I am reading the book "A Treatise on Advanced Calculus" by Philip Franklin. I found this book in our city's central library and liked it at the first reading, am continuing this book as a reference. I got some confusion at the very first topic "Mathematical Induction". I am familiar with mathematical induction pretty much but I could not grasp the way author explained it.
In the link i given $-$ After introduction the first topic starts. It is divided in $8$ small paragraphs. I understand $7$ paragraphs out of these. I have confusion in the $3$rd paragraph which I am quoting here:

"If each of the members of $I$, an infinite collection of positive integers, is less than, or equal to, $N$, every integer of the collection $I$ is equal to some member of the finite collection $1, 2, .\ .\ . N$ Thus there is a finite collection of distinct integers, $F$, such that each member of $I$ is equal to a member of $F$. The greatest integer of $F$ is the greatest of $I$, so that the first collection $I$ has a greatest integer."

The two points on which I have trouble are as follows:

  1. How this paragraph satiates(offsets) the whole article, that is what is the purpose of this paragraph. In other words what is the logic provided by this paragraph which is necessary for an apt understanding of the remaining article. What is the benefit of assuming that $I$ has a greatest integer.
  2. Why the author assumes that $I$, an infinite collection of positive integers, has an upperbound(a greatest number)? We know that $I$ is an infinite set so has not any greatest integer. Moreover how can we disprove this argument, that is does it directly follows from the definition of $I$ that it does not have a greatest integer or it can be proved in a mathematical way?

I understood the remaining $7$ paragraphs. I am not able to understand what the author is trying to convey in this $3$rd paragraph that I quoted.

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    $\begingroup$ Basically the author want to prove the principle of mathematical induction (often assumed as an axiom) as a theorem; for doing so, he states some "more basic" principles which (as argued by Joshua in his answer) he assumes that are "more intuitive" than the theorem he want to prove. $\endgroup$ Commented Mar 3, 2014 at 16:41

1 Answer 1


We didn't assume that the elements in $I$ are all distinct. In essence, this paragraph argues the proposition: any bounded set of positive integers can only contain finitely many distinct elements, without assuming a priori that $I$ is finite.

This has the immediate, significant corollary that every bounded set of positive integers has a greatest element.

This is all fairly intuitive, so such an approach may seem unnecessary. Still, one could argue that it clears up some ambiguities that might crop up along the line. Consider for example a bounded sequence taking values in $\mathbb{N}$ - the above discussion shows that the sequence can only take finitely many values. This is fairly immediate, but arguably non-trivial, as it is related to the properties of the natural numbers - if we consider a bounded sequence taking values in $\mathbb{Q}$ or in $\mathbb{R}$, we cannot draw the same conclusion.

Note that the notions of greatest element and least upper bound are distinct in general. (Consider for example the set $[0,1) \subset \mathbb{R}$. This has a least upper bound, namely 1, but does not have a greatest element.) The fact that they coincide for finite subsets of the natural numbers is an important statement about the nature of the natural numbers, and is closely related to induction.

One additional comment that occurs to me is that the terminology used by the author is highly suggestive. By an infinite set, we would usually understand a set with infinitely many distinct elements, as the definition of "set" usually doesn't allow repeated elements. However, the author of your article was careful to speak of an infinite collection of natural numbers, which does not carry the same connotation.

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    $\begingroup$ I'm not sure whether you're actually making this mistake, but note that you can have an infinite set of distinct integers in $\mathbb{Z}$ that is bounded above. In your article, and indeed in general whenever we start talking about induction, we restrict attention to positive integers, i.e. integers in $\mathbb{N}$. $\endgroup$ Commented Mar 3, 2014 at 16:46
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    $\begingroup$ Ah, I see that you have edited your comment to remove the ambiguity that I commented on above. Cleary you weren't making the mistake I alluded to :). $\endgroup$ Commented Mar 3, 2014 at 16:47
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    $\begingroup$ A sequence $(x_n)$ takes a value $a$ if $x_k=a$ for some $k \in \mathbb{N}$. For example, a constant sequence only takes one value. For your second question, there exist sequences in $\mathbb{Q}$ (and hence in $\mathbb{R}$) that are bounded, yet still take infinitely many different values. Consider for example $x_n = 1/n \in \mathbb{Q}$. For your third question, we do not need this result in order to establish induction. However, it is informative in the sense that it tells us something about the nature of $\mathbb{N}$ that intuitively justifies the induction domino effect. $\endgroup$ Commented Mar 6, 2014 at 12:58
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    $\begingroup$ $N$ is not assumed to be the greatest member of $I$. Instead, we assumed that $I$ is bounded by $N$, which means that all of the elements of $N$ considered individually are smaller than $N$. $\endgroup$ Commented Mar 6, 2014 at 14:01
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    $\begingroup$ Also, we do not conclude that $N$ is the greatest element of $I$, but instead that $I$ has a (non-specified) greatest element. Indeed, the collection $A = \{1, 1, 2\}$ is bounded by $N = 435$, but its greatest element is $2$. $\endgroup$ Commented Mar 6, 2014 at 14:03

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