This is kind of a subjective question, I know; often I find myself failing exams and homeworks because of the way i write down proofs. Either I don't know how to start, or somehow the main point of the proof is lost. I've noticed that in many books there's an "style" but doesn't matter how much i try to mimic it, i can't seem to do it right. I'm more of an algebra person, I love it, but Analysis, well... isn't my strong suit.
If you have any tips I'll appreciate it.

After a while of waiting, I couldn't get the writting that I wanted to post as an example, so here it is a "fresh" example of an excerisize I didn't finish:

  • If $S$ is the set of all the sequences of real numbers, for $\bar x=(x_i), \bar y=(y_i) \in S$ we define: $$d(\bar x, \bar y)= \sum_{i=0}^\infty \frac {|x_i -y_i|}{2^i(1+|x_i-y_i|)}$$ (a) Prove that $d(\bar x, \bar y)$ is a metric in $S$.
    (b) Let $\bar x^k=(x_i^k),\bar x=(x_i)\in S$. Prove that: $$\lim_{k\to \infty} d(\bar x^k,\bar x)=0 \Leftrightarrow \lim_{k\to \infty}x^k_i=x_i \quad \forall \; i\in \mathbb N$$

So I proved (a), and I don't feel there's much to say there. However with (b) I got in trouble very easyly.
($\Leftarrow$) We know that $x_i^k \to x_i$ if $k\to \infty$, that means that $\forall \; \varepsilon>0 \; \exists \; m\in \mathbb N$ such that $\forall k>m \;\; |x_i^k -x_i|<\varepsilon $. On the other hand, what we want to prove is $\forall \; \varepsilon>0 \; \exists \; N\in \mathbb N$ such that $\forall k>N$ $$|\sum_{i=0}^\infty \frac {|x_i^k -x_i|}{2^i(1+|x_i^k-x_i|)}|<\varepsilon$$ So what I tough is that, since we already have that $|x_i^k -x_i|<\varepsilon$ for any $\varepsilon >0$ so I did some "reverse engineering", I took the absolut value that I want to prove and started to operate: $$\mathbf {(1)}\;\;|\sum_{i=0}^\infty \frac {|x_i^k -x_i|}{2^i(1+|x_i^k-x_i|)}| = \sum_{i=0}^\infty \frac {|x_i^k -x_i|}{2^i(1+|x_i^k-x_i|)}<\varepsilon $$ $$\Rightarrow \sum_{i=0}^ n \frac {|x_i^k -x_i|}{2^i(1+|x_i^k-x_i|)}<\varepsilon ,\;\; if\;\; n\to \infty $$ $$\Rightarrow \frac {|x_i^k -x_i|}{2^i(1+|x_i^k-x_i|)}<\varepsilon $$ $$\Rightarrow \frac {|x_i^k -x_i|}{1+|x_i^k-x_i|}<2^i \varepsilon, \;\;where\;\; 2^i\;\;is\;\;constant\;\;with\;\;respect\;\;to\;\; k$$ From here, I'm pretty much frozen. What I wanted was to get to what we already had: $|x_i^k -x_i|<\varepsilon$, but I couldn't, somehow it seems futile. In class we already saw how to do it, and it is completly different from what I tried.

  • 2
    $\begingroup$ Draw pictures! Most ideas can be drawn. Try to read proof from book and draw pictures of what they say in equations. $\endgroup$
    – tom
    Aug 26, 2013 at 20:05
  • 4
    $\begingroup$ This question might be a bit too broad to answer effectively. I think the best way to get helpful feedback for this issue might be to post a particular proof that you've written, where you've "somehow lost the main point", and ask how it could be improved. $\endgroup$ Aug 26, 2013 at 20:07
  • 3
    $\begingroup$ To me, writing proofs has less to do with what field you're working in, and more to do with the way we do mathematics. Perhaps the best way to pick it up (the way most undergrad math majors do in my institution) is taking a bit of logic and/or elementary set theory; working with engineering majors, I feel that's the main gap in the math section of their curriculum. $\endgroup$ Aug 26, 2013 at 20:13
  • 5
    $\begingroup$ I second Omnomnomnom's suggestion. Post a "bad" proof and we'll tell you what's wrong, how to correct it, and what general principles are violated. Otherwise, I'll start my answer with "make sure that each letter is written legibly and is distinguishable from any other one" (no laughing: this is a major problem with many calculus students...) $\endgroup$
    – fedja
    Aug 26, 2013 at 20:18
  • 2
    $\begingroup$ Take several textbooks and compare the proofs of the fundamental theorems. $\endgroup$ Aug 27, 2013 at 7:18

1 Answer 1


I can't shake the feeling that you should try your hand with simpler proofs first. I am trying to address here some of the difficulties that I spotted by making comparisons with a simpler proof, and then discussing the extra difficulties brought about by this proof. This will be long, and not necessarily help you much towards the end. From the point of view of your teacher the most pressing concern is that your proof should show that you understood the need to do things differently.

Let's try the following. In $\mathbb{R}^2$ we use the metric $$ d((x_1,x_2),(y_1,y_2))=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}. $$ The analogue of your task is to prove that if (the $k$ is a superscript, not an exponent) $$\lim_{k\to\infty} x_i^k=x_i$$ for $i=1$ and $i=2$, then $$ \lim_{k\to\infty}(x_1^k,x_2^k)=(x_1,x_2). $$ So you are asked to show that no matter how small $\varepsilon>0$ you are given, you can produce a lower bound $m$ such that the inequality $$ d((x_1^k,x_2^k),(x_1,x_2))<\varepsilon $$ will hold, if $k>m$. Let's take a look at that distance. We want to see that $$ \sqrt{(x_1^k-x_1)^2+(x_2^k-x_2)^2}<\varepsilon $$ for large enough $k$. Getting rid of that pesky square root is easy, so instead we want to show that $$ (x_1^k-x_1)^2+(x_2^k-x_2)^2<\varepsilon^2, $$ again for all large enough indices $k$. Ok. Not too hard as for large $k$ both $|x_1^k-x_1|$ and $|x_2^k-x_2|$ become small. We do our usual "split the $\varepsilon$" business. Our assumptions imply the existence of boundaries $m_1$ and $m_2$ such that A) $|x_1^k-x_1|<\varepsilon/\sqrt2$ whenever $k>m_1$ and B) $|x_2^k-x_2|<\varepsilon/\sqrt2$ whenever $k>m_2$. So if $k>m:=\max\{m_1,m_2\}$ both of these inequalities hold simultaneously, and squaring them and adding them up proves the desired inequality for all $k>m$.

A few remarks about this. There was a bit of "reverse engineering". But the reason for that part was simply to get an idea what kind of upper bounds we must place on the differences $|x_i^k-x_i|$ so that the desired inequality will follow. Note the direction of the flow of logic. The desired inequality must be a consequence of the simpler inequalities. Even though we reverse engineer here, the direction of the flow of implications must not be reversed. Another key point was that there were only finitely many terms contributing to the distance. Therefore we could follow the simple idea of splitting the elbow room provided by that $\varepsilon^2$ equally between the two terms that we can control using our assumptions. Furthermore, the assumptions gave two lower bounds for the index $k$. It was easy to make sure that both of them are satisfied by taking the maximum of the two lower bounds as the lower bound for this new process.

Let's move from $\mathbb{R}^2$ to $\mathbb{R}^n$. No major changes. We can still split that $\varepsilon^2$ evenly between the $n$ terms. Each of the $n$ terms will be below its allotted share $\varepsilon/\sqrt{n}$ from some lower bound $m_i$ onwards. Again we can use $m=\max\{m_1,\ldots,m_n\}$ as the lower bound for the limit process being studied, as we can find the largest among a finite set of numbers.

Let's try something harder still and move to $\mathbb{R}^\omega$, the infinite dimensional space of sequences $(x_i)$. Let's try the metric $$ d((x_i),(y_i))=\sqrt{\sum_{i=0}^\infty(x_i-y_i)^2}. $$ The same task. We assume that $x_i^k\to x_i$ as $k\to\infty$ for all $i$. Does it follow that $d((x_i^k),(x_i))\to0$? Well, there are several problems before we actually get to this point. For the metric to make sense, we need the sequences to be square summable. Ok, so we restrict our space to square summable sequences. Then we need to worry about the square summability of the componentwise limit sequence $(x_i)$. Let's assume that is the case, as it is besides my main points. The main point is that the earlier proof method no longer works. There are infinitely many terms contributing to the distance, so we cannot split the elbow room of $\varepsilon$ evenly among them. Another major obstacle will be that even though we can control any component of the sequence, and find a lower bound $m_i$ such that $|x_i^k-x_i|$ will be as small as we wish, whenever $k>m_i$, but we cannot make all of these to hold simultaneously. This is because there may not be a largest one among the bounds $m_i$ now that there are infinitely many of them.

So what? Well the first obstacle can be overcome. We don't need to split $\varepsilon$ evenly for the reverse engineering part to succeed. We can give to the squared first component the elbow room of $\varepsilon^2/2$, $\varepsilon^2/4$ to the second $\varepsilon^2/8$ to the third and continute in the style of a geometric progression. Each and every component still gets a positive amount of elbow room, which means that we can control its contribution. But the second obstacle cannot be overcome. The theorem is actually false, but I skip the part of giving a counterexample. With this metric we simply do not have this result.

Well, that's why you are hit with this different metric. A key difference is that the contribution of the $i$th components $$ \frac{|x_i-y_i|}{2^i(1+|x_i-y_i|)}<\frac1{2^i} $$ IRRESPECTIVE of the values of $x_i$ and $y_i$. This means that the larger indices will, so to speak, take care of themselves. A proof should reflect that. This gives us a way around the second obstacle above. We give half the elbow room to the infinite tail, and split the remaining elbow room among the finite early parts in some sensible way.

So this proof is somewhat delicate. I fear that your difficulties originate from an earlier course, where you didn't pay due attention to the need to have a way of producing that lower bound $m$ given a fixed $\varepsilon$. If you were not aware of these obstacles along the way, there was zero change for your proof to overcome them. In math there is no monster-truck option of plowing through an obstacle. Your problem is not one of style. My advice is to revisit earlier textbooks and to use the office hour of your instructor. Fixing this will take quite a bit of work. Give yourself some time.

  • $\begingroup$ Thank you so much! I appreciate this. I don't know how aware I was, but I can't deny that I'm a little bummed out. I thought it was an "style" thing because in algebra I don't have this types of problems, I can usually write well and do it well. On the other hand, I'm helping my former calculus teacher (that for some reason belives in me) as some sort of calculus-advisor for his students (first semester) and cont... $\endgroup$
    – Ana Galois
    Sep 1, 2013 at 2:35
  • $\begingroup$ ...cont. and I'm noticing that correcting their mistakes is helping me a lot, but we're just starting and I'm always afraid to do something incorrectly or not noticing their mistakes on time, anyway I'm only an advisor, I'm not forced to "be right all the time" like he says, but still worries me, that I'm an anlysis student, and I'm not able to do things right. Again, thanks a lot. $\endgroup$
    – Ana Galois
    Sep 1, 2013 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.