Aleph-Naught

RonRoss

New member
Joined
Sep 25, 2013
Messages
3
Is Alph Naught the smallest infinity? Initially everyone feels the first set of infinity they encounter is the set of natural numbers. Then speculation sets in and the discovery that many infinities are just that, infinity and do not obey the rules of everyday experience. Rational numbers are countable just like natural numbers and therefore, both infinities are deemed equal. Irrational numbers are not countable and therefore are definitely infinitly more of them, so they garner the term Alph Prime.

Now for my question. If Alph Naught is the smallest infinity (and it may not be, and then my question is moot!), why is that? Prime numbers are a smaller subset. Now I am easily convinced that even numbers = the same infinity as all natural numbers or even that rational numbers also = the same as the natural numbers in being equal in number to infinity due to countability.

However, primes are not technically countable (or are they and I am just unaware of any new break through in math. And if they are NOT countable, I have no one to one correspondence with natural numbers. And as we know, primes are definitely a subset of rational numbers. (and I just stated earlier that I accept that even (or odd or any repeating ratio) is equal to the natural numbers in size @ infinity) The number of primes is approximated by the following function #/(ln #) for any number ie 100/ ln 100 = 21 primes between 1 & 100. This is merely an approximation and not an exact formula and this function merely approximates a discreet value and cannot impose a continuous feature to this function. I state this to dispel the circle within a circle argument ie one set is smaller than another set, but still has the one to one correspondence.

Now many people still insist that the number of primes is infinite and equal in size to alph naught. However, the number of primes diminishes to zero at infinity (look at the limit of the approx function, which I admit, may NOT be the actual function and it is often conjectured that NO function does exist). At close to the origin, Primes are thick as flies. However as one chases after infinity, there are billions, then trillions of numbers between each prime (or perhaps a Mersenne prime pair). Admittedly, you have to be in the google range before the primes begin to thin out, but infinity is still infinity larger than a google.

So, how can the number of primes = to Alph Naught (the smallest infinity?)? Or is Alph Naught not the smallest infinity?
 
Last edited by a moderator:
Is Alph Naught the smallest infinity? Initially everyone feels the first set of infinity they encounter is the set of natural numbers. Then speculation sets in and the discovery that many infinities are just that, infinity and do not obey the rules of everyday experience. Rational numbers are countable just like natural numbers and therefore, both infinities are deemed equal. Irrational numbers are not countable and therefore are definitely infinitly more of them, so they garner the term Alph Prime.

Now for my question. If Alph Naught is the smallest infinity (and it may not be, and then my question is moot!), why is that? Prime numbers are a smaller subset. Now I am easily convinced that even numbers = the same infinity as all natural numbers or even that rational numbers also = the same as the natural numbers in being equal in number to infinity due to countability.

However, primes are not technically countable (or are they and I am just unaware of any new break through in math. And if they are NOT countable, I have no one to one correspondence with natural numbers. And as we know, primes are definitely a subset of rational numbers. (and I just stated earlier that I accept that even (or odd or any repeating ratio) is equal to the natural numbers in size @ infinity) The number of primes is approximated by the following function #/(ln #) for any number ie 100/ ln 100 = 21 primes between 1 & 100. This is merely an approximation and not an exact formula and this function merely approximates a discreet value and cannot impose a continuous feature to this function. I state this to dispel the circle within a circle argument ie one set is smaller than another set, but still has the one to one correspondence.

Now many people still insist that the number of primes is infinite and equal in size to alph naught. However, the number of primes diminishes to zero at infinity (look at the limit of the approx function, which I admit, may NOT be the actual function and it is often conjectured that NO function does exist). At close to the origin, Primes are thick as flies. However as one chases after infinity, there are billions, then trillions of numbers between each prime (or perhaps a Mersenne prime pair). Admittedly, you have to be in the google range before the primes begin to thin out, but infinity is still infinity larger than a google.

So, how can the number of primes = to Alph Naught (the smallest infinity?)? Or is Alph Naught not the smallest infinity?
According to Cantor's definition, two sets have the same number of elements if each of the members of one set can put into one-to-one correspondence with the members of the other set.

Now obviously the number of natural numbers is infinite. Suppose there is a finite number of natural numbers. Then there is a biggest one, but then 1 + plus that number would also be a natural number bigger than any other. Contradiction, ergo the number of natural numbers is not finite and so is infinite. Make sense do far? Cantor originally called the number of natural numbers Aleph.

Now he also proved that there were sets that had more members than Aleph. So Aleph is infinite, but there are bigger infinities.

Now are there infinities smaller than the number of natural numbers? The answer is no. Why not?

Imagine a set A, which is a set of natural numbers that may have a minimum but does not have a maximum. But that means I can put the set of natural numbers into one-to-one correspondence with the members of set A: I match 1 to the smallest member of A, I match 2 to the next smallest member of A, and so on. I never run out of elements in A to match with the natural numbers because A does not have a largest member: there are always more to go.

Now imagine a set B, which is a set of natural numbers that does have have a minimum, s, and a maximum, h. Now I can match 1 to s, 2 to the next smallest member, and when I get to m - s + 1, I have nothing left to match in B. So the number of set B is m - s + 1, which is finite.

So Aleph is the smallest infinity possible. So Cantor called it Aleph Null and then literally drove himself crazy trying to find the next infinity or Aleph One.

The fact is that our intuition, shaped as it is by finite sets, just fails us when dealing with infinite sets.

Now the set of prime numbers has a minimum member, namely 2. But it has no maximum (as was proved no later than Euclid). So the number of primes is infinite and equal to Aleph Null. It's not finite, and certainly there are not more primes than there are natural numbers because primes are defined to be natural numbers.

This is not a terribly rigorous explanation. Daon or pka or maybe lookagain can probably give a more rigorous explanation.
 
Is Alph Naught the smallest infinity?

Now for my question: primes are not technically countable (or are they and I am just unaware of any new break through in math. And if they are NOT countable, I have no one to one correspondence with natural numbers.

First of all it is Aleph-naught. (note the spelling). \(\displaystyle {\aleph _0}\).

Some time after the year 300BCE, Euclid proved that the set of primes is infinite. So it has been known for a very long time that the set of primes is not finite. But it was not until the nineteenth century, the 1880's, that Georg Cantor worked out the philosophy of infinity.

Any infinite countable set is said to be
denumerable.

Here are two theorems which should answer your questions.
1) Any subset of a denumerable set is finite or denumerable.
2) Any super-set of a denumerable set is infinite.

The set of natural numbers,
\(\displaystyle \mathbb{N}\), is denumerable.
The set of prime numbers is an infinite subset of \(\displaystyle \mathbb{N}\) so it is denumerable.
Now \(\displaystyle {\aleph _0}\) is simply the symbol used to denote the cardinality of any denumerable set.

 
First of all it is Aleph-naught. (note the spelling). \(\displaystyle {\aleph _0}\).

Some time after the year 300BCE, Euclid proved that the set of primes is infinite. So it has been known for a very long time that the set of primes is not finite. But it was not until the nineteenth century, the 1880's, that Georg Cantor worked out the philosophy of infinity.

Any infinite countable set is said to be
denumerable.

Here are two theorems which should answer your questions.
1) Any subset of a denumerable set is finite or denumerable.
2) Any super-set of a denumerable set is infinite.

The set of natural numbers,
\(\displaystyle \mathbb{N}\), is denumerable.
The set of prime numbers is an infinite subset of \(\displaystyle \mathbb{N}\) so it is denumerable.
Now \(\displaystyle {\aleph _0}\) is simply the symbol used to denote the cardinality of any denumerable set.


Thank you for this explanation. However, my question is really, if the set of prime numbers is denumerable (countable), then tell me how to systematically count them. Is there some reliable way to count them? If not, are they really countable? Not because there are so many of them, but because there are so few of them (while I readily admit, they are an infinite set). I wholeheartedly buy into every countable set being equal to the set of natural (and rational) numbers and that they all are equal to aleph naught. But I am not so ready to accept a "because I said so!" argument. It is easy to say natural numbers are countable as they ARE! And we accept the fact that Prime numbers are a subset of these natural numbers. But show me how they are counted. I am conjecturing that because they are not countable, they form an infinite subset smaller than the infinite set of natural numbers.

As I stated, I can see trillions of prime numbers, but they are clustered around the origin. As you get further away from the origin, you get to a desert where you have millions, billions, and trillions of natural numbers before you get to a prime. And it only gets worse as you proceed out towards infinity. I sort of imagine it like weeds (primes) in the sand (natural numbers). Your argument is that the weeds are equal in number to the grains of sand. If you have a systematic way to count these weeds, I'll accept your logic. Is there a way to systematically count primes?

And what is infinity (a prime or a composite or simply undefined, but everything divides into infinity)? Just another question as I do not have any formal education in number theory.
 
Aleph-naught

Thanks for the correction on spelling.

Since Primes are denumerable, show me how to count them. That is my question. I don't see any systematic way to count them in a one to one correspondence to natural numbers. I understand that is how rational numbers are considered equal to the natural numbers in that given you can arrange the rational numbers in a systematic order as to count all the rational numbers.

And you may say that since the natural numbers include the subset of primes, that they are also countable. But are they? I am looking for a proof of countability. If I know when the next prime is going to appear, I will buy into the countability explanation. Under ordinary conditions, I would acknowledge countability. I readily accept that every third element or even every millionth element would be countable and therefore equal to the set of natural or rational numbers, because you can determine the next element. Can you do so with Primes?
 
What do you mean by "systematic" way? The obvious numeration is: 1-> 2; 2->3; 3->5; 4->7; 5->11; 6->13; 7->17; etc. Once you have the "nth" prime determined, n+1 is assigned to the next prime. Because there are an infinite number of primes, that cannot end so we will have a prime assigned to each counting number.
 
Thanks for the correction on spelling.

Since Primes are denumerable, show me how to count them. That is my question. I don't see any systematic way to count them in a one to one correspondence to natural numbers. I understand that is how rational numbers are considered equal to the natural numbers in that given you can arrange the rational numbers in a systematic order as to count all the rational numbers.

And you may say that since the natural numbers include the subset of primes, that they are also countable. But are they? I am looking for a proof of countability. If I know when the next prime is going to appear, I will buy into the countability explanation. Under ordinary conditions, I would acknowledge countability. I readily accept that every third element or even every millionth element would be countable and therefore equal to the set of natural or rational numbers, because you can determine the next element. Can you do so with Primes?
Let's number the primes in order of their magnitude.

p1 is 2.

p2 is 3

p3 is 5

p4 is 7

p5 is 11

p6 is 13

p7 is 17.

Got the system?

Now let's say that the number of primes is smaller than the number of natural numbers. So there is some natural number n that represents the last prime.

Now consider the numbers \(\displaystyle \displaystyle m = \prod_{i=1}^np_i,\ and\ k = m + 1.\) Obviously k is a natural number > pn.

\(\displaystyle Given\ integer\ j\ such\ that\ 1 \le j \le n \implies \dfrac{m}{p_j} = h \in \mathbb N.\)

In other words, any prime from p1 through pn divides evenly into m because m consists solely of the product of those primes.

So \(\displaystyle \dfrac{n}{p_j} = \dfrac{m + 1}{p_j} = \dfrac{m}{p_j} + \dfrac{1}{p_j} = h + \dfrac{1}{p_j}.\)

So k is not evenly divisible by any prime from p1 through pn. Now either k is prime, or is it not. If k is prime, then pn is not the largest prime, which is a contradiction. If k is not prime, then it has at least two prime factors. But all of those factors must be larger than pn, which is not the largest prime, again a contradiction. So I can index the primes in order of size by the natural numbers, which gives me a one-to-one correspondence.

As had been pointed out now several times, Euclid developed this proof over 2000 years ago.
 
my question is really, if the set of prime numbers is denumerable (countable), then tell me how to systematically count them. Is there some reliable way to count them? .
The fact is we don't know a rule for listing all the primes.

We know that the set of primes is an infinite subset of the positive integers.
That is enough to say the set of primes is countable even though we don't know the rule.
 
The number of primes is approximated by the following function #/(ln #) for any number ie 100/[ln(100)] = 21 primes between 1 & 100. \(\displaystyle \ \ \) [It's closer to 22 (primes.)]

This is merely an approximation and not an exact formula and this function merely approximates a discreet \(\displaystyle \ \ \) [The needed word here is spelled "discrete."]
(And I tried to be relatively discreet in how I told you this.)


value and cannot impose a continuous feature to this function.
.
 
Top