I'm currently in the following framework: $n \in \mathbb{N}$, $I \subseteq \{1, \dots, n\}$ and $X \subseteq \mathbb{R}$. Now, I need to define a set which contains all vectors $v \in \mathbb{R}^n$ such that the components of $v$ with indices in $I$ are elements of $X$ and the other components are zero. My naive approach to define this set looks like this

$$\{(x_1, \dots, x_n) \in \mathbb{R}^n: x_i \in X, i \in I, x_j = 0, j \in \{1, \dots, n\} \setminus I\}.$$

I don't like this definition as it contains two indices $i, j$ for the components. Also it is not immediately clear that the cases are well defined (each index is covered by a case and no index appears in both terms). In a normal cases environment one could write "else" for the second case. However a cases environment does not look nice within the braces of the surrounding set.

Is there some notation I am missing? How can I write this set more clearly? Of course I could use a description of the set as I did in my introduction of this question.


You can describe it very simply if you do a little preliminary setting up. Let $$Y_i=\begin{cases}X,&\text{if }i\in I\\\{0\},&\text{otherwise;}\end{cases}$$ the desired set is then $\prod\limits_{i=1}^nY_i$.

If you want everything in the curly braces, it’s going to be a bit ugly. One fairly readable version is

$$\{\langle x_1,\ldots,x_n\rangle\in\Bbb R^n:x_i\in X\text{ for all }i\in I,\text{ and }x_i=0\text{ otherwise}\}\,.$$

  • $\begingroup$ Thanks a lot for the answer! The hybrid solution in curly braces seems like a great option for me. The setup above is elegant as well, but would become confusing in the specific context I am working in. $\endgroup$ – user7802048 Feb 5 at 22:49
  • $\begingroup$ @user7802048: You’re very welcome. $\endgroup$ – Brian M. Scott Feb 5 at 22:50

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