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Let's say I have a list of sets $S_i$, for $i=1,\ldots,n$. We often write the cartesian product of all these sets, with the exception of $S_k$ as:

$$S=S_1\times\cdots\times S_{k-1}\times S_{k+1}\times\cdots\times S_n$$

Is there a more succinct way to write it?

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In Wikipedia (https://en.wikipedia.org/wiki/Cartesian_product), I found something, which might be what you are looking for: $\prod_{n=1}^k \Bbb{R} = \Bbb{R}\times \Bbb{R} \times\cdots\times \Bbb{R} = \Bbb{R}^k$. So maybe something like this one is also valid: $$\prod_{\scriptstyle i = 1\atop\scriptstyle i \ne k}^nS_i$$

where $S_i$ is the $i^\text{th}$ set of the list you mentioned.

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    $\begingroup$ Possibly \prod_{i = 1\atop i \ne k}^n would be better? (with \atop separating lines instead of a comma) $\endgroup$ – CiaPan Dec 14 '17 at 11:15
  • $\begingroup$ Yes, you are right. Thank you for the formatting :) $\endgroup$ – ArsenBerk Dec 14 '17 at 11:16
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    $\begingroup$ I changed $\Bbb{R}$ x $\Bbb{R}$ x ... x $\Bbb{R}$ to $\Bbb{R} \times \Bbb{R} \times \cdots \times \Bbb{R}.$ That is proper MathJax usage. $\qquad$ $\endgroup$ – Michael Hardy Dec 14 '17 at 13:13
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    $\begingroup$ In addition to the atop business, another way to notate the indices would be with $\in$. I could see that being especially convenient if these indices are already in a set or are repeated throughout the work. If $E=\{1,\cdots,n\}\setminus\{k\}$ then you can have $$\prod_{i\in E}S_i$$ $\endgroup$ – let's have a breakdown Dec 14 '17 at 13:42
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    $\begingroup$ Another formatting comment: if you look carefully, your subscripts on the product symbol are in a smaller font than the superscript. IMHO it's better to avoid this by using \prod_{\scriptstyle i = 1\atop\scriptstyle i \ne k}^n: compare$$\prod_{i = 1\atop i \ne k}^nS_i\quad\hbox{and}\quad \prod_{\scriptstyle i = 1\atop\scriptstyle i \ne k}^nS_i$$ $\endgroup$ – David Dec 15 '17 at 1:36
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I have seen a notation for this kind of construction during some of my math lectures (but can't find a reference right now). This was mostly in the context of differential forms (e.g. interior product with vector), but can be applied to your case: $$ S_1\times \dotsm \times \widehat{S_k} \times\dotsm \times S_n := S_1\times \dotsm \times S_{k-1}\times S_{k+1} \times \dotsm S_n$$ The hat denotes the factor to be omitted. Note that this is not a universally standard notation, so even the professors that used it defined it at some point early in the lecture.

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    $\begingroup$ This is what I would use, along with a parenthetical remark along the lines of "where the hat denotes the factor to be omitted." I very seldom see anything as formal as in the other answers. $\endgroup$ – Matthew Leingang Dec 14 '17 at 15:00
  • $\begingroup$ Hatcher uses a notation similar to this in his text Algebraic Topology (see page 105 for an example). $\endgroup$ – Xander Henderson Dec 17 '17 at 2:52
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In general, we can write

$$S_1 \times \dots \times S_n := \prod_{i=1}^n S_i$$

and then we can apply all conventions we are used to.

As for your question, this can be written as:

$$\prod_{i = 1 \atop i \neq k}^n S_i$$

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Although it might not be common in set theory, it is common for game theorists to write $S_{-i}$ for $S_1 \times \cdots \times S_{i-1} \times S_{i+1} \times \cdots \times S_n$. See page 15 of chapter one of Osborne and Rubinstein's text on game theory, for example.

That notation is useful in game theory because, if $S_j$ represents the set of strategies available to player $j$, then one often needs to describe how all players except player $i$ have acted. Such a description will be a member of $S_i$. In particular, the notation becomes useful in defining a Nash equilibrium. See https://en.wikipedia.org/wiki/Nash_equilibrium.

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