I came across this set builder definition for the greatest integer function (which is also equal to the floor function) in my Discrete Mathematics course indicated below:
${[[x]]} = {\lfloor{x}\rfloor}$
${\lfloor{x}\rfloor} = {\max{\left\{{{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}
}\right\}}}$
My question is – is this equivalent to the following?:
${\max{\left\{{{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}}\right\}}} \overset{?}= {\max{\left({{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}}\right)}}$
More directly, is this equivalent to the following?:
${\lfloor{x}\rfloor} \overset{?}= {\max{\left({{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}}\right)}}$
Best Answer
Technically, the $\max$ is taken over a set, which is why braces are used in the definition given.
In many situations, where it's clear, the set description is omitted. For instance,
$$\max_{x\in[0,1]} f(x)$$
is used to mean $\max\{f(x)\,|\,x\in[0,1]\}$. Parentheses are not commonly used.