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- 04-15-2015, 06:23 AM #1

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## Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

Different levels of infinity, another Cantor axiom to argue to death...

Cheers

L-zr

- 04-17-2015, 01:06 AM #2
## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

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- 04-18-2015, 02:24 PM #3
## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

Pre-dates Cantor, not really an axiom

Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

- 04-19-2015, 01:05 AM #4

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## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

koeddyrbh:

invisitesimal c. postulate

invisibility action by

because

and

0.999... = 9/10 + 9/100 + 9/1000 + ... =

0.999...9 (n 9s) = 0.999...

Look:

Comprehend ?

2.7

2.97

2.997

2.999...997

Let Sn denote the nth partial sum. Clearly Sn = 1 - 1/10ⁿ

Then for every ε > 0 we have an N = log10 (1/ε) such that:

|Sn - 1| = |1/10ⁿ| < ε for all n > N.

Therefore 0.999... = 1

Let Sn denote the nth partial sum. Clearly Sn = 1 - 1/10ⁿ

Then for every ε > 0 we have an N = log10 (1/ε) such that:

|Sn - 1| = |1/10ⁿ| < ε for n=∞ (all n > N)

Therefore 0.999... = 1

Let S(∞) denote the ∞th partial sum. Clearly S(∞) = 1 - 1/10^∞

Then for every ε > 0 we have an N = log10 (1/ε) such that:

|S(∞) - 1| = |1/10^∞| < ε for n=∞ (all n > N)

Therefore 0.999... = 1

Let S(∞) denote the ∞th partial sum. Clearly S(∞) = 1 - 1/10^∞

Then for ε=inf > 0 we have an Z = log10 (1/ε) also 10^Z = 1/inf such that:

|S(∞) - 1| = |1/10^∞| < ε for n=∞ (all n > Z)

Therefore 0.999... = 1Last edited by 7777777; 10-04-2016 at 02:02 AM.

- 04-19-2015, 12:16 PM #5
## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

Um, no. Bijective cardinality does not mean every member of a set is paired with exactly one member of another set, because not all the "members" are numbers.

So it's not only false, but stupid to say that no elements are left out. After all, most of those mythical objects called "irrational numbers" are not elements of any set. Chuckle.Last edited by john_gabriel; 04-19-2015 at 12:19 PM.

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- 04-21-2015, 03:41 AM #6

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## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

A set is a well defined collection of distinct objects.

Since infinity is undefined 123...and 246.. are not sets.Lies have the stench of death and defeat.

- 04-22-2015, 06:36 AM #7

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## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

Huh...

The set of all natural numbers {1,2,3...} is so well defined and important

that it has its own symbol N. as has other similar sets. It can not be more

well defined then that.

What is at question here and in the .999... thread is how to interpret infiniinty and infinitesimality.

If the set of natural number is finite or not. If line has a finite number of points or not. The answer

to this is not crystal clear and both viewpoints exist. It has however been proven (I can not rember

who did this) that it is not important for other mathematic in general. But it is an intersting subject.

Both viewpoints are attractive for different reasons.

Cheers

L-zrLast edited by Lazer; 04-22-2015 at 08:30 AM.

- 04-22-2015, 10:39 AM #8
## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

- 04-22-2015, 10:59 AM #9

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## Re: Is the infinyty of set {1.2.3...} equal to the infinity of set {1.3.5...} ?

But if there are

__infinite many__natural numbers then__all__natural numbers cannot be defined. So I think the point Astrotech was making is that a "set" is complete, all accounted for. The symbol N is used to describe a "set" that cannot be all accounted for, which really makes no sense.

I liked your post though by the way.

- 04-22-2015, 10:59 AM #10

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