[Tex/LaTex] Square version of \subsetneq

symbols

Does anyone know how to make a 'square version' of \subsetneq?
I tried \sqsubsetneq, but it doesn't work.

Thanks!
Joon

Best Answer

You need the mathabx package:

enter image description here

For such things I highly recommend trying: texdoc symbols

Code:

\documentclass[border=5mm]{standalone}
\usepackage{mathabx}
\begin{document}

  $X \sqsubsetneq B$

\end{document}

Edit: just to complete, here is a print screen from the documentation cited.

enter image description here

Related Solutions

[Tex/LaTex] \Theta with square

A solution with TikZ:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}

\newlength\thetawidth
\newlength\thetaheight

\newcommand*{\SquareTheta}{%
  \begingroup
    \setlength{\mathsurround}{0pt}%
    \settowidth{\thetawidth}{$\Theta$}%
    \settoheight{\thetaheight}{$\Theta$}%
    \begin{tikzpicture}[
      x=\thetawidth,
      y=\thetaheight,
    ]
    \draw[line width=.1\thetawidth, line cap=round]
      (0.125,.05\thetawidth) -- +($(0,1)-(0,.1\thetawidth)$)
      (0.875,.05\thetawidth) -- +($(0,1)-(0,.1\thetawidth)$);
    \draw[line width=.05\thetaheight, line cap=butt]
      (0.125,.025\thetaheight) -- +(0.75,0)
      ($(0.125,1)-(0,.025\thetaheight)$) -- +(0.75,0);
    \draw[line width=.1\thetawidth, line cap=butt]
      (0.275,0.5) -- (0.725,0.5);
    \draw[line width=.0375\thetaheight, line cap=butt]
      ($(0.275,0.5)-(0,.1\thetawidth)$) -- +(0,.2\thetawidth)
      ($(0.725,0.5)-(0,.1\thetawidth)$) -- +(0,.2\thetawidth);
    \end{tikzpicture}%
  \endgroup
}

\begin{document}
\renewcommand*{\arraystretch}{0}
\begin{tabular}{c}
  $\Theta$ \\[1pt]
  $\SquareTheta$ \\[1pt]
  $\Theta\SquareTheta$
\end{tabular}
\end{document}

[Tex/LaTex] How to produce dot between quantifiers

First case: square dot with equal spaceing

The first case is probably \centerdot of package amssymb:

\documentclass{article}
\usepackage{amssymb}
\begin{document}
\[
  \exists n \in \mathbb{N} \centerdot \forall m \in \mathbb{N} \centerdot m < n
\]
\let\oldcenterdot\centerdot
\renewcommand*{\centerdot}{\mathrel{\oldcenterdot}}%
\[
  \exists\,n \in \mathbb{N} \centerdot \forall\,m \in \mathbb{N} \centerdot m < n
\]
\end{document}

\centerdot is a binary operator. In the image in the question, the spacing seems to be slightly larger. Therefore the second line uses

the spacing of a relational operator by redefining \centerdot and
adds a thin space after the quantifiers.

Second case: round dot with asymmetrical spacing

In the second case a bold version of the dot could be used. The spacing seems to be a little special, the example uses

half of a thin space at the left and
a thick space at the right (as for relational operators).

\documentclass{article}
\usepackage{amsmath,amssymb}
\newcommand*{\anddot}{%
  \mathclose{}%
  \nonscript\mskip.5\thinmuskip
  \boldsymbol{.}%
  \;%
  \mathopen{}%
}

\begin{document}
\[
  \forall x \in \mathbb{R} \anddot x^2 \ge 0
\]
\end{document}