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Why are sometimes square brackets used to apply parameters to functions instead of the usual round parentheses?

For instance, in my probability course, they use $\text{P}[X]$ to denote the probability that some event in the set $X$ comes to pass.

$$\text{P}[X] = \sum_{x \in X} p(x)$$

Is there any rule as to when to use square brackets instead of parens or is this arbitrary?

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Who knows? As far as I can tell, it's arbitrary. I would avoid square brackets in general. –  Qiaochu Yuan Feb 26 '13 at 19:14
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I use brackets for stylistic reasons, with no different meaning from parentheses, whenever I judge that too many nested parentheses would look confusing/ugly; for instance I might write $f[g(x)]$ instead of $f(g(x))$. –  user7530 Feb 26 '13 at 19:14
    
I used to distinguish between $\mathbb{P}[\cdots]$ and $\mathbb{P}(\cdots)$: in the former, the description of the set to be measured goes between the brackets, and in the latter, the set to be measured goes between the parentheses. So, for example, $\mathbb{P}[X = x]$ but $\mathbb{P}(\{ x \})$. –  Zhen Lin Feb 26 '13 at 19:16

1 Answer 1

In Jech & Hrbacek's Introduction to Set Theory, the author adopt this notation to avoid confusion about images of sets and images of elements contained in such sets. For instance, is quite common denote $f^{-1}(\{x\}) $ by $f^{-1}(x)$; in the square brackets notation we'd write $f^{-1}[x]$, which is more clean than $f^{-1}(\{x\})$ and not so abusive as $f^{-1}(x)$. Other reason is sets of sets: if we consider a set $A = \{A_1,\dots, A_n\}$ and a function $f:A\to B$ it would not be didactic to write $f(A')$ for some $A'\subseteq A$, for the elements of $A$ is also denoted by capital letters.

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$f^{-1}(\{X\})$ and $f^{-1}[X]$ are different things. $f^{-1}[X]$ is the preimage of a subset $X$ of the codomain and $f^{-1}(x)$ is the preimage of an element of the codomain (only if f is injective). For example, consider a function $f$ from a two-element set $\{a,b\}$ to a two-element set $\{x, \{x\}\}$ such that $f(a) = x$ and $f(b) = \{x\}$. (Such sets are used e. g. used to construct the natural numbers from some axiom systems.) Then $f^{-1}(x) = a$, $f^{-1}(\{x\}) = b$, $f^{-1}[\{x\}] = \{a\}$, $f^{-1}[\{\{x\}\}] = \{b\}$. –  Jaan Jun 3 at 9:32
    
Or if $f$ is a function from $\{a,b\}$ to a three-element set $\{y, \{y\}, \{\{y\}\}\}$ such that $f(a) = \{y\}$ and $f(b) = \{\{y\}\}$, and we denote $x = \{y\}$, then $f^{-1}(x) = a$, $f^{-1}(\{x\}) = b$, $f^{-1}[x] = \emptyset$, $f^{-1}[\{x\}] = \{a\}$, $f^{-1}[\{\{x\}\}] = \{b\}$. (So $f^{-1}(\{x\}) \neq f^{-1}[x]$.) –  Jaan Jun 3 at 9:39
    
There are some authors that write $f^{-1}(\{x\})$ as $f^{-1}(x)$. –  Paulo Henrique Jun 5 at 18:14
    
I guess you meant $f^{-1}[\{x\}]$ by the notation in my comments when you wrote $f^{-1}(\{x\})$. The only consistent notation for preimages that I know of is that in my comments, and Jech & Hrbacek seem to use the same (I looked into the 1999 ed). But outside hard-core set theory one usually doesn't encounter such complex situations and so many authors are indeed a little sloppy and write ordinary brackets instead of square ones and identify one-element sets with their element. Maybe that's even not so bad. But I think that in your answer, $f^{-1}[x]$ should be changed to $f^{-1}[\{x\}]$. –  Jaan Jun 5 at 22:53

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