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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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 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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 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. Tags infinity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shunya Elementary Math 5 July 18th, 2014 09:09 AM sivela Calculus 1 June 25th, 2012 10:04 AM nerd9 Algebra 22 July 18th, 2010 03:47 PM -DQ- Algebra 5 September 14th, 2009 05:13 AM fibonaccilover Applied Math 9 July 26th, 2009 08:07 PM

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