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Jun 22 at 1:44 history edited Marco Ripà CC BY-SA 4.0
Fixing square brackets
Feb 1 at 8:58 comment added Emil Jeřábek A meta post related to this question: meta.mathoverflow.net/questions/5862/a-mathjax-latex-question
Jan 31 at 15:21 history edited Marco Ripà
Added tag "soft question" as already stated at the beginning of the text
Jan 31 at 15:19 vote accept Marco Ripà
Jan 31 at 15:18 comment added Marco Ripà Thank you @Gro-Tsen! I've just taken another look at the definition and I finally agree that knowing the set notation $X^n:= X \times \cdots \times X$ (where $X$ appears exactly $n$ times) can be taken as a minimum requirement for any reader of a graph theory article. Then, if I use $\left[0,4-\sqrt{3}\right] \times \left[0,4-\sqrt{3}\right] \times [0, 2]^{k-2}$ in the abstract, it is more than reasonable to be allowed to write also $[0,2] \times \left[0,4-\sqrt{3}\right] \times \left[0,4-\sqrt{3}\right] \times [0, 2]^{k-3}$ in Section 2.
Jan 31 at 14:04 history edited YCor
edited tags
Jan 31 at 0:15 review Close votes
Feb 13 at 3:06
Jan 30 at 23:43 comment added Gro-Tsen (re my previous comment: apparently MathJax doesn't have \bigboxtimes, but I think you can guess what you should read)
Jan 30 at 23:42 comment added Gro-Tsen For what it's worth, I like the convention that if $\boxtimes$ is some kind of multiplication operator, then the $\boxtimes$-product $\bigboxtimes_{i=1}^n X$ of $n$ copies of the same object $X$ is written $X^{\boxtimes n}$. So for example $E^{\otimes 2}$ for the tensor square, and so on. So if you want to insist that you're taking the $n$-fold Cartesian product of a set, you can write it $X^{\times n}$.
Jan 30 at 23:16 history edited Michael Hardy CC BY-SA 4.0
edited title
Jan 30 at 22:33 answer added Iosif Pinelis timeline score: 4
Jan 30 at 21:42 history asked Marco Ripà CC BY-SA 4.0