# Best way to symbolically express this statement? [closed]

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?

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

• 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. Commented Jan 4, 2023 at 7:37
• 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. Commented Jan 4, 2023 at 7:43
• Yes, I meant the curly notation. Thanks for the heads up! I am editing now. After editing, do you think it is correct now? Commented Jan 4, 2023 at 7:46

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