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July 2nd, 2016, 08:25 AM   #11
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"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.
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Last edited by skipjack; July 2nd, 2016 at 03:40 PM.
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July 2nd, 2016, 08:31 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 View Post
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 $\frac12 n(n+1)$.
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July 2nd, 2016, 08:41 AM   #13
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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 08:58 AM.
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July 2nd, 2016, 12:53 PM   #14
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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.
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July 2nd, 2016, 01:15 PM   #15
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Quote:
Originally Posted by zylo View Post
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 View Post
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!
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Last edited by skipjack; July 2nd, 2016 at 03:19 PM.
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July 2nd, 2016, 04:05 PM   #16
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Quote:
Originally Posted by zylo View Post
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.
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May 1st, 2017, 07:17 AM   #17
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Quote:
Originally Posted by zylo View Post
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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