I am an undergraduate student trying to read the text "Calculus on Manifolds" by Spivak. I have some doubts, in general, about how to write the exercises in this book (or another books) that involves lots of cumbersome notation. Before giving an example, I want to say that as an undergraduate, some teachers asks to students to write very clear proofs and to prove every assertion we made. I think that sometimes, making that very-clear-proofs are very very inefficient or hard. Now, the example that I would like to share is the following (at the end, I will add some comments that I really want to discuss; thanks in advance for taking your time reading this).

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I'll share with you the "$(\Leftarrow)$" part of my proof.

Suppose that for all subrectangles $S_i$ of $P,$ the function $f\vert_{S_i}$ is integrable over $S_i.$ If $S$ is a subrectangle of $P,$ then for all $\varepsilon>0$ we can find a partition $P_S$ of $S$ s.t. $U(f\vert_S,P_S)-L(f\vert_S,P_S) < \varepsilon/N$ (where $N$ is the number of subrectangles of $P$). Lets write $P_S=\{ P_{1,S}, \ldots, P_{n,S} \}.$ Let $Q=\{ \cup_S P_{1,S}, \ldots, \cup_S P_{n,S} \}.$ If we write $A=[a_1,b_1]\times \cdots \times [a_n,b_n],$ is not hard to verify that $\cup_S P_{i,S}$ is a partition of the interval $[a_i,b_i].$ Hence $Q$ is a partition of $A.$ Now, for all subrectangle $S_i$ of $P \quad (1 \leq i \leq N),$ there are subrectangles $S_{1,i}, \ldots, S_{\alpha,i}$ of $Q$ such that $S_i=S_{1,i}\cup \cdots \cup S_{\alpha,i},$ which determine a refinement $Q_{S_i}$ of the partition $P_{S_i}.$ Then we have $U(f\vert_{S_i}, Q_{S_i})-L(f\vert_{S_i},Q_{S_i})<\varepsilon/N.$ Therefore \begin{align*} U(f,Q)-L(f,Q) =& \sum_{R}[M_R(f)-m_R(f)]v(R) \\ =& \sum_{i}\sum_{j} [M_{S_{j,i}}(f) - m_{S_{j,i}}(f)]v(S_{j,i}) \\ =& \sum_{i} \left[ U(f, Q_{S_i})-L(f,Q_{S_i}) \right] = \sum_{i} \left[ U(f\vert_{S_i}, Q_{S_i})-L(f\vert_{S_i},Q_{S_i}) \right] \\ <& \sum_{i} \varepsilon /N =\varepsilon. \end{align*} Then $f$ is integrable.

Note that there are several parts of my proof that are not proper and rigorously proved. For example, in this part: "is not hard to verify that $\cup_S P_{i,S}$ is a partition of the interval $[a_i,b_i].$ Hence $Q$ is a partition of $A.$ Now, for all subrectangle $S_i$ of $P \quad (1 \leq i \leq N),$ there are subrectangles $S_{1,i}, \ldots, S_{\alpha,i}$ of $Q$ such that $S_i=S_{1,i}\cup \cdots \cup S_{\alpha,i},$ which determine a refinement $Q_{S_i}$ of the partition $P_{S_i}$".

I tried to write proofs of that assertions (that, I consider, are intuitive if you draw rectangles in the plane), but I spent tons of time and made very cumbersome notation (that make the proof very unclear and really hard to follow) in proving those statements. So, it would be insanely hard to prove every assertion that I made in a proof.

As a self-learning student, what do you think that I should do in that kind of situations? It should be enough to me having intuitively clear those ideas that requires lots of notation machinery and time? or it should be better that I prove literally any assertion that I made?

What should you recommend? If you have some extra comments, it will be great if you share with me your thoughts.

Thanks a lot if you read all the text I wrote.

  • 1
    $\begingroup$ The thing is that when doing research, authors don't always prove every single detail of everything, as they trust that the readers will be able to fill the details themselves. If one tries to write every single detail to keep it self-contained, then how far do they have to go? Are they expected to write almost a book every time they want to publish a paper? Of course not, right? It is not about writing all details, it is about being able to do it when you really need to. This is what you are being trained for. $\endgroup$
    – Ivo Terek
    Commented Aug 19, 2020 at 3:48
  • $\begingroup$ Yes, it is important to be rigorous, and when learning something for the first time it is even more important to be careful since early misunderstandings will grow. Wet your proof: I don’t follow the logic. What is a subrectangle of the partition $P$? What are $a_i,b_i$? Making guesses, I don’t think the $[a_i,b_i]$ actually are a partition, so your picture might not be clear as you think. $\endgroup$
    – user208649
    Commented Aug 19, 2020 at 3:50
  • $\begingroup$ Terry Tao has a nice old blog post on this: There's more to mathematics thhan rigour and proofs $\endgroup$ Commented Aug 19, 2020 at 4:41
  • $\begingroup$ @CalvinKhor Thank you. I am reading his blog and it is very interesting $\endgroup$
    – rowcol
    Commented Aug 19, 2020 at 5:00
  • $\begingroup$ @CalvinKhor I read the question here, and immediately thought to link to that essay. It might be worth giving a brief summary of the essay as an answer, so that the link is not lost (if you don't get to it, and I find the time, I might do it myself). $\endgroup$
    – Xander Henderson
    Commented Aug 19, 2020 at 14:20

1 Answer 1


I think it is more important that you have the ability to translate such intuitive arguments into very precise ones, if the need arises. But of course to get to this stage you have to spend quite a bit of time in the beginning of your studies in proving almost every assertion with as much clarity as possible. For this, you of course have to be honest with yourself regarding what you're capable of translating into a precise proof vs what you're not, and if you're unable to do so then you should definitely attempt transcribe it into a precise proof.

An exception to this is as @Novice points out: you leave out a few details for now, perhaps because they're too cumbersome/ you do not see how to write it properly. If these are just very technical details, then you can of course skip them, focus on the big picture, and come back to it later (I found that by sometimes moving forward and continuing with the more difficult stuff, I get more used to the difficulty so that later when I return to the earlier stuff, it's not as difficult, and I'm able to solve it).

For example, I studied from both Spivak's Calculus (C) and Calculus on Manifolds (CoM), and what I found is that the integration chapter (atleast the beginning) is very similar in both books, so there were several places where I just read the definitions in (CoM) and moved on because they were soooooo similar to (C). Also, there were a few problems in (CoM) which I wrote very brief proofs for because I had already done detailed proofs in (C), and I was confident that I'd be able to translate the $1$-dimensional argument into a completely rigorous $n$-dimensional argument.

Another example is for induction proofs. Usually a "mantra" that I follow is that "if it is true for two of them, it is true for finitely many by induction", and as such in many cases I do not explicitly write out the full induction proof, because I know that I am definitely able to write up a precise argument should the need arise. Examples of this include assertions like:

  • Suppose $V$ is a vector space over a field. One of the axioms is being closed under addition: if $v,w\in V$ then $v+w\in V$. Prove that for any $n\in \Bbb{N}$, if $v_1,\dots, v_n\in V$ then $\sum_{i=1}^n v_i \in V$ (i.e if $V$ is closed under addition of two elements, then it is closed under finite sums).

  • If $C_1, \dots, C_k$ are closed subsets of $\Bbb{R}^n$ then the union $\bigcup_{i=1}^k C_i$ is also closed. To prove this statement, I would simply prove the case $k=2$, and say that the general case follows by induction, because I know exactly how to write it out explicitly.

For your question specifically, what I would do is to reduce the general case to the case where there are only two subrectangles using induction (again, this follows the mantra of "if it holds for two, it holds for finitely many"). If you want to go a step further, then I would actually suggest trying to do things in increasing difficulty as follows:

  • First start with the $1$-dimensional case as follows. Let $f:[a,b]\to \Bbb{R}$ be a given function, and $c\in [a,b]$. Then, $f$ is integrable on $[a,b]$ if and only if $f|_{[a,c]}$ and $f|_{[c,b]}$ are integrable, in which case $\int_{[a,b]}f = \int_{[a,c]}f + \int_{[c,b]} f$

  • Again in the $1$-dimensional case, (using induction) I'd show that for any partition $P= \{t_0, \dots, t_n\}$ of $[a,b]$, $f$ is integrable on $[a,b]$ if and only if each $f|_{[t_{i-1}, t_i]}$ is integrable, and that $\int_{[a,b]}f = \sum_{i=1}^n \int_{[t_{i-1}, t_i]}f$

Finally, generalize this to the $n$-dimensional case.

Sometimes, the notation and proofs get simplified by doing things from a more abstract perspective and by doing things in the general case right away, but sometimes, things are simplified by looking at the $1$-dimensional case. In this case, you should try to argue how the general case can be reduced to the special case (i.e why there is no loss of generality).

By the way, such "reductions", are very important to utilize in proofs (especially the tedious ones), because sometimes the "crux" of the matter lies in a relatively simple special case, while the general case follows by a few technical arguments (eg using induction or generalizing $1$-dimensions to $n$-dimensions), and you'll see a very clear example of this in Spivak's proof of the Change of Variables theorem later on.

As for your actual question, I think it could benefit from having a slightly different definition of "partition".


Let $A\subset \Bbb{R}^n$ be a closed rectangle. A partition of $A$ is a finite collection $\Pi$ of "almost-disjoint subrectangles". More precisely, $\Pi = \{S_i\}_{i=1}^k$ is a finite set, where each $S_i$ is a closed rectangle contained in $A$, such that $A = \bigcup_{i=1}^k S_i$, and such that for every $i,j\in \{1,\dots, k\}$, if $i\neq j$ then $\text{int}(S_i) \cap \text{int}(S_j) = \emptyset$.

This definition is equivalent to the definition of a partition as given by Spivak (i.e $P = (P_1, \dots, P_n)$ where each $P_i = \{t_{i,0}, \dots, t_{i,n_i}\}$ is a partition of $[a_i,b_i]$), where they are equivalent in the sense that there is a bijective correspondence between the partitions $\Pi$ as I have defined them here and the partitions $P$ as Spivak defines them (and if you really want to be completely rigorous, then write down explicitly what the bijection is... but I think this is unnecessary).

Spivak's definition is specifying the partition by telling you the endpoints of the subrectangles, while this definition is directly telling you the subrectangles, so it's just a small shift in perspective, and the only difference between the two is the manner in which one wishes to formalize a certain intuitive notion.

Now, if $\Pi_i$ is a partition of $S_i$, then $\Pi:= \bigcup_{i=1}^k \Pi_i$ will be a partition of $A$. Why?

  • $\bigcup_{R\in \Pi} R = \bigcup_{i=1}^k \bigcup_{R\in \Pi_i} R = \bigcup_{i=1}^k S_i = A$.

  • If $R,R' \in \Pi$ are distinct (i.e $R\neq R'$) then their interiors are disjoint because of the following two-case argument: case $1$ is $R\in \Pi_i$ and $R'\in S_j$ for some $i\neq j$. Then, $\text{int}(R)\subset \text{int}(S_i)$ while $\text{int}(R')\subset \text{int}(S_j)$ and $\text{int}(S_i) \cap \text{int}(S_j) = \emptyset$ (because $\{S_i\}_{i=1}^k$ is a partition of $A$), so that $\text{int}(R)\cap \text{int}(R') = \emptyset$. The second case is that $R,R' \in \Pi_i$ for some $i$. In this case, by definition of $\Pi_i$ being a partition, it is clear that interiors of $R,R'$ have empty intersection.

Then, from here, a direct calculation shows that $U(f,\Pi) = \sum_{i=1}^k U(f|_{S_i}, \Pi_i)$, and similarly for lower sums.

The reason I prefer this definition of partition for certain situations (like this one) is because to get a partition for $\bigcup S_i$, all you have to do is take the union of the partitions $\bigcup_i \Pi_i$ (as opposed to Spivak's definition for which you have to do things like $Q = \left(\bigcup_S P_{1,S}, \dots, \bigcup_S P_{n,S}\right)$, which set-theoretically is slightly more complicated). On the other hand, if you start with a partition $\Pi$ for $A$, then to get a partition for $S_i$ all you have to do is take $\Pi_i := \{R\cap S_i| \, R \in \Pi\}$. So, from a notational perspective I think it is simpler.

  • $\begingroup$ (I'll comment on your actual proof later on if I have time) $\endgroup$
    – peek-a-boo
    Commented Aug 19, 2020 at 4:13
  • $\begingroup$ Thanks! This is a very nice answer for my question $\endgroup$
    – rowcol
    Commented Aug 19, 2020 at 4:31
  • $\begingroup$ @AbrahamHernández I elaborated more on the particular problem. In this case I'd agree with you that the proof of your assertions while intuitively obvious, are tedious to write out explicitly, and not worth the effort. But once you convince yourself of the equivalence in the definitions of "partition", I think the matter becomes much simpler. $\endgroup$
    – peek-a-boo
    Commented Aug 19, 2020 at 6:04
  • $\begingroup$ (+1) This long and thorough answer hasn’t gotten enough upvotes, IMO. $\endgroup$ Commented Aug 19, 2020 at 12:12

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