# Are there proofs that you feel you did not “understand” for a long time?

Perhaps the "proofs" of ABC conjecture or newly released weak version of twin prime conjecture or alike readily come to your mind. These are not the proofs I am looking for. Indeed my question was inspired by some other posts seeking for a hint to understand a certain more or less well-establised proof, or some answers to those posts. I am interested in proofs at undergraduate levels. Since the question as asked in the title would be too personal, I suggest a longer and hopefully more positive version:

Is there any proof that you feel you didn't understand fully until years later?

The reason that I am interested in this question is that we are currently working on an assessment framework for assessing students' understanding of proof. Reading some previous posts on MO, It occurred to me that perhaps we are too naive in our approach just seeking for understanding logical structure, the key point and so on. It would be very informative if you kindly include in your answer the follow-up of the proof you mention.

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"In mathematics you don't understand things. You just get used to them." -- some wise fellow –  Algernon May 17 '13 at 7:51
define:understand. Do you mean "follow" or "see the thread"? –  darij grinberg May 17 '13 at 10:28
@Algemon: Yet, it would be good if we understand the time we started getting used to the thing. By the way, I think that wise fellow was John von Newmann. –  Amir Asghari May 17 '13 at 13:37
@Amir: Absolutely. I just mentioned the quote as a funny reflection on how we understand things. I don't even completely agree with it myself. Human mind works with making associations, and it is the network of associations that we call understanding. We do get used to things, but in a highly selective fashion, highlighting the relevant connections and forgetting the rest. –  Algernon May 17 '13 at 16:38
... and you are of course right about the name of the wise fellow. I often make a point of not mentioning the names when quoting wise men, as big names have a tendency to bias our personal opinions. –  Algernon May 17 '13 at 16:39

As an undergraduate, I learned the Sylow theorems in my algebra classes but could never retain either the statement or proof of these theorems in memory except for short periods of time (and in particular, for the duration of an algebra exam). I think the problem was that I was exposed to these theorems long before I had internalised the concept of a group action. But once one has the mindset to approach a mathematical object $X$ through the various natural group actions on that object, and then look at the various dynamical features of that action (orbits, stabilisers, quotients, etc.) then all the Sylow theorems (and Cauchy's theorem, Lagrange's theorem, etc.) all boil down to observing some natural action on some natural space (e.g. the conjugacy action on the group, or on tuples of elements on that group) and counting orbits and stabilisers (p-adically, in the case of the Sylow theorems). (Isaacs book on finite group theory emphasises this perspective very nicely, by the way.)

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Thanks! I never understood the Sylow theorems either (except for purposes of exams, of course), but now that you've explained how to look at them, I think that I'll try again. –  Toby Bartels May 18 '13 at 22:00
Almost surely this means that the Sylow theorems are rarely taught properly. –  Todd Trimble Jun 29 '13 at 15:46
You can prove the first Sylow theorem using representation theory: First prove that if a group has a p-Sylow then every subgroup has a p-Sylow. Secondly embed the group into GL_n(F_p) using the regular representation. Thirdly notice that GL_n(F_p) has a p-sylow. This was explained to me an a class taught by David Speyer –  Daniel Barter Jun 29 '13 at 16:44
@DanielBarter: That's not really a proof by representation theory, but rather a proof by linear algebra. –  KConrad Jun 29 '13 at 17:17
In his book Algebra Lang also puts an emphasis on group actions while proving the Sylow theorems. This was also the first time I understood them. –  Martin Brandenburg Aug 16 '13 at 12:29

I hate to sound dumb, but I still don't really understand the Pythagorean theorem as well as I like. I've seen lots of proofs but they all feel too clever for me. To me, understanding a theorem or its proof means being able to see why it's true without having to work out the details of the proof.

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This proof does not seem to "clever" to me. Seems like a natural thing someone might realize if they were given 4 congrunent triangles to play with. takayaiwamoto.com/Pythagorean_Theorem/… –  Steven Gubkin May 17 '13 at 4:15
You lost me at "4 congruent triangles to play with". I do know this proof. I agree that once you're given the 4 congruent triangles, you have a good chance of figuring out this proof. But where does the idea of playing with 4 congruent triangles come from? –  Deane Yang May 17 '13 at 4:17
How about the proof that Einstein is said to have (re)discovered as a teen-ager? It is clear that the areas of similar right triangles is proportional to the squares of corresponding sides---in particular to the squares of their hypotenuses. Dropping a perpendicular from the vertex of the right triangle onto the hypotenuse c divides it into two similar triangles with hypotenuses a and b, hence: kc^2 = ka^2 + k*b^2 so c^2 = a^2 + b^2 –  Dick Palais May 17 '13 at 5:19
@Deane : See also Terence Tao's blog post terrytao.wordpress.com/2007/09/14/pythagoras-theorem –  François Brunault May 17 '13 at 7:43
I don't know why, but the rotational invariance of the dot product is much more intuitive to me, even though the Pythagorean theorem follows from it: $(a,b)\cdot(a,b)=(c,0)\cdot(c,0)$. –  Dustin G. Mixon May 17 '13 at 11:14

The first proof of Tychonoff's theorem I learned, from the Alexander subbase theorem, was completely mysterious to me. I didn't understand it at all. In particular I didn't really understand what the precise role of the axiom of choice was.

Later I learned that there is a much more intuitive proof using ultrafilters or, equivalently, nets. This proof also makes clear the role of the axiom of choice. You want to capture the following intuition: if $X_i$ is a collection of compact spaces and $a : \mathbb{N} \to \prod X_i$ is a sequence, then it should suffice to pick a limit point in each of the projections of $a$ to a sequence in each $X_i$ to show that $a$ itself has a limit point. Unfortunately, sequence convergence doesn't capture the topology of spaces in general, but net convergence and ultrafilter convergence both do, and the above proof works more or less verbatim with "sequence" replaced by either "net" or "ultrafilter."

The axiom of choice enters twice: first a weak version enters in setting up the basic theory of nets or ultrafilters (you need the ultrafilter lemma either way I think), and second the full version enters when picking a limit point in each $X_i$.

This proof also shows that if you only want to prove that a product of compact Hausdorff spaces is compact Hausdorff (often enough for applications), you only need the ultrafilter lemma: the second use of choice doesn't enter into the picture because limit points of ultrafilters are unique if the $X_i$ are Hausdorff! (I assume a similar statement is true for nets but I'm not sure.)

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Fortunately the first proof I have seen used ultrafilters. –  The User May 17 '13 at 14:46
Alexander's subbase lemma can also be proven using ultrafilters. –  Joseph Van Name May 17 '13 at 20:51
You don't need the ultrafilter lemma to set up the basic theory of topology via nets or filters, but you need it to set up the basic theory of topology via universal nets or ultrafilters. And the proof is much cleaner using universal nets or ultrafilters, since you merely need to show that each of either converges. (If instead you try to show that each net has a convergent subnet or that each proper filter has a convergent proper refinement, then you find yourself needing choice again.) And yes, convergence of all nets and proper filters (if convergent at all) is unique in a Hausdorff space –  Toby Bartels May 18 '13 at 16:58
Another way to see the uses of choice in the theorem is to use nonstandard analysis. This works just like ultrafilters, except that you talk about a hyperpoint in the infinitesimal neighbourhood of a standard point p instead of an ultrafilter that converges to p. Like the ultrafilter approach to topology, nonstandard analysis simply doesn't work without the ultrafilter lemma, so that's enough for the Hausdorff case; but the non-Hausdorff case requires an arbitrary choice of p in each space, just as with ultrafilters. –  Toby Bartels May 18 '13 at 22:24
I understood Tychonoff's theorem much better after seeing the version that a countable product of sequentially compact spaces is sequentially compact, which just uses a "diagonal subsequence" trick (and dependent choice). This is sufficient for, say, metric spaces, which in analysis and probability covers almost everything you'll see in real life. And having seen this, the proof of the general case using nets looks analogous. –  Nate Eldredge Jul 31 '13 at 3:38

It took me a very long time to understand the Recursion Theorem. The proof is ridiculously simple: one clear observation about computability (there is a computable total $f$ such that if $\Phi_e(e)\downarrow$ then $\Phi_{f(e)}\cong\Phi_{\Phi_e(e)}$), followed by one line of mysterious symbol-pushing. It only became meaningful to me when I was told to think of it as a diagonal argument that failed (which was also the way it was discovered, if I recall correctly).

Actually, that piece of explanation really changed the way I think about mathematics: it drove home the value of the heuristic principle that if an informal argument doesn't actually work, then there has to be some thing - which will be mathematically interesting - which is actively blocking it. Not always true, but extremely often useful for understanding why math is the way it is (at least for me).

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The diagonal argument that fails is the explanation from pages 36-37 of Soare's book Recursively enumerable sets and degrees, right? I like that explanation, but I find the way Sipser approaches it in Introduction to the theory of computation even more compelling. –  Henry Cohn May 17 '13 at 3:52
The recursion theorem ought to be a corollary of Lawvere's fixed point theorem, right? –  Qiaochu Yuan May 17 '13 at 4:26
It seems plausible, but I don't see it immediately. Perhaps I'm being thick: what is the category we live in, in this case? –  Noah S May 17 '13 at 18:02
I'm not sure (I meant to work this out sometime but haven't gotten around to it). Some Cartesian closed category where the morphisms are computable functions. The point would be that Godel numbering provides something like a surjection $\mathbb{N} \to \mathbb{N}^{\mathbb{N}}$ in such a category, so $\mathbb{N}$ must have the fixed point property. –  Qiaochu Yuan May 17 '13 at 18:28
Hm, I tried it with different categories, but I failed. If you include partial functions, you get into trouble, because the empty set becomes a zero object and the fixed point property becomes trivial. But the set of Gödel numbers of total functions, or of those functions defined on a given recursively enumerable set, is not recursively enumerable. –  The User May 18 '13 at 17:51

It's a bit silly, but I didn't understand the proof of Yoneda's lemma until I taught it (3rd year undergrad). I could follow the proof formally (nodding my head in agreement at each step), but I had no feeling or understanding for why a natural transformation from Hom(-,A) to F should be determined by its value on id_A. It basically boils down to properly appreciating the concept of naturality. As such, finally `understanding' the proof gave me a tremendous sense of accomplishment!

I think that my difficulty in understanding it was that the proof is so short and clean, "almost a tautology", just following a simple diagram about, that it all goes by "too fast" and I never struggled with it enough to develop an idea for why Yoneda's lemma should be true or why the steps in the proof were natural and sensible.

Beyond that, it was difficult for me to grasp the "diagram-chase" without a "philosophical meaning" (see this MO question) for the lemma. I think that having a strong mental "visual" image for a proof, however imprecise, is virtually a prerequisite to human understanding of mathematical proofs- my current mental model for the Yoneda Lemma proof is as a sort of system identification, with identity morphisms playing the role of Dirac delta functions to be "fed into" the system.

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In my copy of Mac Lane's book, I drew a large, very large "?" on the page with the Yoneda Lemma. Later I realized I already understood special and typical cases of Yoneda's Lemma and this helped a lot to finally understand it. For example, if $M$ is an $R$-module, then a homomorphism $R \to M$ is the same as an element of $M$ (this is the (enriched) Yoneda Lemma applied to functors $R \to \mathsf{Ab}$): If you know the image of $1 \in R$, the rest is determined by linearity. The same holds in general: When you know $\hom(A,-) \to F$ on $\mathrm{id}_A$, the rest is determined by naturality. –  Martin Brandenburg Aug 16 '13 at 12:45

In high-school we were told that a continuous function on $[0,1]$ must attain a maximum value. I recall seeing an "epsilon-delta style" proof of this, without really understanding it. Not only was it hard to see the main ideas of the proof, but without a proper definition of the real numbers the argument necessarily becomes shaky at some point.

A couple of years later I learned the topological definitions of continuity and compactness in terms of open sets and finite open covers. Suddenly the steps of the proof suggest themselves naturally, and everything boils down to showing that $[0,1]$ is compact by repeatedly cutting subintervals in half. This in turn pins down exactly what properties we need of the real number system for the argument to work.

To me this was a demonstration of the amazing power of good definitions.

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Instead of cutting subintervals, you can define $x \in [0,1]$, for a given cover, to be the least upper bound of the set of $y$ for which there is a finite subcover of $[0,y]$. Then by the existence of an open set containing $x$, it is not hard to show that $x=1$. Plus: explicit use of least upper bound property. Minus: large quantifier load. –  S. Carnahan May 17 '13 at 16:09
Johan: you were taught the epsilon-delta definition of limits, continuity, etc. in high school? was this a regular class? –  Ori Gurel-Gurevich May 18 '13 at 21:11
Ori: We were taught the theorem, not the proof. And continuity was explained only by examples. When I told the teacher (tongue-in-cheek) I didn't believe the theorem, she laughed and gave me a calculus book (I liked Martin Gardner's books better though). –  Johan Wästlund May 18 '13 at 21:56
I learnt the epsilon-delta definition of limits in 1990 in the regular calculus class at a regular high school in Nebraska. (I can't recall if we learnt the definition of the Riemann integral or did all of the proofs, but these were in the book.) I didn't learn that this was unusual until years later. –  Toby Bartels May 18 '13 at 22:29

The proof that the trace is well defined for square matrices looked like symbol pushing to me. Many years later I realized that the proof is nonsense if you live in certain infinite dimensional worlds. Then I finally understood what was going on in finite dimensions.

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Bill, that one still bothers me so it seems I have yet to understand what is going on in finite dimensions. Is the answer somewhere or should I ask on MathOverflow? –  François G. Dorais Jun 29 '13 at 17:07
@FrançoisG.Dorais Do any of these answers help? mathoverflow.net/questions/13526/… –  Yemon Choi Jun 29 '13 at 20:00
@YemonChoi: Interesting. Is your suggestion that, in the finite dimensional case, there is a Haar measure that gives unit ball mass 1? Ultimately, local compactness is the essential thing? –  François G. Dorais Jun 29 '13 at 20:14
@FrançoisG.Dorais my own answer was just suggested as something geometric, so it isn't really the way I would suggest thinking about the trace (esp over positive characteristic). But I think some other answers suggest the following, which I learned from n-lab people (1) in finite dimensions, every operator is a finite sum of rank-one operators (2) there is a natural-in-category theory sense way to associate a number to a rank-one operator. Putting these together gives ${\rm End}(V) \cong V\otimes V^* \to k$. I don't know if this is what Bill had in mind –  Yemon Choi Jun 29 '13 at 20:19
I was aware of that argument too but I didn't feel it was enlightening. We'll have to wait and see what Bill had in mind. –  François G. Dorais Jun 29 '13 at 20:31

The Monty Hall problem. Here's an excerpt from the Wikipedia article if anyone hasn't heard of it:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Edit: As Douglas points out in the comment, there is the implicit assumption, as in the TV show, that the host always opens a door different from the one chosen by the player and always reveals a goat by this action.

I "understood" the proof in the sense that there seems no logical flaw. But I didn't in the sense that it didn't really convince me. It wasn't convincing enough until I heard the trick of considering the case when there are thousands of doors. Probability theory is evil.

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That statement of the Monty Hall problem is incomplete, as is mentioned in the Wikipedia article. –  Douglas Zare May 17 '13 at 5:34
@Douglas Thanks. I edited the post. –  Yuichiro Fujiwara May 17 '13 at 5:59
For me, the best way to see this is to measure the probability of winning given a strategy. Consider the "keep"-strategy, in which you do not switch. In this case, you need to hit the correct door to win, i.e., 1 in 3 chance. In the "switch"-strategy, you need to miss the correct door to win, which leads to a 2 in 3 chance of winning. –  Pål GD Jul 29 '13 at 8:28
In order to put the confusion to rest completely, one needs to take the usual false argument that says fifty fifty and then find a step (implicit or not) in that argument and say "Found it. This step right here is the wrong step". Ivan Li does just that in this quora answer, crux of which is realizing that information lies not just in the outcome, but also in how that outcome is generated. –  Jisang Yoo Nov 29 '13 at 18:41

The first proof of quadratic reciprocity I read in a book was using some figure, with several lines in it, and some lattice points. I did not really understand the argument and had the feeling that the proof might not be correct at all.

Later I saw proofs which were much clearer to me (for a discussion about proofs of quadratic reciprocity, see here: What's the "best" proof of quadratic reciprocity?).

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Are you refering to Eisenstein's wonderful geometric proof? math.nmsu.edu/~history/eisenstein/eisenstein.html –  Martin Brandenburg Aug 16 '13 at 12:54

I remember not understanding the proof of the fundamental theorem of calculus. My teacher, who was otherwise very good, didn't cover the proof; she told us we could look at it ourselves if we were curious. (This was a high school class.) I did take a look, but I couldn't make heads nor tails of it.

It so happens that I didn't encounter this proof again until I was a postdoc teaching second quarter calculus. I was relieved to learn that it was now quite trivial!

If I were teaching calculus at the high school level I wouldn't leave the students entirely on their own, but I also don't think I would want to tell them anything like a real proof of the fundamental theorem. I would probably be satisfied with trying to get across the general idea in an intuitive way.

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To prove the FTC, you first need a definition of the integral, and some textbooks don't even have that! –  Toby Bartels May 18 '13 at 21:58
It is possible to give an intuitive and very simple proof of FTC if you are willing to bullshit slightly: ms.unimelb.edu.au/~ram/Teaching/2006Fallmath221/LecNotesF2000/… –  Frank Thorne Jul 31 '13 at 2:51
@Frank: that's exactly how I explain it when I teach calculus. –  Nik Weaver Jul 31 '13 at 4:23

Castlenuovo's Contractibility criterion: Let S be a (smooth, projective, algebraic, complex) surface. Suppose by a minor miracle that S contains a smooth curve E, isomorphic to $\mathbb{P}^1$, which satisfies $E^2=-1$. Then $E$ may be blown down to get a $\textit{smooth}$ surface.

I tried learning this during my master's. I was fine with why $E$ could be blown down - you explicitly build the map by modifying a vample divisor. What I couldn't grasp for the life of me was why the resulting surface is smooth - I still don't know why (though I haven't look at it since)!

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Does this count as a proof at the undergraduate level? –  Qiaochu Yuan May 17 '13 at 4:30
No it does not - alas I am caught red handed not reading the entire original message! –  Robert Garbary May 17 '13 at 13:28

I did not initially understand the first isomorphism theorem. It was not until my first year of graduate school, when my wife wanted me to teach her some group theory, that I really hit on a natural sequence of ideas which made the first isomorphism theorem "click" for me. I talk about it here:

Does any textbook take this approach to the isomorphism theorems?

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I believe the only way to understand the proof of a seemingly trivial theorem is to consider a somewhat similar case where its statement fails. I didn't really understand the first isomorphism theorem for groups until I considered the case of semigroups, where it is formally wrong, and how its statement can be fixed. In the same spirit, calculus proofs of statements about real numbers become much more clear once you start working with p-adic numbers and general metric spaces. –  Anton Fetisov Aug 14 '13 at 21:36

I didn't feel I had really grokked Euler's magnificent identity

$$e^{ix} = \cos(x) + i \sin(x)$$

for a very long time. The first time I saw this formula, I think I was fourteen, and it was one of the most exciting discoveries of my young teenage life, answering many questions roiling around in my mind in one fell swoop.

It wasn't much later where I could drone my way through a proof using power series, but somehow a deeper intuition (connected with the exponential map from a Lie algebra to a Lie group, 1-parameter subgroups, logarithmic spirals, and the definition $e^x = \lim_n (1 + \frac{x}{n})^n$) didn't come until much, much later. I still think this is one of the most fantastic and beautiful formulas in mathematics.

Not a proof exactly, but a very basic concept that I feel I've only just begun to properly appreciate is -- wait for it -- the concept of equality. This is a humbling thing to admit. But the idea that equality, something we've all dealt with our entire lives, can be understood as an inductive notion: this came as a pretty big revelation to me, and this understanding plays a pretty important role in intensional type theory and homotopy type theory. I recommend Mike Shulman's MathCamp Notes (for high school students!) as a nice basic introduction for those who are curious. It's both simple, and incredibly deep.

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"it was one of the most exciting discoveries of my young teenage life" in general or within mathematics? ;) –  Martin Brandenburg Aug 16 '13 at 12:57
@MartinBrandenburg Both. –  Todd Trimble Aug 16 '13 at 13:49
I would like to note that I have just realized that $e^x = (e^{\frac{x}{n}})^n \approx (1+\frac{x}{n})^n$ for large $n$, by using the tangent line approximation at $0$. This was really the one part of the whole $e^{ix}$ story which I didn't quite have a firm grasp of. Additionally, one can view $(1+\frac{x}{n})^n$ as the result of Euler's method to DE $y'=y$, $y(0)=1$, on the interval $[0,x]$ with $n$ equal partitions. –  Steven Gubkin Sep 29 at 17:11
@StevenGubkin Yes, precisely! In terms of pictures, successive powers of $1 + \frac{ix}{n}$ make me think of a growing snail shell that for large $n$ hews fairly close to a circle. Once upon a time people thought of $e^{i\pi} + 1 = 0$ as something almost mystical, but such insights dispel the mystery and make it seem close to obvious. I still think it's wonderful! –  Todd Trimble Sep 29 at 17:53
@StevenGubkin Here's a pretty good gif, if you want to explain this to students one day: commons.wikimedia.org/wiki/File:ExpIPi.gif. –  Todd Trimble Sep 29 at 18:06

The Poincaré lemma. One can find an elementary proof in many places, but I've always found it to be rather mysterious. On the other hand, I think the proof becomes rather transparent if it is broken up into steps that directly use the following basic but important properties of cochain complexes and their cohomologies: (a) cochain maps induce maps in cohomology, (b) cochain homotopic maps induce the same map in cohomology, (c) a short exact sequence of cochain complexes induces a long exact sequence in cohomology. Of course, I became familiar and comfortable with these ideas much later than my first exposure to the elementary proof.

The role of Cartan's magic formula, $\mathcal{L}_X = d\iota_X + \iota_X d$, becomes very clear. It simply shows that the Lie derivative $\mathcal{L}_X$ is a cochain map that is homotopic to the zero map. This is enough to break up the long exact cohomology sequence induced by $\mathcal{L}_X$ and its kernel into a bunch of short exact sequences. Ultimately, this shows that the cohomology of the de Rham complex on $\mathbb{R}^n$ is concentrated in the lowest degree and represented by locally constant elements.

Most importantly, it is clear from the non-elementary presentation that the strategy of this proof is generalizable to other cochain complexes, which is not very obvious from its elementary presentation.

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I'd also like to contribute an answer on the fundamental theorem of calculus. The standard proof (to my knowledge, the only proof) of the "first fundamental theorem"

$$\int_a^b F'(x) dx = F(b) - F(a)$$

(when $F'$ is Riemann-integrable) looked and looks like a huge cheat (the proof is given, unnecessarily, on Wikipedia. Specifically, it cheats by computing the right-hand side rather than the left-hand side; it doesn't give much insight into how the antiderivative arises. When I saw this in high school I spent a month or two trying to put together alternative, direct proofs to remedy this; they were all wrong.

At some point I forgot the hypothesis that $F'$ is merely Riemann-integrable and proved the "second fundamental theorem"

$$\frac{d}{dx} \int_a^x f(t) dt = f(x)$$

for continuous $f$ in a very satisfactory way (probably basically the same as the one on Wikipedia as well); of course, this implies the first one under that more restrictive hypothesis. It gives no insight either into how the definition of the Riemann integral gives rise to an antiderivative, since it just uses very abstract linearity and locality properties of the integral.

I still don't really understand what is really going on with this contrast. Both theorems generalize to Lebesgue integrals and the second one has the same proof; the first one gets increasingly technical as you try to pin down the correct hypotheses. They both are based on good intuition but it baffles me why a minor adjustment in hypotheses entails a complete change in direction.

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I feel like I made a big breakthrough in my understanding of the fundamental theorem once I realized that when you use Euler's method on y'=f(x) given y(a)=0 and try to find y(b), the expression given by Euler's method is the riemann sum for the definite integral of f(x) from a to b. This shows why finding antiderivatives is linked to definite integration for me. –  Steven Gubkin May 17 '13 at 4:09
That's actually a great point, since Euler's method is the physically intuitive interpretation of the fundamental theorem, with $f'$ = velocity and $f$ = position. –  Ryan Reich May 17 '13 at 4:14
I think of the fundamental theorem of calculus as breaking up the interval into smaller subintervals and applying the tangent line approximation on each successive subinterval. –  Deane Yang May 17 '13 at 4:22
Or vice versa, using the difference quotient to estimate the derivative on each subinterval and observe that if you multiply each difference quotient by the size of the subinterval and add them all up, you get the difference between the values of the function at the endpoints of the full interval. –  Deane Yang May 17 '13 at 4:25
I had learnt the ‘first’ and ‘second’ fundamental theorems under the opposite names, and I'd almost forgotten that I once read a calculus textbook that had them switched, until I read your post now that also has them switched (switched compared to how I learnt them, I mean). I wonder how that came about? –  Toby Bartels May 18 '13 at 22:03

This may look silly, but while I'm capable to prove it by heart, with variations, I still don't fully understand what makes the Galois correspondence work.

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The one inclusion is trivial in each case, equality comes from degree considerations. Isn't that enough? –  Martin Brandenburg Aug 16 '13 at 13:00

During my undergraduate career, I did not quite grasp understand the proofs of results from classical cardinal arithmetic (even though I could confirm that they were logically correct). For example, the following result due to König:

Let $I\neq\emptyset$ and suppose that for every $i\in I$, $\lambda_i$ and $\kappa_i$ are cardinals such that $\kappa_i<\lambda_i$. Then, $\sum_{i\in I} \kappa_i<\prod_{i\in I}\lambda_i$,

Part of the issue was that I had yet to fully wrap my head around the idea of looking at $\prod_{i\in I} \lambda_i$ as a set of functions from $I$ into the ordinals with $f(i)\in\lambda_i$. I understood that this was how one viewed these structures, but I didn't have much intuition in that regard. However, I recently started reading through Abraham and Magidor's chapter in the Handbook and doing the exercises in there. When I looked back through Jech's Set Theory to brush up on classical cardinal arithmetic, I realized that many of these proofs are very natural.

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"Edgar Allan Poe:" "Any cryptosystem that human invented are breakable." So, any proofs are understandable. However until now, the zodiac 340 letter (or 340 symbol code) is not cracked.

For me, these two problems were very outstanding:

$1)$ Suppose there are infinite number of points in a plane and the distances between each two point is an integer, then they are all on a single straight line. This problem proposed by Sylvester and the solution by Paul Erdős is very strange for me.

$2)$ For $n\geq 2r$, the chromatic number of Kneser graph $K_{(n:r)}$ is $n-2r+2$. When I studied the proof of this theorem again and again (and also I understood it), I found that I can not understand it much more.

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Rosser’s trick.

It is the same idea as Gödel’s original proof​​​​​, we construct kind of a “liar paradox”, but while Gödel’s proof uses the obvious formal conditions which are needed to make the “paradox” formally work, Rosser’s trick drops all the intuitive arguments why the “liar paradox” is a “paradox” and replaces them by a formal, syntactic trick.

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Could you please add a few lines for explanation. –  Amir Asghari Aug 16 '13 at 8:36
@AmirAsghari Done. –  The User Aug 16 '13 at 11:40