[Tex/LaTex] \bigcdot symbol in LaTeX

Question

In my diploma thesis I need to LaTeX a concatenation over a set of sets.

I use the \cdot symbol for concatenation of two sets as in A \cdot B. (Please correct me if this is a wrong symbol for concatenation).

I would need to type a concatenation over a set of sets, using a \bigcdot symbol or an equivalent one.

It should be like when you use the \bigcup symbol to do a union over a set of sets, as in \bigcup_{i=1}^n A_{i}.

My LaTeX installation, however, does not know about such a symbol and I haven't found a clue on Google search either.

Answer 1

Besides what has been said in this post. You can also define a new "bigcdot" operator using scalerel package, for example:

\documentclass{article}
\usepackage{amsmath,scalerel}

\DeclareMathOperator*{\Bigcdot}{\scalerel*{\cdot}{\bigodot}}

\begin{document}

\[\Bigcdot_{i=1}^\infty A_i=A_1+A_2+\cdots\]

\end{document}

gives you:

EDIT:

I considered some alternative solutions so I will post them here.

As one can see, a problem with the scalerel solution is that it stretch the \cdot too wide such that the symbol become less appeal (at least to me). So how do we get a smaller dot with bigger "size"? For instance, let's say we want to have the dot in \bigodot (we can use \boldsymbol{\cdot} for the dot only) without the circle, how do we do it?

Well, there are at least two ways: first, we want to declare this new symbol as a math operator, such that superscript and subscript will go directly above and below it, then we either enclose the bold cdot with a box of total height of \bigodot, or we insert a invisible character of the size of \bigodot after the bold cdot, as illustrate by the code (both do not need any package other than amsmath):

\newsavebox\dotbox
\sbox{\dotbox}{\(\displaystyle\bigodot\)}
\DeclareMathOperator*{\bigcdot}{\raisebox{0pt}[\ht\dotbox][\dp\dotbox]{\(\boldsymbol{\cdot}\)}}

and

\newsavebox\dotbox
\sbox{\dotbox}{\(\displaystyle\bigodot\)}
\newlength{\dotheight}
\setlength{\dotheight}{\ht\dotbox}
\addtolength\dotheight{\dp\dotbox}
\DeclareMathOperator*{\bigcdot}{\boldsymbol{\cdot}\rule[-\dp\dotbox]{0pt}{\dotheight}}

That said, we can also redefine other "small" symbol, such as \bullet and \|, in similar ways. In the end, let's compare the results:

\[
\Bigcdot_{i=1}^\infty A_i=\bigccdot_{i=1}^\infty A_i=\bigcdot_{i=1}^\infty A_i=\bigodot_{i=1}^\infty A_i=\sum_{i=1}^\infty A_i
\]