How would the set $\mathbb{N}= \{1, 2, 3, \dots \}$ be a finite set according to Definition 1.3.1 from Bartle's *Introduction to Real Analysis*?

Question

This is from Sherbert and Bartle's Introduction to Real Analysis.

1.3.1 Definition

(a) The empty set is said to have $0$ elements.

(b) If $n\in \mathbb{N}$, a set $S$ is said to have $n$ elements if there exists a bijection from the set $\mathbb{N}_n:= \{1, 2, \dots, n\}$ onto $S$.

(c) A set $S$ is said to be finite if it is either empty or it has $n$ elements for some $n\in \mathbb{N}$.

(d) A set $S$ is said to be infinite if it is not finite.

I'm trying to understand the following from the textbook:

Also it is conceivably possible that the set $\mathbb{N}= \{1, 2, 3, \dots \}$ might be a finite set according to this definition.

I was thinking that one would think that $\mathbb{N}= \{1, 2, 3, \dots \}$ is a finite set from (c), since we could informally say that $\mathbb{N}$ has $n$ elements for some $n \in \mathbb{N}$. But how would this make sense with part (b) of the definition? I don't see how I could find a bijection from $\mathbb{N}_n$ onto $\mathbb{N}$ for some $n \in \mathbb{N}$.

Answer 1

In the introduction to this section ($\S 1.3$ Finite and Infinite Sets) the authors say that

The notions of "finite" and "infinite" are extremely primitive, and it is very likely that the reader has never examined these notions very carefully.

In other words, the authors assume that this is the first time that the reader has seen a definition of a finite set and an infinite set. So it is quite possible that in the mind of a novice reader, upon seeing these definitions, they are unable to immediately conclude that $\mathbb{N}$ is indeed an infinite set (based on how finite and infinite sets have just been defined). The purpose of the sentence

Also, it is conceivably possible that the set $\mathbb{N} := \{ 1,2,3,\dots \}$ might be a finite set according to this definition.

seems to be to reassure the reader that it is natural to feel at first glance that $\mathbb{N}$ might be a finite set under these definitions. They also say this in the previous sentence:

From the definitions, it is not entirely clear that a finite set might not have $n$ elements for more than one value of $n$.

Essentially, the authors anticipate these type of questions to arise in the mind of the reader, and by pointing out that "it is not entirely clear" and that "it is conveivably possible" that such strange things can occur if we were to work with Definition 1.3.1, they are not only acknowledging these questions but also indicating that they will be dealt with very carefully later on.

So, if it is clear to you that $\mathbb{N}$ is indeed an infinite set as per Definition 1.3.1, then that's great! Go ahead and try to prove it yourself, and then check your work with that given in the textbook. It will be a worthwhile exercise to test your understanding in this manner.

Answer 2

Note that this question asks how it could be "conceivably possible" that this would be the case - since, of course, in "actuality", it isn't. To understand this, we thus have to examine a sort of mathematical "parallel universe" if you will where this definition did consider $\mathbb{N}$ to be finite. This is a standard idea in logic, related to modal logic: "X is possible" means "there exists a 'possible world' where X is true", and thus to show that possibility we should attempt to build such a possible world.

To that end, suppose we define an "alternative universe" version of $\mathbb{N}$ called $\mathbb{N}'$ that looks something like this

$$\mathbb{N}' := \mathbb{N} \cup \{ \omega + n, n \in \mathbb{N} \} = \{ 0, 1, 2, \cdots, \omega, \omega + 1, \omega + 2, \cdots \}$$

where $\omega$ is a new element intended to have the effect that it is "larger than every usual natural number". Namely, we put to this set the ordering that all elements of the form $\omega + n$ are larger than those of the form $m$, where $m, n \in \mathbb{N}$, while elements of the same type are ordered in the way one would expect, as suggested by the presentation of the set on the far right.

Next, suppose we were to repeat the Definition 1.3.1 you give from the book only using this $\mathbb{N}'$ in place of $\mathbb{N}$. If now we take $n = \omega$ we have the bijection

$$\begin{align} 0 &\leftrightarrow 1\\ 1 &\leftrightarrow \omega\\ 2 &\leftrightarrow 2\\ 3 &\leftrightarrow \omega + 1\\ 4 &\leftrightarrow 3\\ 5 &\leftrightarrow \omega + 2\\ &\cdots\\ \omega &\leftrightarrow 0 \end{align}$$

where we've just paired up the elements of the set $\mathbb{N}'_\omega = \{ 0, 1, 2, \cdots, \omega \}$ on the left with those of $\mathbb{N}'$ on the right by putting every even natural with the sequence of naturals plus 1, and we've put every odd natural with the sequence of extended naturals. Finally, we save 0 to be paired up with $\omega$ itself. Thus by this definition, "in this parallel universe" "$\mathbb{N}$", i.e. $\mathbb{N}'$, is "finite" with size $\omega$. Even worse, we can actually do the same for the other, larger "sizes" as well and thus we get the even weirder result that the "size" is really any $\omega + n$ for $n \in \mathbb{N}$ (the "real" $\mathbb{N}$, that is).

So we have to eliminate this somehow, that is, to show that in "reality" $\mathbb{N}$ is not like this set $\mathbb{N}'$, i.e. it contains no naturals who are in bijection with itself. And that's why more work is needed from this point on.