspondence of the DIGITS of the strings (2:17) to the Set of Counting Numbers seems to me to be incorrect.

[danishdidgeridoo]

Thank you for a great series of videos. They've been incredibly helpful in my current math/philosphy project dealing with infinity.

Regards from Denmark.

[intermender]

Speaking as a non-mathematician and layman, i found this series of vids amazing and wonderful,a gently progressive introduction, yet suprisingly effective in conveying a level of mind-boggling, eye-widening stuff i never thought i'd have access to in my lifetime.

And so elegantly clever! Definently food for thought and followup.

Thank you very much for this most accessible insight into the maths world. Its really heartwarming to find such quality contemplation and springboard given for free.

[DavidsonSSCC]

Thank you for taking time for these very kind comments!

[intermender]

you're welcome and thank you for your time too. I am currently trying to figure out more maths now i have a taste for it and enjoying it very much.

we need a carl sagan of mathematics and you should apply for the job.

[nowenterpsie]

Get your precepts out for all to see:

Mathematicians such as this seem to treat Infinity as a 'number', or a 'size' , and this is the basic and fundamental flaw in all of this logic. Infinity is a non-number, closely related to Zero or None. By definition a 'number' or any 'size' is bounded by finite-ness.

We are waking up.

[ILoveYouToThe9999999]

My function is both injective and surjective.

[x1101011x]

Suppose (x_1, x_2, ..., x_k) are positive rational numbers and p is a positive irrational number such that (x_1)^p + (x_2)^p + ... + (x_k-1)^p = (x_k)^p. Let r be any positive rational number. Multiplying both sides by r^p yields (r * x_1)^p + (r * x_2)^p + ... + (r * x_k-1)^p = (r * x_k)^p.

Therefore, (r * x_1, r * x_2, ..., r * x_k) is another k-tuple of positive rationals that maps to the same element p, and the function is not injective.

[ILoveYouToThe9999999]

I know but that is not my function. What i meant was that the function is injective and surjective with the set of natural numbers and not with the k-tuple.

[x1101011x]

Here's a proof that your mysterious function doesn't exist:

There are uncountably many real numbers, by Cantor's diagonal argument. There are countably many rational numbers by Cantor's enumeration. The irrationals are S = R \ Q and have cardinality |S| = |R| - |Q|, which is uncountable because, if |S| were countable, then R = S U Q would be countable, a contradiction.

If you're going to try to convince us that Mathematics is inconsistent, I'm going to actually have to see your function...

[x1101011x]

6:30

2^aleph0 = aleph1 is equivalent to the continuum hypothesis, which Cantor most definitely did not prove.

The cardinality of the natural numbers is aleph0 = beth0. The cardinality of real numbers is 2^aleph0 = beth1.

The aleph numbers are the different sizes of infinity. The beth numbers are the size of the power sets of the previous size, starting from countably infinite. beth(i) = 2^beth(i-1).

The generalized continuum hypothesis is that aleph(i) = beth(i).

[ILoveYouToThe9999999]

My function is both injective and bijective.

[musicappreciate]

What is the philosphical problem with one

monolithic infinity. You can always conceive of adding more elements to a set.

[sean0101a]

musicappreciate, yes you could always add more elements to a set. But that would be a countable set, { 1,2,3,4,....,infinity }. There are uncountable set of size aleph-1 which are bigger than the natural numbers.

[sherlockfury]

so... you can create various decimals into infinity, and not create all numbers; and those numbers that aren't created exist only in larger infinities?

So, are the subsets of larger infinities, which are infinitely large, also equal to the larger set in size (cardinality?)?

[Symbioticism]

these videos are very interestingand clearly explained thankyou

[ILoveYouToThe9999999]

The union of any number of countable sets is countable.

[x1101011x]

The union of any COUNTABLE number of countable sets is countable.

[ILoveYouToThe9999999]

Let P be the set of all irrational numbers p such that there exist nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2. Let Q be the set of all rational numbers. Then The set P is countable since the set (Q × Q × Q) ∪ (Q × Q × Q × Q) ∪ (Q × Q × Q × Q × Q) ∪ (Q × Q × Q × Q × Q × Q) ∪ ... is countable, where ∪ is the union sign.

[ILoveYouToThe9999999]

The probability that two distinct irrational numbers on the number line can be close enough to exclude any rational numbers is higher than the probability that two distinct rational numbers on the number line can be close enough to exclude any irrational numbers.

[Davidson1956]

To use probability to prove anything you would probably need to define a probability density function, prove it fits the situation, and then apply Chebyshev's Theorem.

Consider two arbitrarily close irrational numbers. One could find a sufficiently large integer N such that the product of N and each irrational produces two numbers with an integer in between them. Divide that integer by N and you've found the rational number between the two irrationals.

[ILoveYouToThe9999999]

The uncountability of the set of all irrational numbers means that the set of all irrational numbers is more numerous than the set of all rational numbers, and this means that there are at least two distinct irrational numbers on the number line such that there exist no rational numbers between them, but, this is a contradiction to what we know that between any two distinct real numbers on the number line there exists a rational number. Thus, the set of all irrational numbers is countable.

[Davidson1956]

Interesting argument, but the term "numerous" is vague or undefined in a mathematically precise way, too vague to claim that it would imply two irrational numbers could be close enough to exclude any rational numbers.

Consider that between any two irrational numbers there is a countable number of rational numbers, but between any two rational numbers there is an uncountable number of irrational numbers.

[ILoveYouToThe9999999]

Let S be the subspace of the Euclidean space R of all real numbers such that S consists of all the points p such that p is an irrational number and such that there exist nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2. Then for every point p that belongs to S, p is not an isolated point of S.

[ILoveYouToThe9999999]

There exists at least one irrational number p such that there exist no nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2.

If the above statement is false, then the set of all irrational numbers is countable.

[ILoveYouToThe9999999]

Let [a,b] be an arbitrary closed interval on the real-number line where a and b are distinct real numbers. Let P be the set of all irrational numbers p greater than a and smaller than b such that there exist no nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2. Then by the uncountability of the set of all irrational numbers, the set P is uncountable.

[ILoveYouToThe9999999]

Let [a,b] be an arbitrary closed interval on the real-number line where a and b are distinct real numbers. Let P be the set of all irrational numbers greater than a and smaller than b such that there exist no nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2. Then by the uncountability of the set of all irrational numbers, the set P is uncountable.

[ILoveYouToThe9999999]

Let P be the set of all irrational numbers p such that there exist nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2. Then the set P is countable.

[ILoveYouToThe9999999]

Let [a,b] be an interval where a and b are real numbers. Let P be the set of all irrational numbers greater than a and smaller than b such that there exist no nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2. Then by the uncountability of the set of all irrational numbers, the set P is uncountable.

[ILoveYouToThe9999999]

Let P be the set of all irrational numbers that have the property: If p is an irrational number, then there exist nonzero rational numbers x(1), x(2), ..., x(n-1) and x(n) satisfying the equation x(1)^p + x(2)^p + ... + x(n-1)^p = x(n)^p where n is an integer greater than 2. Then by the uncountability of the set of all irrational numbers, the set P is countable.

[dasklavierleben]

Haha! Man this is trippy.

I have studied the different orders of infinity before, but a thought just now occured to me. There are infinitely many orders of infinity, but that sort of infinity is countable. Therefore, the infinity of the orders of infinity is of the aleph null order. . . does that mean anything? Does that even make sense? I don't know. Just thought I'd share.

[ILoveYouToThe9999999]

I found a surjective function from the set of natural numbers to the set of irrational numbers, and this makes the set of irrational numbers countable.

[Davidson1956]

Please pardon my skepticism, but this would refute 140 years of well-established mathematics in the area of transfinite numbers.  What is your function, or how could you describe it?

[ILoveYouToThe9999999]

If n is a positive irrational real number, then the equation x^n + y^n = z^n has solutions in positive rational real numbers x, y and z.

[theboombody]

@Davidson1956 I haven't found anything like that. I only found a diagonalization argument showing that even whole numbers can form that aleph 1 set. Of course the whole numbers have infinite digits, but at least there's nothing to the right of the decimal point, so I still call them whole. You can see my video "Does Counting Exist?" for further information.

[x1101011x]

Unfortunately, you need more than a surjection to prove equal cardinality; you need a bijection.

For instance, let f : R -> N be such that f(x) is the greatest integer less than or equal to x. This function is surjective, but it does NOT prove that there are countably many real numbers because it is not injective. Many elements from the reals are mapping to the same natural number.

Each element in the domain must map to a single distinct element in the range to show equal cardinality.

[x1101011x]

I misread the direction in the original post; I was thinking the function was from the irrationals to the naturals.

Still, having surjectivity only guarantees that the range has the same or smaller number of elements. You can only say equality because countably infinite is the smallest infinite.

However, producing such a function would yield a contradiction and prove every theorem in the system. The irrationals would be both countable and uncountable, so I wouldn't really say they were equal.

[nortexoid]

Wow, another horrible video with awful misinformation, especially the part that claims that the proof of the independence of the continuum hypothesis is a consequence of Godel's incompleteness theorems! Learning is great, but spreading misinformation is better!

[Davidson1956]

There are those who produce things and then there are those who simply live to criticize. Why don't you edify us with your own video?

Better yet, look up the work of Gödel (1939-40) or the work of Paul Cohen (1963) related to the Zermelo-Fraenkel set theory.

I'll grant you this: Mathematicians are still wrestling with the Continuum Hypothesis. However, Gödel and Cohen did settle it within the standard ZF system.

[nortexoid]

I *have* looked it up. Can you give a sketch of the proof from Godel's incompleteness theorems? The mathemematical/logical community would like to see.

[Davidson1956]

I wish I could but I don't have that level of expertise in logic. The relatively few people who do probably don't do anything with YouTube. I have a bare bones understanding of Gödel's original 1931 theorem and I plan to produce a video on it over the Christmas break if I can find the time. My goal with YouTube is to present great math at a layman's level of understanding. It's been a fun challenge.

[robidogg]

ah my other comment didnt post. What I basically said was obviously we can't start with 0.00000......1 because we'de never get there, but what if we started with ( in binary)

0.1

Then 0.01, 0.11

..

0.001, 0.010, 0.011, 0.101, 0.111

..

Why is this not continuable ad infinitum?

[Davidson1956]

Great try at a counterexample! I had to think about this over lunch.

You are correct that the numbers you listed are countable. One could list these in a systematic way in a one to one correspondence with the counting nos. But these numbers are all terminating decimals, which makes them rational numbers, a subset of the fractions.

In Cantor's example he assumes that he has also listed all the irrational numbers in that binary form, those which do not terminate or have a repeating pattern.

[robidogg]

Ahhhhhhh yes. That makes sense. Theres such subtleties in proof isnt there! Thanks for spending the time to explain that though, Are you a maths prof. then? I'm due to embark on a maths degree in a couple of days, any advice!?

[Davidson1956]

I thought I had replied, but perhaps it didn't post.

Math is an overwhelmingly large subject which no human could possibly master any more. There are two sides, pure and applied. If you are drawn to math due to the great ideas and abstractions, then you'll be drawn to the pure side. And if you are inspired to share it with others through teaching then, like me, you'll end up with the greatest job you could possibly and reasonably want.

[robidogg]

An excellent set of videos, concise and clear but also far-reaching.

When I was thinking about the infinity of the reals being larger than the infinity of the intergers, I had one problem. I thought I could put the reals in a countable fashion. Please point out the flaw in this as I'm sure there must be one but I can't find spot it!

Suppose we start with 0.0000000000.

Obviously we can't start counting at 0.000000......................­..1 because we'de never get there, but what if we started

[Davidson1956]

Thanks for your kind comments about this video series!

I find it fascinating that there is no smallest positive number, as you basically have indicated. In calculus we sort of call it a differential, and in earlier days it was called a fluxion, but it's purely symbolic, not concrete.

Cantor's proof with the 1's and 0's establishes that the reals are uncountable since the set of all decimals with 1's and 0's is a subset of the reals.

[CogitoErgoCogitoSum]

How do you prove that the new diagonalized number isnt in the original list?

[Davidson1956]

The new number differs in at least one decimal place from every other number since every number on the list was used to construct it.

[sniperister]

Very nice series. I would like to see more videos on infinity and incompleteness theory. From what I have been told, Cantor drove himself crazy because he kept proving and then disproving his continuum hypothesis over and over as if his brain got stuck in an infinite loop.

[Davidson1956]

Thank you for the kind comments and your interest. I may add another video on the Cantor set, and then I'll do a few videos to tie up the loose ends from Zeno's Paradox and calculus issues arising from infinity and infinitesimal.

Gödel's Incompleteness Theorem is pretty amazing. I hope to do a video on it this fall, as well.

Poor Cantor did go nuts from his obsession. It didn't help that he was overly sensitive to critics such as Kronecker and was probably bipolar.