1

I'm trying to create an enumerated list that has four items in math mode, three of which are enclosed by a large right curly brace.

I found a semi-solution that uses the multirow package, but it works for only two items in the curly brace section.

I'm hoping to find a solution that involves the enumerate or enumitem packages, because I want the ability to change the labeling of the enumeration to roman numerals. Some answers use the tikzmark package, but I was hoping for something less of a big machine.

If you have any elegant solution to my problem outside the ideas I have, thats fine too!

Here is a MWE for you to test (ignore most of the preamble):

\documentclass[12pt]{report}
\usepackage[margin=1in]{geometry}
\usepackage[most]{tcolorbox}
\usepackage{amsmath , amsthm , amssymb, mathtools}
\usepackage{multirow}
\ExplSyntaxOn

\NewDocumentCommand{\betternewtcbtheorem}{O{}mmmm}
{
    \newtcbtheorem[#1]{#2inner}{#3}{#4}{#5}
    \NewDocumentEnvironment{#2}{O{}}
    {
        \keys_set:nn { hushus/tcb } { ##1 }
        \hushus_tcb_begin:nVV {#2inner} \l__hushus_tcb_title_tl \l__hushus_tcb_label_tl
    }
    {
    \end{#2inner}
}
\cs_if_exist:cF { c@#5} { \newcounter{#5} }
}


\cs_new_protected:Nn \hushus_tcb_begin:nnn
{
\begin{#1}{#2}{#3}
}
\cs_generate_variant:Nn \hushus_tcb_begin:nnn { nVV }
\keys_define:nn { hushus/tcb }
{
    title .tl_set:N = \l__hushus_tcb_title_tl,
    label .tl_set:N = \l__hushus_tcb_label_tl,
}


\ExplSyntaxOff


\betternewtcbtheorem[number within = chapter]{dfn}{Definition}%
{
    enhanced,
    before title = {\stepcounter{dfn}},
    colback=blue!10,
    colframe=blue!35!black,
    fonttitle=\bfseries,
    top=3mm,
    attach boxed title to top left={xshift = 5mm, yshift=-1.5mm},
    boxed title style = {colback=blue!35!black}
}{dfn}


\newcommand\rdot[1][.5]{\mathbin{\vcenter{\hbox{\scalebox{#1}{$\bullet$}}}}}
\begin{document}


\begin{dfn}[title=Subring]
    A subring $S$ of a ring $R$ is a subgroup that is closed under multiplication. That is $S\subset R$ if $\forall a,b \in S$, [
    \begin{tabular}{ll}


    (1) $a+b\in S$ \quad (closure under $+$) & \multirow{3}{*}{{\LARGE \}} $S$ is a subgroup} \\
    (2) $0\in S$                                                 \\
    (3) $-a\in S$                                                    \\
    (4) $a\rdot b\in S$ (closure under $\rdot $)

\end{tabular}
\]



\end{dfn}


\end{document}

enter image description here

Hushus46
  • 211
  • A shameful "solution" in the absence of anything else: \multirow{3}{*}{{\scalebox{1}[4]{\}}} $S$ is a subgroup} – Ivan Feb 05 '21 at 00:27
  • You might do that with the \listliketab and bigdelim packages (Unrelated: needless to load amsmath if you load mathtools). – Bernard Feb 05 '21 at 00:32

1 Answers1

2

Set the construction using a nested tabular: The first 3 rows with a \left....\right\} with the last row as part of the outer/main tabular.

enter image description here

\begin{dfn}[title=Subring]
    A subring $S$ of a ring $R$ is a subgroup that is closed under multiplication. That is $S\subset R$ if $\forall a,b \in S$, \[
    \begin{tabular}{ l }
        $\left.\kern-\nulldelimiterspace
        \begin{tabular}{@{} l @{}}
            (1) $a + b \in S$ \quad (closure under $+$) \\
            (2) $0 \in S$                               \\
            (3) $-a \in S$
        \end{tabular}\right\} S \text{ is a subgroup}$ \\
            (4) $a \rdot b \in S$ (closure under $\rdot$)
        \end{tabular}
    \]
\end{dfn}
Werner
  • 603,163