My Math Forum Power Set of the Natural Numbers is Countable

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 July 2nd, 2016, 07:25 AM #11 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90 "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. Thanks from manus Last edited by skipjack; July 2nd, 2016 at 02:40 PM.
July 2nd, 2016, 07:31 AM   #12
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Nobody argues that what you say applies for finite sets of size $n$. But as you say
Quote:
 Originally Posted by zylo There is no such thing as n=infinity.
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: 1,155 Thanks: 90 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: 18,034 Thanks: 1393 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. Thanks from topsquark and manus
July 2nd, 2016, 12:15 PM   #15
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Quote:
 Originally Posted by zylo 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.
No. Induction is used to show that something is true for every natural number $n$ greater than or equal to the base case. Induction has nothing to do with the infinite.

Quote:
 Originally Posted by zylo If it is shown that P(n) is countable for every n, why does it not follow that P(N) is countable?
Because $n$ is a natural number, a member of the set of natural numbers. $\mathbb N$ is the set of natural numbers. It is not a member of the set of natural numbers. Your proposition doesn't even make sense for $n=\mathbb N$. (Not surprisingly, since $n=\mathbb N$ doesn't make any sense either). $S_{\mathbb N} = \{1,2,\ldots \mathbb N\}$. What does that mean? Nothing. It is nonsense.

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
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Quote:
 Originally Posted by zylo P(N) = P({1,2,…}) is countable if P({1,2…,n}) is countable for any natural number n. . . . Are there any mathematical objections?
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.

May 1st, 2017, 06:17 AM   #17
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Quote:
 Originally Posted by zylo 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.
The set Sn also has a largest member, no matter what n is. In fact, it is n. So, by your reasoning, the set of natural numbers must also have a largest member. What do you think that is?

Or maybe there is something wrong with this inference, from a finite set to an infinite one?

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