Proper use of indicator function

Question

Given a set $X$ and a subset $A \subseteq X$ the indicator function $\boldsymbol{1}_{A} : X \rightarrow \{0,1\}$ of $A$ is defined as $$\boldsymbol{1}_{A}(x) = \begin{cases} 1 & \text{if } x \in A \\ 0 & \text{otherwise} \end{cases}$$

However, in papers I often see it used as

$$\boldsymbol{1}(x) = \begin{cases} 1 & \text{if x is true} \\ 0 & \text{otherwise} \end{cases}$$

For example $\boldsymbol{1}(2 \leq 3) = 1$. I understand that this is very convenient. In a paper I'd like to write

$$\boldsymbol{1}(x_{1}^{i} \leq x_{1}, \ldots, x_{d}^{i} \leq x_{d})$$ so that it equals $1$ if $x_{1}^{i} \leq x_{1}, \ldots, x_{d}^{i} \leq x_{d}$.

I'm concerned that this might be sloppy and technically incorrect since it is no longer really a set function. Would something like

$$\boldsymbol{1}_{]-\infty, x_{1}] \times \cdots \times]-\infty, x_{d}}(\boldsymbol{x})$$ where $\boldsymbol{x} = (x_{1}^{i}, \ldots, x_{d}^{i})^{T}$ be "better" and "more" correct (It does seem to agree with the above definition as a set function)? It does seem notationally more burdensome.

Answer 1

As long as you clearly define what the notation means and it doesn't contradict the conventions of your field, this is unlikely to cause an issue. At worst, the referee would probably advise that you change the notation or write out the statement in words (or as a predicate) and resubmit the paper.

Answer 2

Why wouldn't it be a set function? Fix a (first-order) language, i.e. a set of variable symbols, function symbols, relation symbols, constant symbols, and then some logical symbols (like commas, the logical connectives, equality symbol, quantifiers, and parentheses) and then take as a set all the well-formed formula in that language. This is a set.

Additionally, the solution you come up with makes the subset that the indicator function is defined over depend on the input, so it isn't actually a function taking in $(x_1^i,\ldots, x_d^i)^\intercal$.

If you're really worried about the set of well-formed formula being a set (and don't want to deal with the rigorous logical stuff), then you could abandon the indicator function stuff altogether and just define $$f(x_1,\ldots, x_d)= \begin{cases} 1 & \text{if $x_j^i\leq x_j$ for each $j$} \\ 0 & \text{otherwise} \end{cases}$$ This has domain $X^d$, and so is a set, meaning that $f$ is actually a set function.