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April 10th, 2016, 09:52 PM   #11
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Quote:
Originally Posted by v8archie View Post
It does have an infinite number of elements. But that just means that there isn't a last one. There's no need to add another number to make that happen.


No, I'm just saying that for the function you are trying to describe, the range is not the natural numbers, it's the ordinal numbers.


No, it's a cardinal number which means it describes the size of a set. An ordinal describes the ordering of elements of a set.

Perhaps this video will help. (With thanks to whoever posted it this morning).
Cool video, and thanks for your help, but I am not quite sold.
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April 11th, 2016, 01:56 AM   #12
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It is said that there is an infinite number (aleph 0) of elements in the set of natural numbers.
Here is your 'contradiction', particularly the words in bold.

You didn't say which definition of the natural numbers you are using (there are several all equivalent ones).
But whichever definition you are using you have to follow through from its starting premises and deduce the rest of applicable number theory.

As far as I am aware, there are no definitions that include either the phrase 'infinite number' or introduce an aleph.

But a consequence of the axioms does show that all the natural numbers are finite.

The scheme runs

First state what you mean by a natural number eg

Axiom 1) 1 is a natural number or 1 is a member of the set of natural numbers

Then state a way to get more natural numbers

Axiom 2) If x is a natural number then so is (x+1) or if x is a member then so is (x+1)

You need some more axioms to complete the scheme but these are enough to show that

Lemma The set of all natural numbers is an infinite set.

This does not tell you how many members the set has, just that the list never ends.

That is what is meant by an infinite set in this case.

Note this scheme has still not said anything about how many members there are about either the full set or some subset of N.

Archie has shown that the number of members of the full set is not a member of N,
So at that point the scheme could introduce the aleph as a name for this number, if you wish, but before this it is better to develop the scheme to show the other point archie made, which is that the set is open at the top, or has no largest member.

Another consequence of these two axioms is the one I mentioned earlier, that all members of N are finite.
Can you show this as an exercise?

Appreciating this is because it is the beginning of a heirarchy whereby we develop more and more complete sets of numbers as a result of consequences discovered by exploring relationships between the members of each set.

Note that theis axiom set is old and simple, but not favoured today because it relies on definitions of '1' and '+'.
However it has the great advantage of highlighting the line of thinking, so enabling understanding.
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Last edited by studiot; April 11th, 2016 at 02:12 AM.
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April 11th, 2016, 02:33 AM   #13
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Quote:
Originally Posted by Mathbound View Post
It is said that there is an infinite cardinal number ($\aleph_0$) of elements in the set of natural numbers.
The two uses of "number" are different (I've put the word "cardinal" in the above statement to clarify that). As $\aleph_0$ isn't a natural number, there's no reason for it to match a member of the set.

What works for finite sets doesn't necessarily work in exactly the same way for infinite sets.
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April 11th, 2016, 11:17 AM   #14
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The video brought up something very interesting, at least to me. It seemed to imply that aleph 0 is something like a limit. The natural numbers approach it but never reach it kind of like the sum of 1/n, as n --> infinity, approaches 2 but never reaches it, at least on the real number line without the extension to infinity. Now if n = aleph 0, then it would reach 2, but in our case n must be a natural number.

Then the power set of aleph 0, 2^(aleph 0) blows way past the "aleph 0 limit" and onto a currently unknown infinity aleph 1.

So maybe it should be said the set of natural numbers approaches an infinite number of elements like a limit but never really reaches it. Does anyone agree with this?
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April 11th, 2016, 12:55 PM   #15
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Originally Posted by Mathbound View Post
The natural numbers approach it but never reach it kind of like the sum of 1/n, as n --> infinity, approaches 2 but never reaches it
Actually it doesn't. You probably mean $\sum \limits_{n=0}^\infty {1 \over 2^n} = 2$. But that's not really the point of the post.

When we talk about limits, we are talking about a rigorously defined process (and it is a process) of examining successive expressions and determining where they are going.

When we talk about sets, we are referring to static objects. The natural numbers are not a process. They are a static collection of objects, and there are $\aleph_0$ of them.

Also, $\aleph_0$ is a cardinal number, not a natural number or a real number. So as you said, we don't refer to $\aleph_0$ as the "limit" in $\sum \limits_{n=0}^\infty {1 \over 2^n} = 2$ because we are using natural numbers. Furthermore, the notation $\sum \limits_{n=0}^\infty {1 \over 2^n} = 2$ doesn't actually refer to any infinity. It is an abbreviation for $\lim \limits_{n \to \infty} \sum \limits_{k=0}^n {1 \over 2^k} = 2$ which is itself an abbreviation for the full $\epsilon-\delta$ definition of the limit which talks not about infinity, but about what happens to the summation for all $n$ greater than some natural number $N$.

So there may be some very loose similarity with the concept of limits if you happen to be counting as in the video, but in terms of practical or theoretical rigour, I don't think so.

Although this may partly be because in my mind $\aleph_0$ is completely divorced from numbers. I would argue that it is not even clear that the set of real numbers is in any meaningful sense "larger" than the set of rationals (and thus the natural numbers). It just happens that sets of cardinality "larger" than $\aleph_0$ do not have whatever property it is that would enable us to list them. This is all a personal philosophical discussion though and certainly not to be taken as mainstream.
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Last edited by skipjack; April 12th, 2016 at 01:32 PM.
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April 11th, 2016, 01:16 PM   #16
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I'm sorry ,

did I manage to slip in post#12 without the OP noticing?
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April 11th, 2016, 03:46 PM   #17
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Quote:
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I'm sorry ,

did I manage to slip in post#12 without the OP noticing?
I read it, but I do not have time to start multiple conversations.
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April 11th, 2016, 03:51 PM   #18
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Quote:
Originally Posted by v8archie View Post
Actually it doesn't. You probably mean $\sum \limits_{n=0}^\infty {1 \over 2^n} = 2$. But that's not really the point of the post.

When we talk about limits, we are talking about a rigorously defined process (and it is a process) of examining successive expressions and determining where they are going.

When we talk about sets, we are referring to static objects. The natural numbers are not a process. They are a static collection of objects, and there are $\aleph_0$ of them.

Also, $\aleph_0$ is a cardinal number, not a natural number or a real number. So as you said, we don't refer to $\aleph_0$ as the "limit" in $\sum \limits_{n=0}^\infty {1 \over 2^n} = 2$ because we are using natural numbers. Furthermore, the notation $\sum \limits_{n=0}^\infty {1 \over 2^n} = 2$ doesn't actually refer to any infinity. It is an abbreviation for $\lim \limits_{n \to \infty} \sum \limits_{k=0}^n {1 \over 2^k} = 2$ which is itself an abbreviation for the full $\epsilon-\delta$ definition of the limit which talks not about infinity, but about what happens to the summation for all $n$ greater than some natural number $N$.

So there may be some very loose similarity with the concept of limits if you happen to be counting as in the video, but in terms of practical or theoretical rigour, I don't think so.

Although this may partly be because in my mind $\aleph_0$ is completely divorced from numbers. I would argue that it is not even clear that the set of real numbers is in any meaningful sense "larger" than the set of rationals (and thus the natural numbers). It just happens that sets of cardinality "larger" than $\aleph_0$ do not have whatever property it is that would enable us to list them. This is all a personal philosophical discussion though and certainly not to be taken as mainstream.
Interesting, but I think I am just going to have to accept it unless I can prove that there is a contradiction, and that is something that I am currently not able to properly show.

Last edited by skipjack; April 12th, 2016 at 01:37 PM.
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April 11th, 2016, 04:34 PM   #19
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Since so many folks are so rude on this forum I don't suppose I shall bother to help anyone again.

Goodnight all.
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April 12th, 2016, 11:50 AM   #20
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Since so many folks are so rude on this forum I don't suppose I shall bother to help anyone again.

Goodnight all.
Thank you for taking the time, but it just doesn't make sense to me to start two discussions on a specific topic simultaneously with two different people.

From other help forums that I have seen, only one person takes on a student, at least until somebody else sees something wrong with how that person is helping the student or just wants to add some fresh ideas if the other method is obviously not working. And of course, one may jump in if the other just simply gives up on that student.

We are in the "Wild West" of forum etiquette. There is no known universal way to behave on these forums yet, at least not that I can tell. And worst of all, the communication relayed is not quite complete as there is no body communication. For example, if we were all at a coffee shop discussing this, and you see my eyes locked on v8archie's as he/she tries to help me, you may get the impression that extra help is not necessary. But if you saw me huffing and puffing and looking up at the ceiling, and there is a pause, you may just instinctually know to jump in and help guide me into what v8archie is trying to convey.

So please, trust me, people are grateful for your help, and nobody wants to be rude to someone like you who wants to help.
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