I am reading Keisler's Elementary Calculus (which can be downloaded here). I am having trouble understanding his proof sketch of Extreme Value Theorem and how he is applying the Transfer Principle.

For reference, he defines the "Transfer Principle" as:

Every real statement that holds for one or more particular functions holds for the hyperreal natural extension of these functions.

On page 164 (using left corner numbering) of the book he provides the following "sketch":

Extreme Value

I understand the counter examples and I am able to understand the issues with them using standard tools. I don't understand, however, how one can immediately utilize the Transfer Principle. It is not immediately obvious to me that "there is a partition point $a + K\delta$ at which $f(a + K\delta)$ has the largest value."

To elaborate, the proof seems circular. In trying to to "expand" the sketch to be more precise. I ended up writing instead of:

By the Transfer Principle, there is a partition point $a + K\delta$ at which $f(a + K\delta)$ has the largest value.


Applying the Transfer Principle to the Extreme Value Theorem we see that the Extreme Value holds for hyperreals as well. Hence, there is a partition point $a + K\delta$ at which $f(a + K\delta)$ has the largest value.

But this relies on a proof of the Extreme Value Theorem for reals.

Hopefully what I am saying makes sense, please ask for any clarification.

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    $\begingroup$ +1 I also tried to read this proof from Keisler and was not satisfied with it. In fact any of the theorems which depend on completeness of real numbers are not proved satisfactorily in the book. $\endgroup$ – Paramanand Singh Sep 21 '18 at 7:51

For me the problem here is that the statement Keisler gives of the transfer principle doesn't quite fit with how it is being used. You might want to look at a more formal source to clarify exactly what transfer says (try Goldblatt, or https://en.wikipedia.org/wiki/Transfer_principle and the references it cites).

Here transfer is being applied to the statement "if $n$ is a natural number and $s_1,\ldots, s_n$ are reals then $\{s_1,\ldots,s_n\}$ has a maximum." Transfer (in its full form - not necessarily in the way Keisler states it) tells you that this applies for $n \in \mathbb{N}^*$ too, which is exactly what is needed in the proof.

This can seem confusing if you think of nonstandard natural numbers as "infinitely large," because it is certainly not true that an infinite subset of $\mathbb{R}^*$ has to be bounded. This application of transfer tells us only that if $\nu$ is any natural number, even a nonstandard one, then every sequence $s_1,\ldots, s_\nu$ is bounded.

It is helpful to work through an example of a discontinuous unbounded function on a compact interval, to see why Keisler's argument wouldn't apply to that. Let's take $f(0)=0$ and $f(x)=1/x$ for $x>0$, so that $f: [0,1] \to \mathbb{R}$ is unbounded. We start by picking a partition of $[0,1]$ with evenly spaced points $0,1/H,2/H,\ldots, (H-1)/H, 1$, where $H$ is an "infinitely large" natural number. There is indeed a partition point at which $f$ is maximal, namely $f(1/H)=H$. The standard part of $1/H$, which gets called $c$ in Keisler, is 0. But you can't get any relationship betweeen $f(c)$ and $f(1/H)$, even though $c$ and $1/H$ are infinitely close, because $f$ is not continuous at 0. This means the argument breaks down, as it must.

  • $\begingroup$ While I don't have much knowledge of non-standard analysis, the statement about maximum of the set with infinitely many elements does sound a bit hard to believe. $\endgroup$ – Paramanand Singh Sep 24 '18 at 14:00
  • $\begingroup$ And the argument presented in Keisler seems to imply that range of any function on $[a, b] $ has a maximum. $\endgroup$ – Paramanand Singh Sep 24 '18 at 14:15
  • $\begingroup$ @ParamanandSingh "Infinitely many" is misleading here: they are indexed by a nonstandard natural number (I prefer "illimited"). The argument does not tell you anything about an arbitrary function - how would you get a relation between $f(c)$ and $f(a+K\delta)$ without continuity? $\endgroup$ – Matthew Towers Sep 24 '18 at 14:22
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    $\begingroup$ @ParamanandSingh Keisler's argument does tell you that $f(a+K\delta)$ is finite: it is infinitely close to $f(c)$ (by continuity!). $\endgroup$ – Matthew Towers Sep 24 '18 at 14:46
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    $\begingroup$ @m_t_ Thank you for the answer, I have been busy and unable to look closely for a while. For reference, your last comment to ParamanandSingh helped me the most as it explicitly mentions the relation to continuity and the transfer principle. $\endgroup$ – Dair Sep 30 '18 at 0:33

Let me propose a different interpretation of transfer (in the formulation suggested by m_t_) for the Intermediate Value Theorem.

Consider the following "standard" argument about ordinary real numbers. For all $n \in \mathbb{N}$, it is possible to partition $[a,b]$ into $a, a+\frac{b-a}{n}, \ldots, a+n\frac{b-a}{n}=b$. Since there is a finite number of partition points (they are $n+1$), it is well-defined the maximum among the values $f(a), f\left(a+\frac{b-a}{n}\right), \ldots, f(b)$.

Recall that, by transfer, finite and $^\ast$finite sets satisfy the same properties. As a consequence, you can always pick the biggest element of a $^\ast$finite set. Hence, transfer entails also that for all $n \in\ \! ^\ast\mathbb{N}$ the number $$\max\left\{ f(a), f\left(a+\frac{b-a}{n}\right) , \ldots, f\left(a+n\frac{b-a}{n}\right)=f(b) \right\}$$ is well-defined for every partition of $^\ast[a,b]$ into $n$ equal parts. In other words, you are proving the Extreme Value Theorem by applying transfer to the statement "every finite set has a maximum element", and not to the Extreme Value Theorem for real numbers.

Let now $n \in\ \! ^\ast\mathbb{N}$ be infinite, and let $$f\left(a+K\frac{b-a}{n}\right) = \max\left\{ f(a), f\left(a+\frac{b-a}{n}\right) , \ldots, f\left(a+n\frac{b-a}{n}\right)=f(b) \right\}.$$ (What I call $\frac{b-a}{n}$ is the infinitesimal $\delta$ in the original proof by Keisler). Define $c=\ \!^\circ\left(a+K\frac{b-a}{n}\right)$. By continuity of $f$ you have that $^\circ\left(^\ast f\left(a+K\frac{b-a}{n}\right) \right) = f(c)$. From here onward, it is possible to follow the original proof by Keisler.

  • $\begingroup$ Thank you @LultZ for the advice. $\endgroup$ – Emanuele Bottazzi Nov 11 '18 at 17:45
  • $\begingroup$ What is the cardinality of *finite set? Does it match the cardinality of $\mathbb{N} $? $\endgroup$ – Paramanand Singh Nov 12 '18 at 4:11
  • $\begingroup$ To me the argument is a translation of the following idea in standard analysis in the language of non-standard analysis : partition the interval into $n$ parts and choose the division point where $f$ has maximum value and call it $x_n$. The sequence $x_n$ has a limit point $c$ and $f(c) $ is the maximum value of $f$. $\endgroup$ – Paramanand Singh Nov 12 '18 at 4:16
  • $\begingroup$ @ParamanandSingh from a set-theoretical point of view, *finite sets that are not finite are never countable. However, every *finite set has a well-defined "number of elements", i.e. its internal cardinality is some $n \in\ ^\ast\mathbb{N}$. $\endgroup$ – Emanuele Bottazzi Nov 18 '18 at 13:54
  • $\begingroup$ I am not sure (or lets say i don't know) what you mean by internal cardinality. Is it something different from the usual cardinals aleph null or $\mathfrak{c} $? Since you mention they are not countable, I guess their cardinality should be $\mathfrak{c} $. $\endgroup$ – Paramanand Singh Nov 18 '18 at 14:12

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