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I am struggling to understand the notation i.e. what does the bold 1 refer to?

$\mathbf{1}_{A\cap B}(x)=\mathbf{1}_{A}(x)\times\mathbf{1}_{B}(x)$

2 Answers2

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It's presumably the indicator function \begin{align*} {\bf 1}_S(x) &= \begin{cases} 1 & \text{if $x\in S$}; \\ 0 & \text{if $x\not\in S$}. \end{cases} \end{align*} It's also (and more commonly, at least in my experience) denoted by $\chi_S$.

anomaly
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  • You are correct that it is more commonly referred to as $\chi_S$ (and in those contexts called the "characteristic function"). However IMHO it's a very good thing that the $\textbf{1}_S$ notation is gaining popularity, as I find it much more intuitive and easier on the eyes. – Ben W Jan 13 '21 at 22:23
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    @BenW: what's unintuitive about using the Greek for "ch" as the symbol for the characteristic function? My first reading of $\mathbf{1}_S$ would be the identity function on $S$ and I had to do a double take and think about the OP's equation to see that $\chi_S$ was intended. If you don't like $\chi_S$, wouldn't $\mathbf{I}_S$ be better? – Rob Arthan Jan 13 '21 at 22:39
  • @RobArthan it's a subjective thing I guess, but I have found that reading papers with $\textbf{1}_S$ is simply easier than with $\chi_S$. Not sure why that is, but that is what I have found. – Ben W Jan 13 '21 at 23:01
  • @BenW: fair enough: each to their own $\ddot{\smile}$. – Rob Arthan Jan 13 '21 at 23:06
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It's the indicator function. $$ \textbf{1}_A(x)=\left\{\begin{array}{ll}1&\text{ if }x\in A\\0&\text{ if }x\notin A\end{array}\right. $$

Ben W
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