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**Ivar Sand****Member**- Registered: 2013-01-22
- Posts: 14

In short. This post presents the idea that there are two kinds of mathematicians, the exploring mathematicians who pioneer the results, and the smart mathematicians who follow and produce the elegant proofs that survive to the text books. A theorem with a proof that is considered smart is chosen from a text book, and then a more elementary proof is presented.

Background. One day, not so long ago, I started wondering about how to prove that a continuous function defined on a closed interval is bounded. I remembered that I had been thinking about the same problem many years earlier and that I had not been able to find a proof. I remembered that I had a runaround experience while trying to locate a point where the function approached infinity. So I gave up trying to construct a proof and looked up the proof in a text book. I found that the proof was smart, a proof that I could not be depend upon being able to find myself.

In spite of this, I continued to think that a straightforward proof ought to exist for this theorem. I fantasised about there being two kinds of mathematicians, the exploring mathematician and the smart mathematician; the exploring mathematicians being the pioneer who is the first one to prove a theorem. The exploring mathematicians would apply engineering-like methods to construct a proof, which might prove to be long and complicated. Then a smart mathematician would follow and replace the exploring mathematician's proof with a smarter and more elegant proof. The new proof would stand the test of time better than the original proof and survive to the text books. Whether this distinction has anything to do with reality, I don't know.

Nevertheless I wanted to try to construct a more elementary proof of the theorem. So I did (or tried to), and I posted this article to demonstrate how such an exploring mathematician's proof might look like. Below, Proof 1 is the "smart" proof and Proof 2 is my proof.

The theorem¹. A continuous function defined on a closed interval is bounded.

Proof 1. I'll only sketch the proof here. One assumes the contrary that the continuous function is unbounded. One divides the interval into two equal parts. One observes that the function has to be unbounded on at least one of the two parts. One chooses one of the parts, divides that interval into two equal parts, chooses the part where the function is unbounded, and so on. The lengths of the chosen intervals approach zero, and the leftmost endpoints of the intervals approach a point in the function's interval of definition. One uses the concept of supremum to prove that the sequence of endpoints converges. The function's continuity at that point is then shown to imply that the function cannot be unbounded on an interval around that point after all.

Comment. One needs to verify that the unboundedness property and the continuity property collide. I think a difficulty lies in the fact that the unboundedness property is the property of an area or interval whereas the continuity property is the property of a point. (The unboundedness property is not the property of a point in the interval of definition because if a function were unbounded or infinite at a point it would be undefined there.) I guess the smartness of Proof 1 lies in the fact that the function's unboundedness property is included in every step of the process. So to speak, one keeps the unboundedness property in hand all the way while one searches for a point in order to use the function's continuity there.

Proof 2. Let the function's name be f and its closed interval of definition be [a,b]. As in Proof 1, we assume the opposite that f is unbounded, and aim at a contradiction.

We start looking for f's unboundedness at a. Since f is defined at a it is finite there, so a is not the point we are searching for. If a = b f cannot be unbounded, so we deduce that b must be greater than a.

Now let the argument of f, call it x, increase continuously from the point a while we look for a point where |f(x)| = |f(a)| + 1. Since f is continuous, f(x) varies smoothly as x increases there are no jumps in the function value. The first point we encounter that satisfies |f(x)| = |f(a)| + 1 we call x[sub]1[/sub]. We proceed from x[sub]1[/sub] moving continuously towards higher values of x looking for a point where |f(x)| = |f(x[sub]1[/sub])| + 1 and call the first such point x[sub]2[/sub]. Proceeding in this way we get a sequence of points {x[sub]1[/sub], x[sub]2[/sub], }.

Suppose we reach b, and we find that our sequence contains a finite number of elements. If the sequence is empty, f is bounded on [a,b] by |f(a)| + 1, otherwise f is bounded by |f(x[sub]n[/sub])| + 1 where x[sub]n[/sub] is the last element in {x[sub]1[/sub], x[sub]2[/sub], }. In both cases f is bounded. We conclude that the number of points in {x[sub]1[/sub], x[sub]2[/sub], } is infinite.

The function value at a point x[sub]i[/sub] in {x[sub]1[/sub], x[sub]2[/sub], } for an i > 1 satisfies |f(x[sub]i[/sub])| = |f(x[sub]i-1[/sub])| + 1. We find |f(x[sub]i[/sub])| = |f(a)| + i for i ≥ 1 which implies that |f(x[sub]i[/sub])| approaches ∞ as i increases.

Since the monotonic sequence {x[sub]1[/sub], x[sub]2[/sub], } is bounded (by b), it converges². Call the point of convergence x[sub]0[/sub].

Choose an ε > 0. We have:

1. Since f is continuous at x[sub]0[/sub], there is a δ > 0 so that |f(x) - f(x[sub]0[/sub])| < ε for all x satisfying |x - x[sub]0[/sub])| < δ. Now

|f(x) - f(x[sub]0[/sub])| < ε ⇒

|f(x) - f(x[sub]0[/sub])| + |f(x[sub]0[/sub])| < ε + |f(x[sub]0[/sub])| ⇒

|f(x) - f(x[sub]0[/sub]) + f(x[sub]0[/sub])| ≤ |f(x) - f(x[sub]0[/sub])| + |f(x[sub]0[/sub])| < ε + |f(x[sub]0[/sub])| ⇒

|f(x)| ≤ |f(x) - f(x[sub]0[/sub])| + |f(x[sub]0[/sub])| < ε + |f(x[sub]0[/sub])| ⇒

|f(x)| < ε + |f(x[sub]0[/sub])|

We conclude:

|f(x)| < |f(x[sub]0[/sub])| + ε whenever |x - x[sub]0[/sub])| < δ (1).

2. Since {x[sub]1[/sub], x[sub]2[/sub],
} converges towards x[sub]0[/sub], there is an integer m1 so that

|x[sub]i[/sub] - x[sub]0[/sub]| < δ whenever i > m1 (2).

3. Since |f(x[sub]i[/sub])| approaches ∞ with i, there is an integer m2 so that

|f(x[sub]i[/sub])| > |f(x[sub]0[/sub])| + ε whenever i > m2 (3).

Now choose m = max (m1,m2). This gives us new versions of (2) and (3):

|x[sub]i[/sub] - x[sub]0[/sub]| < δ whenever i > m (2')

|f(x[sub]i[/sub])| > |f(x[sub]0[/sub])| + ε whenever i > m (3').

The continuity condition (1) expresses that f is bounded by |f(x[sub]0[/sub])| + ε on the open interval (x[sub]0[/sub] - δ, x[sub]0[/sub] + δ) whereas the unboundedness or divergence condition (3') expresses that |f(x[sub]i[/sub])| is greater than this bound for an i satisfying i > m, call it j. Now since x[sub]j[/sub] lies in (x[sub]0[/sub] - δ, x[sub]0[/sub] + δ) according to the convergence condition (2'), we deduce from (1):

|f(x[sub]j[/sub])| < |f(x[sub]0[/sub])| + ε

and from (3'):

|f(x[sub]j[/sub])| > |f(x[sub]0[/sub])| + ε.

These two inequalities are inconsistent, and this contradiction allows us to conclude that our original assumption of f being unbounded is false. QED.

References.

1. Tom M. Apostol, Calculus, Volume 1, One-Variable Calculus, with an Introduction to Linear Algebra, Second edition, 1967, p. 150.

2. Ibid., p. 381.

I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.

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**cmowla****Member**- Registered: 2012-06-14
- Posts: 57

I think I'm more of the experimental type myself. I guess "smart" could mean conceptual and "exploring" could mean experimental. Being an experimentalist by nature, I had a very difficult time with upper level math courses in college such as advanced calculus and abstract algebra before I got my B.S. in math. Now for abstract algebra at least, I'm working on a Rubik's cube project which I believe is teaching me a thing or two about the real idea behind abstract algebra that I was aware of but didn't see any application for.

I am sorry to say that I can't comment on your proof (because I honestly don't care enough about calculus at the moment to have interest), but a little while back I too looked at a topic from advanced calculus which I wanted to construct a proof which I could understand better for some of the basic limit laws. (What do you think about those? Are they valid?). Hopefully someone will comment on your proof, I was certainly lucky to get at least one response on mine.

I think proofs which explain more and show every step of the mind are more elegant than compact proofs which assume the reader has "mathematical maturity".

I am many things, but incapable of understanding complicated concepts is not one of them (even though I still debate that sometimes for no good reason). For example, I created this "parity algorithm" for the 4x4x4 Rubik's cube from scratch. I explained the entire process that went through my mind, different paths I tried before, and the logic behind the entire process in a video about an hour and a half long (in 4 parts).

Part I.

The reason why that algorithm is so significant is not just because it scrambles the cube so much (see the beginning of this video to see what a

"normal" parity algorithm

looks like. Rather, the minimum number of 90 degree slice turns to "flip an edge" before I came around was 25 (like that "normal" algorithm has). Mine uses only 19...

One of the lines of bobbym's signature says that "90% of mathematicians do not understand 90% of currently published mathematics." I wonder why that is? Hmmm? **[1]** I do realize that there is so much information published now that it would be impossible for any mathematician to be able to understand a significant portion of published topics because he/she could not possibly read it all in his or her lifetime. But I am assuming that bobbym's quote is saying that, if a mathematician is presented with published math at random, he could not understand what he reads 9 times out of 10. **[2]** However, I theorize that the main problem is poor communication. That is, when most mathematicians write their papers, they assume that their readers understand every elementary concept which is prerequisite knowledge to whatever it is they are writing about. There are so many ways to say the same thing, and we have to be careful how to write something down in technical language, especially if people's lives depend on it, but I still don't like how, say, the language in which the majority of wikipedia's math articles are written in, for example. I thought taking discreet math would have helped me to understand proofs because I learned what the basic symbols mean, but even if we assume that a particular proof does not build upon a proven theorem or lemma, one word or term in a proof that the reader isn't familiar with might require that the reader must read another topic to become familiar with that word, and this continues...I have come to many dead ends that I cannot possibly understand what someone is trying to say because even the prerequisite knowledge is written in a language that I cannot understand nor care to spend a lifetime trying to learn.

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**Ivar Sand****Member**- Registered: 2013-01-22
- Posts: 14

Thanks for your post, cmowla.

To be honest, I guess I used the idea of there being smart mathematicians and exploring mathematicians as a way of marketing my proof, hehe. However, a Physics professor at the university where I studied did say to me once that he liked an A- (or B+) student, or not quite first-rate student, better than an A student because an A- student is more innovative than an A student, who is conservative. An A- student is familiar with making mistakes and is not afraid of making them. An A student, on the other hand, is afraid of making mistakes and does not dare say things that might prove to be false.

I have posted a comment on your post about the basic limit laws after having spent some time trying to understand your post. I hope you appreciate my comment although I understand that you are not into calculus at the moment.

Well, I think there is such a thing as mathematical maturity. I remember taking a course in linear algebra once. It was a spring course, and I had the feeling of understanding nothing before Easter and everything after Easter.

I cannot comment on Rubik's cube as I am not into that area.

Regarding your item [1]: I guess we have to accept that mathematics is difficult; there seems to be general acceptance of this. This causes papers to be poorly understood, which is sad. However, maybe someone someday will develop a mathematics language that would make mathematics easy to understand.

Regarding your item [2]: I think an idea behind listing in a paper the prerequisite knowledge is to suggest to the unqualified reader to stop reading the paper so as to avoid wasting his or her time reading it. By the way, I have now included the text "(calculus)" in the name of this thread so that readers not interested in calculus can skip the thread. (Oops, no, I didn't it is not possible.)

By the way, perhaps there is a simple contribution in my post #1 in regard of the effort of making papers easy to read. There, I try to align in successive lines the same texts in mathematical expressions. That way, it is easy for the reader to see what is unchanged in a pair of lines so as to find these most interesting parts of the lines fast.

I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,523

Some physics professor that must have been. This year the class average for the midterm has been 64% for which my teacher was extremely mad at. The weird thing is, he behaved slightly better with the students in the mid-70%s rather then my friends and I who were in the 90%s range. That would mean he preferred the experimental learners then the conceptual ones.

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