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 May 28th, 2017, 09:12 PM #41 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 Is it possible, that |N|=|P(N)| for the domain [-∞,∞], as both |N| and |P(N)| representing infinite "quantitative parameters" sets, along a line?
May 29th, 2017, 06:17 AM   #42
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 Originally Posted by Microlab Is it possible, that |N|=|P(N)| for the domain [-∞,∞], as both |N| and |P(N)| representing infinite "quantitative parameters" sets, along a line?
No. See one of my previous posts where I went through a bit of a longer exposition than needed to show that $|\mathbb{N}| < |P( \,\mathbb{N}) \,|$ (I was trying to walk you through the logic behind the proof).

Simply, let's assume there is a function $f$ that is bijective (or surjective) from $\mathbb{N}$ onto $P( \,\mathbb{N}) \,$. We denote it like this:

$$\exists f : \mathbb{N} \xrightarrow{\text{1-to-1, onto}} P( \,\mathbb{N}) \,$$

Now, remember my 'sneaky' set $A$?:

$$\text{Let } A = \{ n \in \mathbb{N} : n \notin f( \,n) \, \}$$

In English, the above definition says "let $A$ be the set of natural numbers $n$ such that $n$ is not an element of $f( \,n) \,$." For example, if 1 was an element of $f( \,1) \,$, then 1 would not be an element of $A$ by definition. The same goes for 2, 3, 4, ... (all natural numbers).

We know that $A$ is a set of natural numbers (or the empty set, denoted $\emptyset$), therefore, $A$ is an element of $P( \,\mathbb{N}) \,$. Because $A$ is in $P( \,\mathbb{N}) \,$, there must be a natural number $k$ such that $f( \,k) \, = A$ should our assumption that $f$ is bijective be correct. No natural number $k$ can exist, however:

$k \in A \implies k \notin f( \,k) \, = A$ (a contradiction: "$k$ is in $A$ implies $k$ is not an element of $f( \,k) \, = A$)
$k \notin A \implies k \in f( \,k) \, = A$ (again, a contradiction: "$k$ is not in $A$ implies that $k$ is an element of $f( \,k) \, = A$).

Therefore, there does not exist a natural number $k$ such that $f( \,k) \, = A$. Where no natural number $k$ can map to $A$, the function $f$ is not bijective (or, likewise, surjective).

Where no bijective or surjective function can exist from $\mathbb{N}$ onto $P( \,\mathbb{N}) \,$, we have proven:

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

May 29th, 2017, 06:30 AM   #43
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 Originally Posted by AplanisTophet No. See one of my previous posts where I went through a bit of a longer exposition than needed to show that $|\mathbb{N}| < |P( \,\mathbb{N}) \,|$ (I was trying to walk you through the logic behind the proof). Simply, let's assume there is a function $f$ that is bijective (or surjective) from $\mathbb{N}$ onto $P( \,\mathbb{N}) \,$. We denote it like this: $$\exists f : \mathbb{N} \xrightarrow{\text{1-to-1, onto}} P( \,\mathbb{N}) \,$$ Now, remember my 'sneaky' set $A$?: $$\text{Let } A = \{ n \in \mathbb{N} : n \notin f( \,n) \, \}$$ In English, the above definition says "let $A$ be the set of natural numbers $n$ such that $n$ is not an element of $f( \,n) \,$." For example, if 1 was an element of $f( \,1) \,$, then 1 would not be an element of $A$ by definition. The same goes for 2, 3, 4, ... (all natural numbers). We know that $A$ is a set of natural numbers (or the empty set, denoted $\emptyset$), therefore, $A$ is an element of $P( \,\mathbb{N}) \,$. Because $A$ is in $P( \,\mathbb{N}) \,$, there must be a natural number $k$ such that $f( \,k) \, = A$ should our assumption that $f$ is bijective be correct. No natural number $k$ can exist, however: $k \in A \implies k \notin f( \,k) \, = A$ (a contradiction: "$k$ is in $A$ implies $k$ is not an element of $f( \,k) \, = A$) $k \notin A \implies k \in f( \,k) \, = A$ (again, a contradiction: "$k$ is not in $A$ implies that $k$ is an element of $f( \,k) \, = A$). Therefore, there does not exist a natural number $k$ such that $f( \,k) \, = A$. Where no natural number $k$ can map to $A$, the function $f$ is not bijective (or, likewise, surjective). Where no bijective or surjective function can exist from $\mathbb{N}$ onto $P( \,\mathbb{N}) \,$, we have proven: $$|\mathbb{N}| < |P( \,\mathbb{N}) \,|$$
Thanks.
I have to work on it. I am missing a lot.
Can you suggest any literature I can pull up to read and study.

May 29th, 2017, 07:36 AM   #44
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 Originally Posted by Microlab Thanks. I have to work on it. I am missing a lot. Can you suggest any literature I can pull up to read and study.
I went through a few books trying to find one that I think will suit you well. Try this one:

http://www.math.toronto.edu/weiss/set_theory.pdf

Jech does a wonderful job too, but can be a little more difficult... check out this one that he co-authored:

http://www.unalmed.edu.co/~jmramirez...set-Theory.pdf

May 29th, 2017, 07:39 AM   #45
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Quote:
 Originally Posted by AplanisTophet I went through a few books trying to find one that I think will suit you well. Try this one: http://www.math.toronto.edu/weiss/set_theory.pdf Jech does a wonderful job too, but can be a little more difficult... check out this one that he co-authored: http://www.unalmed.edu.co/~jmramirez...set-Theory.pdf
Thanks for all your support!

May 29th, 2017, 07:57 AM   #46
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 Originally Posted by Microlab Thanks for all your support!
You're welcome! Enjoy!!

 May 29th, 2017, 11:09 AM #47 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 Does anyone has any idea? How its possible to endlessly splitting PRECISED matter into any |P(N)| size particles that will never make same size matter again? Look like I am trying to fill 10 gallons tank with light.
May 29th, 2017, 11:12 AM   #48
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Quote:
 Originally Posted by Microlab Does anyone has any idea? How its possible to endlessly splitting PRECISED matter into any |P(N)| size particles that will never make same size matter again? Look like I am trying to fill 10 gallons tank with light.
Math $\neq$ physics.

May 29th, 2017, 11:28 AM   #49
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 Originally Posted by Maschke Math $\neq$ physics.
Ok, ok, sorry. But still ...
I think, something is missing out there. Something big!

May 29th, 2017, 02:38 PM   #50
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 Originally Posted by Maschke Math $\neq$ physics.
What about coordinates along of precised line - | P ( ab ) |?
Where (a) - starting point and ( b ) - ending.
This is a math!

Last edited by Microlab; May 29th, 2017 at 02:43 PM.

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