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The purpose of this thread is to focus the attention of the community on posts that may require reopen and undeletion votes. A request should be posted as an answer below (one request per answer).

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    $\begingroup$ @BillDubuque The moderation team made that change to the instructions several years ago. This is a request from the moderators. Please do not change it again. $\endgroup$
    – Xander Henderson Mod
    Commented Aug 5 at 16:43

99 Answers 99

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Please reopen this question and this question.

Both questions was closed because of lacking context, however, the reason I want to reopen both of these questions is that both questions has a score of positive $4$ and $5$, which means it was well-received, even if the OP didn’t show any effort.

I think we must only close these questions once these questions received an answer.

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  • $\begingroup$ At that time , questions were loosely upvoted. This did not mean that they were suitable for this site. The questions are definitely off-topic for modern standards , so it would be wrong to reopen them. $\endgroup$
    – Peter
    Commented Mar 24 at 13:42
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    $\begingroup$ My suggestion would be that if you want to see Answers for such Questions, you ask them again with the required context. After all, your researched interest in them (beyond whatever vote scores they had) should supply a motivation for Readers to respond helpfully. Of course linking to the old Questions and checking that they have not in meantime been duplicated is a good practice. $\endgroup$
    – hardmath
    Commented Mar 24 at 16:15
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    $\begingroup$ @Peter: The close votes on those two questions took place seven and eight years ago, respectively. So I think we don't need to doubt the application of "modern standards" in their closure. $\endgroup$
    – hardmath
    Commented Mar 24 at 16:24
  • $\begingroup$ The two questions have now been deleted. It would appear that this was an unintended consequence of drawing attention to them. $\endgroup$
    – Gerry Myerson
    Commented Mar 24 at 21:13
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Please Consider This Question I had posted on Mathematics Stack Exchange Website. I was confused with this question and could not understand what to do and how to approach in this type of problems. I edited it many times But repeatedly they ask for providing additional information. I had provided context upto my knowledge but if they want more they might ask me for that in the comments what else they need. I will try to provide that. But none tells anything just closed it and deleted it.

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    $\begingroup$ I feel like the "apparent" lack of effort is a problem here. Apart from writing the definitions of $f([x])$ and $f(f([x]) + 1)$ and saying "it doesn't yield anything", you haven't done anything. If you're clueless, take a look at the question here. Bullet points 2 and 3 are applicable : you haven't tried a specific $f$, and "lower settings" indicate a simple $f$. Maybe try $f(x) = x$? $f(x) = x+1$? Report your observations. Make people interested in your question. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Apr 2 at 13:40
  • $\begingroup$ Helpful I will keep this in mind next time. Thank you. $\endgroup$
    – user1304613
    Commented Apr 2 at 14:25
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I request the reopening of the question "How to derive the formula $(I-BA)^{-1} = I - B(I+AB)^{-1}A$", which was closed as a duplicate. I answered the question.

OP explicitly acknowledges that he is aware of the formula and its standard proof (which is the only proof detailed in the answers to the question linked as duplicated). I believe the question is essentially different as it asks for approaches to finding the inverse without knowing the formula beforehand.

New contributor
Vladimir Lysikov is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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  • $\begingroup$ The linked duplicate is, itself, closed as a duplicate (I've added that dupe target to the list). Note, also, this comment (math.stackexchange.com/questions/2054109/…), which leads to another answer which seems to address the question. $\endgroup$
    – Xander Henderson Mod
    Commented 2 days ago
  • $\begingroup$ Ok, sorry, I didn't notice that. I still believe that the question is different and may attract different interesting answers. $\endgroup$
    – Vladimir Lysikov
    Commented 2 days ago
  • $\begingroup$ I am not sure that I see how it is different. The question asks how one might discover the formula, and I linked to an answer which explicitly describes how the formula might be discovered. Does that not answer the question? $\endgroup$
    – Xander Henderson Mod
    Commented 2 days ago
  • $\begingroup$ Well, this is probably not the only way to discover the formula, so other answers may explain other ways. I'm not that familiar with the moderation style, so maybe it's not the intention to have questions like this. $\endgroup$
    – Vladimir Lysikov
    Commented 2 days ago
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    $\begingroup$ But other answers could easily be added to the linked duplicate, rather than to the new question. The goal is to consolidate, not disperse. $\endgroup$
    – Xander Henderson Mod
    Commented 2 days ago
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Please consider reopening Proving that $\langle x, y \rangle$ is a not principal ideal of $\mathbb{Q}[x, y]$

Although related, the current linked question, which is closed too, asking for an example, is not a duplicate. Moreover, the post is linked to a chain of questions that are all closed.

If this question is to be closed, there is supposed to be a better "duplicate" target: in the all closed chain of questions, NONE new answer is allowed to added.

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  • $\begingroup$ This proof in the linked dupe does exactly what the OP seeks (put $c = y\in\Bbb Q[y]$) so it is not clear what you mean by "better dupe target". There are many possible dupe targets (chase the links) for this common exercise, and nothing at all novel in the (duplicate) answers posted. [Your profile shows zero activity in abstract algebra. Experience in a field is usually necessary in order to properly assess duplicates]. $\endgroup$
    – Bill Dubuque
    Commented Aug 24, 2022 at 1:38
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    $\begingroup$ It seems that you have a very different definition of "dupe" from mine, whatever it is. Experience in a field $\Large{\neq}$ activities in your mathSE profile. "better dupe" means one does not need to "chase the links", not to mention that it is linked to a closed question that may be voted to be deleted by users. $\endgroup$
    – user1046533
    Commented Aug 24, 2022 at 1:41
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    $\begingroup$ In my experience it is - esp. for most users whose tags don't go much beyond calculus. Please explain what you mean by a "better dupe target". We can easily add more dupe links. $\endgroup$
    – Bill Dubuque
    Commented Aug 24, 2022 at 1:43
  • $\begingroup$ Yes, usually, maybe. $\endgroup$
    – user1046533
    Commented Aug 24, 2022 at 1:47
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    $\begingroup$ Your recent comments (and votes) on main made it clear that you are strongly against abstract dupes. But this is only a very slight abstraction - viz. specializing the nonzero nonunit $c$ to $y\in \Bbb Q[y]\,$ in the Lemma in the dupe. It would be greatly obfuscatory to have separate questions for various choices of nonunits $c$ in motley domains. A primary goal of abstract algebra is to prove theorems at their natural level of abstraction - so they apply as widely as possible. $\endgroup$
    – Bill Dubuque
    Commented Aug 24, 2022 at 2:02
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    $\begingroup$ "Your recent comments (and votes) on main made it clear that you are strongly against abstract dupes." Please don't put labels on me Bill. I do not want to discuss anything in general under this thread.// Comments are not supposed to be answers in main; one should address the issue Arturo mentioned in meta if one wants to handle abstract dupe: (cont.) $\endgroup$
    – user1046533
    Commented Aug 24, 2022 at 2:27
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    $\begingroup$ (cont.) *It would require the "survivor variant" to have answers that address both the general strategy (so as to be useful for future minor variants) and specific application to the question at hand (which shows how the general strategy applies to a specific problem). * $\endgroup$
    – user1046533
    Commented Aug 24, 2022 at 2:30
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    $\begingroup$ This is probably a more exact dupe target (containing the "survivor variant"). $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Aug 24, 2022 at 10:08
  • $\begingroup$ @Sarves But that is already true for the existing linked dupe (case $K = \Bbb Q$; the simple proof works for coefficient field $K$ and no simplification arises when $K = \Bbb Q$) $\endgroup$
    – Bill Dubuque
    Commented Aug 24, 2022 at 15:35
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    $\begingroup$ @BillDubuque You're right, the same proof works with $K$ replaced by $\mathbb Q$, but I think you should leave a comment mentioning this , however obvious it might seem. Heck, I'll do it. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Aug 24, 2022 at 15:51
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    $\begingroup$ @Sarvesh Any student studying ideal theory in rings that doesn't see immediately how to specialize a theorem on fields to the field $\Bbb Q$ has seriously put the cart before the horse. This goes back to my original comment - it is crucial for those assessing dupes to have significant experience in the topic in order to understand which inferences can be assumed trivial and left to the reader. $\endgroup$
    – Bill Dubuque
    Commented Aug 24, 2022 at 16:25
  • $\begingroup$ This post has been deleted. $\endgroup$
    – The Amplitwist
    Commented Nov 20, 2022 at 13:35
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I answered this question around 3 years ago.

7 months ago, it was commented, in bold no less, that there was an error. This brought my answer from a +4 score down to +2. I reviewed the comment and the supposed error did not feature in my answer. I explained as much in a comment back. This morning I woke up and my answer score is -1 and my answer has been deleted. Another comment was left yesterday, this time decrying my comment as handwaving and claiming that I'm making the same error (to reiterate: I'm not) and showing it doesn't work with a counterexample against the wording of the question (which is specifically about a set of consecutive integers, like my answer).

I've since edited my answer to flesh out the details a bit more and really spell it out, but it feels very unfair to have someone declare and insist upon an error without really trying to understand my answer.

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    $\begingroup$ I'm just reading your answer, and the first issue I have is that there is no quantifier attached to the $a$ in the first bit of working. I presume it is an $\exists$, and comes after the initial $\forall$? I'm then unsure precisely what the $k$ is, but I don't think we can "match" $1\cdot2\cdots n$ with the sequence $(k+1)(k+2)\cdots(k+n)$, because we have no control over the $a$ - unless the $a$ is actually quantified by a $\forall$. $\endgroup$
    – user1729
    Commented May 2, 2023 at 8:37
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    $\begingroup$ Another problem, which I think is the one BillDubuque is pointing out, is that your answer seems to only really rephrase the fact that "one in every $t$ consecutive integers is divisible by $t$." This is true, but more is needed here, as you will be double-counting these sequences. So you need to flesh out the claim "thus the same is true across the whole product". $\endgroup$
    – user1729
    Commented May 2, 2023 at 8:48
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    $\begingroup$ (For example, for $4!$ we need sequences of $4$, $3$ and $2$ integers in our product, and without further argument these sequences need to be disjoint, i.e. we need to rewrite $4!$ as the product of $4+3+2=9$ consecutive integers, but this is impossible!) $\endgroup$
    – user1729
    Commented May 2, 2023 at 8:50
  • $\begingroup$ I agree, $a$ ought to have it's own $\exists$. $a$ is decided by $k$ and $n$. If we are multiplying, as an example, $123\cdot \ldots \cdot 128$, then $n=6$ as there are six numbers, we pick $a=21$ as $21\cdot 6=126$ is in our set. Then our numbers go from $123\equiv 3 \pmod 6$ to $128\equiv 2 \pmod 6$. $\endgroup$
    – Rhys Hughes
    Commented May 2, 2023 at 9:07
  • $\begingroup$ Regarding the second problem, I can see the concern, however it works out with the same observation that we are cycling the residues $\mod n$. I fleshed out the details in what I added earlier, but in essence, if $n$ has a divisor $d$, there will be a number $x$ in our set with $x\equiv d\pmod n$. Take the same set I used earlier, we have $6=2\cdot 3$ and $123\equiv 3\pmod 6$ while $128=2\pmod 6$. $\endgroup$
    – Rhys Hughes
    Commented May 2, 2023 at 9:15
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    $\begingroup$ Instead of fleshing out the details below, you should edit your answer so that it is clearer in a first reading. $\endgroup$
    – user1729
    Commented May 2, 2023 at 14:02
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    $\begingroup$ The argument is either incorrect or seriously incomplete. It is impossible to decide which since you have not supplied rigorous justification of the huge gap, i.e. precisely how the OP's claim is inferred from "one in every $t$ consecutive integers is divisible by $t$". I doubt that anyone has any idea how you propose to do so based on what you have written. $\endgroup$
    – Bill Dubuque
    Commented May 2, 2023 at 18:05
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Undeleted by users then deleted by bot, then undeleted again by users, then deleted by post author on Jan. 30, 2024

Please undelete and reopen this:

https://math.stackexchange.com/q/4618260/104041

The question is mine.

I put an awful lot of effort into it. The downvotes confuse me and I have no clue why it was deleted (beyond the automatic deletion when there's enough downvotes with no answer).

The question is relevant to my research and it's important to have it easily accessible.

I believe it to be high quality.

The main feedback objections seems to be a matter of style, not substance: the notation used and the layout of the thing. It reads better on a mobile than on the main site, I believe; but what does it matter?

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    $\begingroup$ FYI this question was never closed, so there's no need to ask for a reopening. And the deletion was automatic, as per this criteria. $\endgroup$
    – KReiser
    Commented Nov 25, 2023 at 18:25
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undeleted by vote of users, then redeleted by OP

https://math.stackexchange.com/questions/4965439/requirements-for-a-subset-to-have-a-best-approximation-element-in-a-metric-space was closed, improved but left closed, and finally self-deleted after receiving a satisfactory answer in comment. I think the question was interesting and should be undeleted and reopened, and that @SassatelliGiulio's comment should be posted as an answer.

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    $\begingroup$ The answer is quite clearly false for a variety of metrics. Say the discrete metric. $\endgroup$
    – Arctic Char
    Commented Sep 2 at 6:34
  • $\begingroup$ Oops! Right. This (and the downvotes) makes me change my mind. Sorry for the inconvenience. Should I delete my request? $\endgroup$
    – Anne Bauval
    Commented Sep 2 at 12:26
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    $\begingroup$ For transparency I would prefer to keep it. In general I don't think any request in this thread should be deleted. $\endgroup$
    – Arctic Char
    Commented Sep 2 at 14:26
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Please reopen Introductory Group theory textbook.

It's illogical for this to be closed as "opinion-based", when introductory textbook recommendations remain open.

Books on Number Theory for Layman
Good Book On Combinatorics
What is the best book to learn probability?

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  • $\begingroup$ The three questions you link were all asked twelve years ago, which is like four generations on math.stackexchange. And you've now made them targets for closure.... $\endgroup$
    – Gerry Myerson
    Commented Jun 2, 2023 at 6:48
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    $\begingroup$ Now I've noticed the Group Theory one is also twelve years old (and it was posted by a fellow who was enrolled in my Group Theory course at the time!). $\endgroup$
    – Gerry Myerson
    Commented Jun 2, 2023 at 6:55
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    $\begingroup$ While I do think book recommendation questions have a place at MSE, but I am unsure about this specific one. The highest-voted answer is about Rotman's book, and I do not think this is a good answer (of course, this is my opinion - see the close reason! - but it does have some extremely complicated content, and for example the pretty standard "classification of f.g. abelian groups" comes after less-standard stuff on Classical and Mathieu groups). Moreover, this question is not going anywhere - it is just closed rather than deleted, and questions can still be closed as a duplicate of it. $\endgroup$
    – user1729
    Commented Jun 2, 2023 at 11:26
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Deleted (after failed Reopen review)

My own question

https://math.stackexchange.com/q/4962377/688539

I find the closure completely senseless. What is wrong with this question?

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    $\begingroup$ You've already gotten some feedback in a Comment on your Question. A couple of suggestions, without meaning to be repetitious. State or give a link to what you mean by "supremum axiom" for the real numbers and carefully restate how in your opinion it should be imposed on the rationals $\mathbb Q \subset \mathbb R$. At present the few lines in the body of your Question and the adjuration that answers should not be of the form "it has a supremum when it has a supremum" leaves it to the Reader to guess what positively is wanted. $\endgroup$
    – hardmath
    Commented Aug 24 at 14:29
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