我注意到，當我從電腦現代切換到拉丁現代時，分隔符號的大小發生了變化。在 12pt 字體大小的文件中，這會導致內聯數學\bigl(太大而無法容納，並擴展了該行佔用的垂直空間。這對現代計算機來說不是問題。這是一個錯誤嗎？
\documentclass[12pt]{article}
\usepackage{lmodern}
\usepackage{amsmath}
\begin{document}
An equivalence class~$[a]_{\sim} \in A / {\sim}$ consists of all the elements
in $A$ that are mapped to $b = f(a)$. By the axiom of choice, there exists a
choice function~$c \colon A / {\sim} \to A$ which selects a representative
element of each equivalence class. There exists a
function~$h \colon B \to A / {\sim}$, so that
$h(b) = h\bigl(f(a)\bigr) = [a]_{\sim}$. This allows us to construct the
function~$g = c \circ h$, which is in fact a right-inverse of $f$.
\[
\Biggl(\biggl(\Bigl(\bigl((X)\bigr)\Bigr)\biggr)\Biggr)
\]
\end{document}
計算機現代：
拉丁現代：
在拉丁現代案例中，第四行上方有一個間隙，而在電腦現代案例中則沒有。
編輯：
這個帖子egreg 解決了同樣的問題。
答案1
由於歷史原因，可縮放分隔符號使用固定大小，即使對於 cm，這顯然不是一個好主意（請參閱exscale包），但在這裡它根本不好，如果您以其自然縮放大小使用字體，那麼... …
\documentclass[12pt]{article}
\DeclareFontFamily{OMX}{lmex}{}
\DeclareFontShape{OMX}{lmex}{m}{n}{%
<->lmex10%
}{}
\usepackage{lmodern}
\usepackage{amsmath}
\begin{document}
\show\big
\showthe\baselineskip
\setbox0\hbox{$\big($}\showthe\dimexpr\ht0+\dp0
\showoutput
\showbox0
An equivalence class~$[a]_{\sim} \in A / {\sim}$ consists of all the elements
in $A$ that are mapped to $b = f(a)$. By the axiom of choice, there exists a
choice function~$c \colon A / {\sim} \to A$ which selects a representative
element of each equivalence class. There exists a
function~$h \colon B \to A / {\sim}$, so that
$h(b) = h\bigl(f(a)\bigr) = [a]_{\sim}$. This allows us to construct the
function~$g = c \circ h$, which is in fact a right-inverse of $f$.
\[
\Biggl(\biggl(\Bigl(\bigl((X)\bigr)\Bigr)\biggr)\Biggr)
\]
\end{document}
現在尺寸顯示為
> 14.5pt.
l.17 \showthe\baselineskip
?
> 14.40013pt.
剛好\big小於\baselineskip
您可以重新定義\big大小，使其保持在內部\baselineskip，從而避免\lineskip在段落內使用膠水。
\documentclass[12pt]{article}
\usepackage{lmodern}
\usepackage{amsmath}
\begin{document}
\makeatletter
\renewcommand{\big}{\bBigg@{0.92}}
\makeatother
\show\big
\showthe\baselineskip
\setbox0\hbox{$\big($}\showthe\dimexpr\ht0+\dp0
\showoutput
An equivalence class~$[a]_{\sim} \in A / {\sim}$ consists of all the elements
in $A$ that are mapped to $b = f(a)$. By the axiom of choice, there exists a
choice function~$c \colon A / {\sim} \to A$ which selects a representative
element of each equivalence class. There exists a
function~$h \colon B \to A / {\sim}$, so that
$h(b) = h\bigl(f(a)\bigr) = [a]_{\sim}$. This allows us to construct the
function~$g = c \circ h$, which is in fact a right-inverse of $f$.
\[
\Biggl(\biggl(\Bigl(\bigl((X)\bigr)\Bigr)\biggr)\Biggr)
\]
\end{document}
\show文件頂部顯示
> 14.5pt.
l.13 \showthe\baselineskip
?
> 13.24792pt.
<to be read again>
顯示\big(小於 14.5pt 基線 kip。
根據經驗選擇的 0.92：0.93 讓字體跳到下一個可用尺寸，這裡太大了。
答案2
您可以透過強制使用字體來解決問題cmex。我將使用exscale類似的設定amsfonts。
\documentclass[12pt]{article}
\usepackage{lmodern}
\usepackage{amsmath}
\DeclareFontFamily{OMX}{lmex}{}
\DeclareFontShape{OMX}{lmex}{m}{n}{%
<-7.5>cmex7%
<7.5-8.5>cmex8%
<8.5-9.5>cmex9%
<9.5->cmex10%
}{}%
\begin{document}
% \showoutput
An equivalence class~$[a]_{\sim} \in A / {\sim}$ consists of all the elements
in $A$ that are mapped to $b = f(a)$. By the axiom of choice, there exists a
choice function~$c \colon A / {\sim} \to A$ which selects a representative
element of each equivalence class. There exists a
function~$h \colon B \to A / {\sim}$, so that
$h(b) = h\bigl({f}(a)\bigr) = [a]_{\sim}$. This allows us to construct the
function~$g = c \circ h$, which is in fact a right-inverse of $f$.
\[
\Biggl(\biggl(\Bigl(\bigl((X)\bigr)\Bigr)\biggr)\Biggr)
\]
\end{document}
