July 2nd, 2016, 07:25 AM  #11 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 911 Thanks: 72 
"Mathematical induction can be used to prove that the following statement, which we will call p(n), holds for any natural number n. 0+1+2+...n=n(n+1)/2 The proposition p(n) gives a formula for the sum of the natural numbers less than or equal to number n. The proof that p(n) is true for each natural number n proceeds as follows.." Quoted from (with my underlining): https://en.wikipedia.org/wiki/Mathematical_induction Of course, if you can show that p(n) is true for each/any/all n (see OP and Post #5), you don't need induction. To summarize, OP shows P(n) countable is true for each/any/all n and so P(N)=P(1,2,....) is true. If you disagree, which I suspect some respondents do, how would you show a proposition is true for all the natural numbers? Note distinction between P(n), powerset, and p(n), proposition. Last edited by skipjack; July 2nd, 2016 at 02:40 PM. 
July 2nd, 2016, 07:31 AM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,444 Thanks: 2115 Math Focus: Mainly analysis and algebra 
Nobody argues that what you say applies for finite sets of size $n$. But as you say So what you say does not apply to sets of infinite size: infinite sets. Just as the infinite sum 1+2+3+... does not have a value defined by the formula $\frac12 n(n+1)$. 
July 2nd, 2016, 07:41 AM  #13 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 911 Thanks: 72 
The set of all natural numbers does not contain the number n=infinity. That's why you use induction or show a proposition is true for every n. Edit: If it is shown that P(n) is countable for every n, why does it not follow that P(N) is countable? Last edited by zylo; July 2nd, 2016 at 07:58 AM. 
July 2nd, 2016, 11:53 AM  #14 
Global Moderator Joined: Dec 2006 Posts: 16,590 Thanks: 1199 
Because you have not produced a 1 to 1 correspondence between ℕ (the set of all natural numbers) or a subset of ℕ and the powerset of ℕ. For each natural number n, n + 1 is a natural number, and hence finite. It doesn't follow by induction that ℕ is a finite set. Similarly, the countability of the powerset of each finite set doesn't imply the countability of the powerset of an infinite set, such as ℕ. 
July 2nd, 2016, 12:15 PM  #15  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,444 Thanks: 2115 Math Focus: Mainly analysis and algebra  Quote:
Quote:
As usual, your post shows that you know nothing of what you are talking about. Learn some mathematics! Last edited by skipjack; July 2nd, 2016 at 02:19 PM.  
July 2nd, 2016, 03:05 PM  #16 
Global Moderator Joined: Dec 2006 Posts: 16,590 Thanks: 1199  Objection: the statement "P({1,2,…}) is countable" doesn't use "n", so it isn't a particular case of the statement that P({1,2…,n}) is countable for any natural number n.


Tags 
countable, natural, numbers, power, set 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
The Real Numbers are Countable  zylo  Topology  49  June 25th, 2016 08:52 PM 
The Real Numbers are Countable  zylo  Math  24  February 29th, 2016 09:46 AM 
Question on the countability of the power set of natural numbers (Cantor's theorem)  Polaris84  Number Theory  10  February 24th, 2016 09:39 PM 
The real numbers are countable  zylo  Topology  35  January 29th, 2016 07:23 PM 
power set of the natural numbers  blabla  Real Analysis  2  February 25th, 2012 02:31 PM 