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May 27th, 2017, 08:51 AM   #31
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How can infinity exist inside of presized location.
0 - infinity - 1
0 - infinity - 0.1
0 - infinity - 0.001
As I am still able to pull frames of infinite from each interval.
0 - ( 0.09... 0.099... 0.0999... ) - 0.1
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May 28th, 2017, 05:22 AM   #32
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Absolutely gorgeous!
There is only one aspect.
If I have way smaller interval compare to larger interval, like:
( 0 to. 0.001 ) compare to ( 0 to 1 ), I am running to " spooky action ".
The set of infinite numbers, in the smaller interval equals to set of infinite numbers in the larger interval. The infinity is constant no matter where you look for it.
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Any continuous interval of real numbers will have the same cardinality, yes.

We can have a finite set of real numbers: e.g. {0.5, 0.75}.

We can have a set of real numbers with cardinality $\aleph_0$: e.g. {0.1, 0.11, 0.111, 0.1111, ... }.

We can also have a set of real numbers with cardinality $2^{\aleph_0}$: e.g. (0, 1).

Now here is a million dollar question for you. Can we have a set of real numbers (or, any set really) that has a cardinality X such that $\aleph_0 < X < 2^{\aleph_0}$? A solution to this question would prove whether or not the continuum hypothesis is true.

https://en.wikipedia.org/wiki/Continuum_hypothesis
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May 28th, 2017, 09:44 AM   #33
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I am ready to sink in to black hole.
Can I ask for exact representation for N0 and 2N0 and I will work on X.
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May 28th, 2017, 10:39 AM   #34
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I am ready to sink in to black hole.
Can I ask for exact representation for N0 and 2N0 and I will work on X.
Thanks
Yes.

The cardinality of $\mathbb{N}$ is $\aleph_0$.

The cardinality of both $P( \,\mathbb{N}) \,$ and $\mathbb{R}$ is $2^{\aleph_0}$.

See my previous posts for what $\mathbb{N}$ and $P( \,\mathbb{N}) \,$ are if you are unsure (the natural numbers and the powerset of the natural numbers, respectively). The set of real numbers is denoted $\mathbb{R}$.

When referring to the cardinality of a set, you may enclose the set in "| |". For example, $|P( \,\mathbb{N}) \,| = |\mathbb{R}| = 2^{\aleph_0}$.

Don't go sinking into a black hole, as there is no need for that.

Infinite sets can have properties that are counter intuitive. Trying to solve the countinuum hypothesis (working "on X" as we put it) is probably too much, as the hypothesis has gone unsolved for a couple hundred years. Just learning the basics is a good start.

Maybe I should have shown you something simpler to start. For example, we can show that the cardinality of the set of positive even integers is the same as the cardinality of the set of natural numbers by showing that a bijective function exists between the two sets. To go from the positive even numbers to the natural numbers, our function could be defined as follows:

$$\text{Let } f : \{ x \in \mathbb{N} : x \text{ is even} \} \rightarrow \mathbb{N}$$
$$f( \,x) \, = \frac{x}{2}$$

This happens to be a bijective function because:

$$\{1 = \frac{2}{2}, 2 = \frac{4}{2}, 3 = \frac{6}{2}, ... \}$$

Knowing that there exists a bijective function $f$ from the set of positive even integers onto the natural numbers shows that the two sets' cardinalities are equal, so we write:

$$|\{ x \in \mathbb{N} : x \text{ is even}\}| = |\mathbb{N}|$$
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May 28th, 2017, 11:19 AM   #35
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Thank you and I will study more farther, I really will.
I would like so any reader of this topic can understand it easy.
Little help here:
You asked to name number or set of numbers that is more than one set of numbers and less then another set. Am I right? Like:
N0<X<2N0
Where each set of numbers represents "infinity"?

Last edited by Microlab; May 28th, 2017 at 11:27 AM.
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May 28th, 2017, 11:42 AM   #36
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Originally Posted by Microlab View Post
Thank you and I will study more farther, I really will.
I would like so any reader of this topic can understand it easy.
Little help here:
You asked to name number or set of numbers that is more than one set of numbers and less then another set. Am I right? Like:
N0<X<2N0
Where each set of numbers represents "infinity"?
I asked if there is a set with a cardinality that is greater than the cardinality of the natural numbers but less than the cardinality of the powerset of natural numbers.

More specifically, does there exist a set $S$ such that:

$$|\mathbb{N}| < |S| < |P( \,\mathbb{N}) \,| = |\mathbb{R}|$$

If we denote the cardinality of $S$ as simply $X$ (i.e., let $|S| = X$), an equivalent way of asking this question is to ask if there exists a cardinal number $X$ such that:

$$\aleph_0 < X < 2^{\aleph_0}$$

where (to help you with notation)

$$\aleph_0 = |\mathbb{N}| < X = |S| < 2^{\aleph_0} = |P( \,\mathbb{N}) \,| = |\mathbb{R}|$$
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May 28th, 2017, 12:12 PM   #37
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I have to go through all this tonight!
Please bear with me.
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May 28th, 2017, 12:18 PM   #38
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Sorry, where can I get the key for math symbols?
Thanks again
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May 28th, 2017, 12:33 PM   #39
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Sorry, where can I get the key for math symbols?
Thanks again
This has quite a few of them:

https://oeis.org/wiki/List_of_LaTeX_...atical_symbols

Is there some symbol in particular that you would like clarification on? Tip: If I've used the symbol, you can quote my post to see how I wrote it, then ask me what it means.

Last edited by AplanisTophet; May 28th, 2017 at 12:35 PM.
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May 28th, 2017, 12:45 PM   #40
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Thank you!
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