1. Let $a$ and $b$ be parameters.

How can we say formally that if $d_{ij}\in [a,b]$, then $c_{ij}=f(d_{ij})$ and if $d_{ij}\notin [a,b]$, then $c_{ij}=0$?

  1. I want to define $M_{i}$ as the combination of $j$ indexes while $d_{ij}< b$. Is the following notation correct?

$$M_{i}=\{j \in J :d_{ij}< b\}$$

  • 4
    $\begingroup$ Depends on what you mean by "formal". I would not be against simply $$c_{i,j} = \begin{cases} 0, & d_{i,j} \not \in [a,b] \\ f(d_{i,j}), & d_{i,j} \in [a,b] \end{cases}$$ or, if one wants to use the language of characteristic functions, $$\mathbf{1}_A(x) := \begin{cases} 1, & x \in A \\ 0, & x \not \in A \end{cases}$$ then you could say $$c_{i,j} = f(d_{i,j}) \cdot \mathbf{1}_{[a,b]}(d_{i,j})$$ These are ultimately just the same thing; I wouldn't overcomplicate it unless you have a particular need in mind. $\endgroup$ Commented Jan 4, 2023 at 7:37
  • $\begingroup$ What do you mean when you say "combination"? If you're meaning a set then you should be using curly braces for set-builder notation. $\endgroup$
    – Lemmon
    Commented Jan 4, 2023 at 7:43
  • $\begingroup$ Yes, I meant the curly notation. Thanks for the heads up! I am editing now. After editing, do you think it is correct now? $\endgroup$
    – madetolast
    Commented Jan 4, 2023 at 7:46

1 Answer 1

  1. You can use an indicator function, (also known as a characteristic function). Then, we would notate this as

$$c_{ij} = \chi_{[a,b]}(d_{ij}) f(d_{ij})$$

(Edit: I agree with PrincessEev that for most uses a definition by cases would tend to be more readable.)

  1. Yes, that notation is perfect. You have defined $M_i$ as the set of all $j$ in $J$ such that $d_{ij}$ is less than $b$.

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