July 1st, 2016, 11:49 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,081 Thanks: 87  Power Set of the Natural Numbers is Countable
If Sn = {1,2,....n} P(Sn) has (2^n)1 members, no matter what n is. Therefore the power set of the natural numbers is countable. Comments: P(A) is the set of all subsets of A. If A has n members, the number of subsets of A is nC1+nC2+...nCn=(2^n)1. (1+1)^n=nC0+nC1+nC2+...nCn. Also, n elements are countable if n is a natural number. Last edited by skipjack; July 2nd, 2016 at 02:52 PM. 
July 1st, 2016, 12:17 PM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,570 Thanks: 613 Math Focus: Wibbly wobbly timeywimey stuff. 
This is verging on being spam... Dan 
July 1st, 2016, 02:51 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
It is spam. Zylo is apparently incapable of learning anything about infinite sets.

July 1st, 2016, 07:24 PM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,573 Thanks: 667 
Equivalently: 8 is greater than 7. Therefore the power set of the natural numbers is countable. Or: today is Friday. Therefore the power set of the natural numbers is countable. 
July 1st, 2016, 09:00 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,081 Thanks: 87  https://en.wikipedia.org/wiki/Power_set P(N) = P({1,2,…}) is countable if P({1,2…,n}) is countable for any natural number n. P({1,2..,n}) is countable because the 2^n subsets (including empty set) of {1,2,..,n} are unique and so correspond 1:1 with the natural numbers (up to 2^n). Are there any mathematical objections? Last edited by skipjack; July 2nd, 2016 at 03:05 PM. 
July 2nd, 2016, 02:30 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,707 Thanks: 1356 
As n is a natural number, it is finite. Hence it doesn't follow that the power set of a countably infinite set, such as the set of all the natural numbers, is countable.

July 2nd, 2016, 05:16 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,081 Thanks: 87  There is no such thing as n=infinity. You show something is true for ALL n. All n and any n are the same thing.

July 2nd, 2016, 05:50 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
This demonstrates that your reasoning applies only to finite sets. It says nothing about infinite sets. Learn some mathematics. 
July 2nd, 2016, 05:58 AM  #9 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,081 Thanks: 87 
x^n=xxxx...x, n times, for any/all n. Any person with a licence can drive is the same as all persons with a licence can drive. The semantics is just a way to distract from truth of OP. None of the replies so far invalidates the OP, on the contrary. 
July 2nd, 2016, 07:22 AM  #10  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra  Quote:
Whereas the truth is that there are no persons of infinite size. In this case, there are sets of infinite size, but your deductions do not apply to sets of infinite size because you have specified the size of the set to be $n$. And, as you say Learn some mathematics! You do not have competence to say whether any disproves your ridiculous claims because you don't understand basic logic or mathematics. Last edited by v8archie; July 2nd, 2016 at 07:25 AM.  

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