-5
$\begingroup$

I can not well understand that the natural set $\mathbb{N}$ is an infinite set (which contains infinite many elements) while each natural number is finite? I already find the same question in Why set of natural numbers is infinite, while each natural number is finite?, but I still can not well understand.

  1. Review: This conclusion comes from the following reasons:

    • The reason all natural numbers are finite. By the Peano axioms (of natural number), all naturals are constructed from $0$ by the successor operations or by the operations of "+1". As $0$ is finite, and if any number $n$ is finite, so is the successor $n+1$. Repeating this '+1' process step by step, we may find all successors are finite, or all naturals are finite.
    • The reason the natural set $\mathbb{N}$ is an infinite set. This is simple, as by the axiom of infinity (set), the infinite set is in fact also defined in above successor way. Thus a natural set is an infinite set.
    • The natural set $\mathbb{N}$, or the infinite set defined by axiom, contains infinite number of elements. This question is not trivial as until now we even don't has a clear definition of the infinite number.
      • infinity: In the reason all natural numbers are finite, we show all numbers constructed from $0$ by operation "+1" are finite numbers, or all naturals are finite numbers. We may define infinite number as the opposite - those can never be reached from $0$ by operation "+1" step by step. Or, as said by wiki, it can be treated as the number which is larger than any natural number. While the infinity is defined, we find that there is a gap between finite numbers and infinite numbers, we can never get an infinite number from a finite number by finite many of "+1" operation.
      • The number of elements in natural set $\mathbb{N}$.
        • We may set a counter $t$ with the construction process of natural set. For the beginning, the first element $0$, the counter $t=1$; then for each successor process "+1", we also ask $t=t+1$. With the help of the counter, we immediately get any set $S$ build by the step-by-step successor process "+1" is a finite set, because the counter $t=1+\sum_{i=1}^{M}1$ is built from "$0$" by step-by-step "+1" process, so that it is a finite number.
        • From above illustration, we immediately have the conclusion that any set $S$ build by the step-by-step successor process is a finite set, and is the subset of natural set $S\subset\mathbb{N}$. And, any finite number (can always be written as $t=1+\sum_{i=1}^{M}1$ is always corresponding to one subset $S$, thus can never count (or say smaller than) the number of element of $\mathbb{N}$. As the number of element of $\mathbb{N}$ $t_0$ greater than any finite number $t=1+\sum_{i=1}^{M}1$, by definition, it must be infinity.
  2. My question: The above statements do make sense. However problem may occur when we think inversely. As from "$0$" by the step-by-step successor ("+1") process we can only access finite set $S\subset\mathbb{N}$, there must exist some elements in the infinite set $\mathbb{N}$, which can never be contained by any finite set $S$. The question is: what are these elements? Are they finite numbers? Should be no, as any finite number must be contained in some finite set $S$. Then they must be infinite numbers. We then get the contradiction conclusion that $\mathbb{N}$ contains some infinite numbers, which can never be construct from $0$ by the step-by-step successor process! So what is the problem with my deduction?

  3. EDIT - the answer: after discussions in the following I know where is the wrong.

    • The key point is when can we introduce infinite. It need to be careful as we can never get infinite from finite by "+1" operation. So the question is how to introduce infinite while avoid the 'gap' between finite and infinite. The answer is in fact really simple, the infinite is introduced only when any finite number can not describe the problem.

    • Examples. For any member of natural set, it is not infinite, as it is always in the form $t=\sum_{i=1}^{M}1$. As compare, if we allow $t=\sum_{i=1}^{\infty}1$, i.e. allow infinite many of addition, we may get a infinite number. However, such a number can never be get by step-by-step successor operation from $0$, as it is always a finite addition. While for the number of natural set, as any finite number can not describe it, it must be an infinite number.

    • My argument there must exist elements in the infinite set $\mathbb{N}$, which can never be contained by any finite set $S$ is not accurate as I should compare $\mathbb{N}$ with the union of all finite sets. Only when there are difference between them, can we have a discussion. However, now we have infinite number of finite sets. Here again we introduce infinite because for any finite number of finite sets there may still exist maximum elment $n$, we then can find a bigger $n+1$ is not included by this unions. The infinite union of finite sets $S$ can be a infinite set, thus there is no difference between two sets.

$\endgroup$
  • 1
    $\begingroup$ Your error is about quantifiers: any element of the natural numbers is indeed contained in some finite subset. There are no "weird numbers" that cannot be accessed by some (perhaps large) finite number of successor operations. At the same time, you can always produce a natural number not in any given finite subset, so they are infinite. $\endgroup$ – Ian Jul 21 '18 at 5:32
  • $\begingroup$ So, what is the difference in elements between the finite subset $S$ and the infinite set $\mathbb{N}$? @Ian $\endgroup$ – X liu Jul 21 '18 at 5:33
  • $\begingroup$ Whatever elements you didn't put into the finite subset. For instance $n+1$ is not in $\{ 0,1,\dots,n \}$. Put now you can put $n+1$ in to have that one be in a finite subset, but now you are still missing $n+2$, etc. $\endgroup$ – Ian Jul 21 '18 at 5:35
  • $\begingroup$ Again, it is in another bigger subset, which is still not the infinite set $\mathbb{N}$. In this way you can build arbitrary big subset as you like, however no subset can including everything. $\endgroup$ – X liu Jul 21 '18 at 5:36
  • 1
    $\begingroup$ Sure. But any element is in some finite subset., even though all finite subsets miss some elements. These statements are compatible because they are quantified differently. $\endgroup$ – Ian Jul 21 '18 at 5:39
3
$\begingroup$

Infinite sets are weird, and relying on a pre-formed intuition about infinity can be dangerous. In the second section, you make the error of assuming that since any subset of $\mathbb{N}$ formed by finite applications of the successor process is finite, there must be some elements in $\mathbb{N}$ that cannot be accessed from this process. Using finitely many steps, you can form an arbitrarily large set $S \subset \mathbb{N}$ of the form $\{1, ..., n\}$. $\mathbb{N}$ is the set of all things that can be formed from $1$ with finite applications of this process, hence by definition any element of $\mathbb{N}$ can be found in some finite set $S$ of this form. Now, the question is whether or not the set of all finite numbers, $\mathbb{N}$, is finite. A simple argument shows that this is false.

Suppose for contradiction that there are only finitely many finite numbers formed by the successor taking process. Since any finite subset of $\mathbb{N}$ has a greatest element, there must be some maximal $n$ that can be formed by the process of taking successors. However, the successor of $n$ is clearly a strictly larger natural number, which contradicts our hypothesis. So there must be infinitely many finite numbers.

The key feature here is that you can always increase a natural number by $1$ while remaining finite. Since $\mathbb{N}$ is closed under the taking of successors, this forces $\mathbb{N}$ to be infinite while still only consisting of finite numbers. This is a bit strange at first, but, as our argument shows, perfectly logically sound.

$\endgroup$
  • $\begingroup$ Yes. This statement is already included in my first section. I can following this logically. The problem is when you think inversely, what is the difference between infinite set $\mathbb{N}$ and finite set $S$, or the union of all possible finite set $U=\cup S_{i}$, we get the problem. @Alex Nolte $\endgroup$ – X liu Jul 21 '18 at 6:08
  • 1
    $\begingroup$ @XLiu The difference between $\mathbb{N}$ and $S$ is infinite (you have infinitely many elements, and you take finitely many away). In your union, you are taking a union of infinitely many finite sets, and you get an infinite set. $\endgroup$ – Morgan Rodgers Jul 21 '18 at 6:12
  • $\begingroup$ The same argument resolves your issue with the inverse process. Just ask if the union of all finite sets so formed is finite. If it is, the same argument yields a strictly larger set of finite elements of $\mathbb{N}$ which is a contradiction. $\endgroup$ – Alex Nolte Jul 21 '18 at 6:16
  • 1
    $\begingroup$ It is proven false and $\{1\}\cup \{2\} \cup \{3\} \cup .....=\mathbb N$ and $\mathbb N$ is infinite. The is no reason to claim that the infinite union of sets must be finite and we can show that there are times it is not. So it is false. And you have no conflict. " I just want to show there exists conflicts." Well, there aren't. What there are is that you have misunderstandings. I don't want to be mean but it's important you recognize that the math is good. It's your understanding that is weak. $\endgroup$ – fleablood Jul 21 '18 at 7:13
  • 1
    $\begingroup$ "I am talking about {1}∪{1,2}∪{1,2,3}∪..." {1}∪{1,2}∪{1,2,3}∪.. is also equal to $\mathbb N$. It is an infinite union of finite sets. "I agree". As you said quote "And until now I can not find a way to prove it to be true or false" unquote. That means you do not agree and Alex has claimed (correctly) that he HAS proven it to be false. $\endgroup$ – fleablood Jul 21 '18 at 7:55
0
$\begingroup$

" The above statements do make sense. However problem may occur when we think inversely. As from "0" by the step-by-step successor ("+1") process we can only access finite set S⊂N,"

It can only access finite sets if you do it finite times. The conclusion is you can not use the step by step successor process to access an infinite set in any finit number of steps.

there must exist some elements in the infinite set N, which can never be contained by any finite set S

That does not follow at all!

What follows is for any finite set there exists a bigger finite set. And that although each may occur in some different finite sets, there is no finite set containing them all.

"Are they finite numbers?" All numbers are finite.

"Should be no, as any finite number must be contained in some finite set S". And every one of them ARE contained in some finite set. But the finite sets they are contained in are different sets.

Let's put it this way suppose I asked if you and I were siblings. And you answered No. And I said but every-one has a mother and father. And you said, yes, every-one has a mother and father, but that doesn't mean everyone has the same mother and father. And I said but either everyone are siblings and they have the same mother and father, but no couple can be the parent to everyone. Therefore there must be people who have no parents! Who are these people? Are they zombies? Did they pop up out of nowhere?

Every number, has a finite value and can be a member of a finite set in many many different ways. But for every finite value there are finite values that are larger. And each finite set, in and of itself can not contain all the numbers. But that's okay, because the numbers that are not contained in the set, can be contained in other set${ }$s (not all of them in the same other set, but different other set${}$S; an infinite number of other sets).

$\endgroup$
  • $\begingroup$ This analogy is interesting though not accurate. However, I think I know what I am missing after a long discussion. Can you help me to check the following? If its right, I will add to main post. The key point is when can we introduce infinite, it need to be careful as we can never get infinite from finite by "+1" operation. So the question is how to introduce infinite while avoid the 'gap' between finite and infinite. The answer is in fact really simple, the infinite is introduced only when any finite number can not describe the problem. To be continued. @fleablood $\endgroup$ – X liu Jul 21 '18 at 9:13
  • $\begingroup$ For any member of natural set, it is not infinite, as it is always in the form $t=\sum_{i=1}^{M}1$. As compare, if we allow $t=\sum_{i=1}^{\infty}1$, i.e. allow infinite many of addition, we may get a infinite number. However, such a number can never be get by step-by-step successor operation from $0$, as it is always a finite addition. While for the number of natural set, as any finite number can not describe it, it must be an infinite number. To be continued. @fleablood $\endgroup$ – X liu Jul 21 '18 at 9:13
  • $\begingroup$ My argument there must exist elements in the infinite set $\mathbb{N}$, which can never be contained by any finite set $S$ is not accurate as I should compare $\mathbb{N}$ with the union of all finite subsets to find the difference. However, now we have infinite number of finite sets. Here again introduce infinite because for any finite number of finite sets there may still exist maximum number $n$, we then can find a bigger $n+1$ is not included by this unions. The infinite union of finite sets $S$ can be a infinite set, thus there is no difference between the two sets. @fleablood $\endgroup$ – X liu Jul 21 '18 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.