July 2nd, 2016, 07:25 AM  #11 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 830 Thanks: 67 
"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: 6,344 Thanks: 2083 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: 830 Thanks: 67 
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: 16,191 Thanks: 1147 
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: 6,344 Thanks: 2083 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: 16,191 Thanks: 1147  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.


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countable, natural, numbers, power, set 
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