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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 $n \in \mathbb{N}$.

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    $\begingroup$ Which textbook? $\endgroup$ – Lord Shark the Unknown Nov 17 '18 at 8:10
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    $\begingroup$ It is conceivably possible until you rule it out, by proving that there is no bijection from $\mathbb N_n$ to $\mathbb N$, for any $n$. $\endgroup$ – Bungo Nov 17 '18 at 8:12
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    $\begingroup$ Well, the book doesn't say it's possible that $\bf N$ is finite, it says it is conceivably possible that $\bf N$ is finite. Sort of the same way that it's conceivably possible that unicorns exist. Anyway, does the book elaborate on this at all? The book doesn't just leave it hanging there, and go off to some entirely unrelated topic, does it? $\endgroup$ – Gerry Myerson Nov 17 '18 at 8:14
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    $\begingroup$ @GerryMyerson: Eventually the book proves that $\mathbb{N}$ is an infinite set. $\endgroup$ – K.M Nov 17 '18 at 8:22
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    $\begingroup$ Then I guess that settles it, right, K.M.? $\endgroup$ – Gerry Myerson Nov 17 '18 at 8:32
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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.

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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.

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  • $\begingroup$ So in summary, the set $\mathbb {N'} $ is chosen as $\mathbb {N} \cup \{\omega + n , n \in \mathbb {N}\} $ because it's too intuitive that $\mathbb {N} $ is an infinite set? So we show a bijection between the positive integers in $\mathbb {N'} $ onto $\mathbb {N_m} $ which can't be shown false yet until it is later proven that there's no bijection from $\mathbb {N}$ onto $\mathbb {N_m} $, as you've stated at the end? $\endgroup$ – K.M Nov 17 '18 at 21:31
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    $\begingroup$ K.M. The point is that the set $\mathbb{N}'$ would work in the definition as a suitable "counting set" which would cause the counting set itself to be rendered as finite despite being "intuitively" infinite, thus showing how it is "conceivably possible" $\mathbb{N}$ "might be a finite set according to this definition". We have to show, then, that $\mathbb{N}$ is, in fact, not like this set and thus that our definition "captures the intuitive notion of infinity" requires more work. $\endgroup$ – The_Sympathizer Nov 17 '18 at 23:16
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    $\begingroup$ Namely, we have to show that there are no naturals which are, themselves, "intuitively infinite" which would then cause the naturals to wrongly accept themselves and some other sets as being finite. $\endgroup$ – The_Sympathizer Nov 17 '18 at 23:20

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