tcbtheorem 转至下一页

Question

breakable由于某种原因，尽管我已经启用并且有足够的空间让定理适合第一页，但我的 tcbtheorem 仍会转到下一页，但当我在定理之前输入随机内容时它就会起作用。

\documentclass{article}
\usepackage{amsmath,enumitem,parskip}
\usepackage[most]{tcolorbox}
\usepackage[bottom=3cm, right=3cm, left=3cm, top=3cm]{geometry}

\makeatletter
\newcommand\RedeclareMathOperator{%
  \@ifstar{\def\rmo@s{m}\rmo@redeclare}{\def\rmo@s{o}\rmo@redeclare}%
}
\newcommand\rmo@redeclare[2]{%
  \begingroup \escapechar\m@ne\xdef\@gtempa{{\string#1}}\endgroup
  \expandafter\@ifundefined\@gtempa
     {\@latex@error{\noexpand#1undefined}\@ehc}%
     \relax
  \expandafter\rmo@declmathop\rmo@s{#1}{#2}}
\newcommand\rmo@declmathop[3]{%
  \DeclareRobustCommand{#2}{\qopname\newmcodes@#1{#3}}%
}
\@onlypreamble\RedeclareMathOperator
\makeatother
\newenvironment{subtheorems}
 {\itemize[nosep,font=\normalfont\bfseries,
   leftmargin=3em,itemindent=-1em,align=left]}
 {\enditemize}

\newtcbtheorem{axm}{Axm}% 
{colback=red!10!white,colframe=pink!100!,fonttitle=\bfseries\color{black},separator sign none,description delimiters parenthesis,terminator sign={.},enhanced,breakable
}{th}

\newcommand{\R}{\mathbf{R}}

\title{18.012 Calculus Volume I by Thomas M. Apostol}
\author{Yamo Bueno the Sketchy Cafe}

\begin{document}

\maketitle

\section*{Introduction}
\subsection{A Set of Axioms for the Real-Number System}
\subsubsection{The field axioms}
\begin{axm}{Real Number Field Axioms}{}
    \begin{subtheorems}
        \item[(\(\R\)A0) Closure under addition] \[\forall x,y\in\R\left(x+y\in\R\right)\]
        \item[(\(\R\)A1) Associativity of addition]
    \[\forall x,y,z\in\R\left[\left(x+y\right)+z=x+\left(y+z\right)\right]\]
        \item[(\(\R\)A2) Commutativity of addition]
    \[\forall x,y\in\R\left(x+y=y+x\right)\]
        \item[(\(\R\)A3) Identity element for addition]
    \[\exists0\in\R\forall x\in\R\left(x+0=x=0+x\right)\]
        \item[(\(\R\)A4) Inverse elements for addition]
    \[\forall x\in\R\exists\left(-x\right)\in\R\left(x+\left(-x\right)=0=\left(-x\right)+x\right)\]
        \item[(\(\R\)M0) Closure under multiplication] \[\forall x,y\in\R\left(x\cdot y\in\R\right)\]
        \item[(\(\R\)M1) Associativity of multiplication]
    \[\forall x,y,z\in\R\left[\left(x\cdot y\right)z=x\left(y\cdot z\right)\right]\]
        \item[(\(\R\)M2) Commutativity of multiplication]
    \[\forall x,y\in\R\left(x\cdot y=y\cdot x\right)\]
        \item[(\(\R\)M3) Identity element for multiplication]
    \[\exists1\in\R\forall x\in\R\left(x\cdot 1=x=1\cdot x\right)\]
        \item[(\(\R\)M4) Inverse elements for multiplication]
    \[\forall x\in\R\exists\left(x^{-1}\right)\in\R\left[x\cdot \left(x^{-1}\right)=1=\left(x^{-1}\right)\cdot x\right]\]
        \item[(\(\R\)D) Distributivity of multiplication over addition]
    \[\forall x,y,z\in\R\left[x\cdot\left(y+z\right)=x\cdot y+x\cdot z\right]\]
    \end{subtheorems}
\end{axm}
\end{document}