通常在一段文本中编写内联数学公式时，行距保持不变，但请考虑以下示例
使用以下代码生成
\begin {document}
\textbf{$c_0$ is a complete metric space.} Suppose $x $ is a limit point of $c_0$. Then there exists $(x^{(k)} )_{k \in \mathbb{N}} \in c_0$ and such that $x^{(k)} \to x$. Let $\epsilon>0$ and pick $N$ such that $\|x^(k)-x\|_\infty < \frac{1}{2} \epsilon$ for $k \ge N$. Since $x^{(N)} \in c_0$, there is some $N'$ such that $|x_i^{(N)} | < \frac{1}{2} \epsilon$ for $i \ge N'$. Then $|x_i| \le |x_i^{(N)} | +|x_i-x_i^{(N)} | \le |x_i^{(N)} | +\|x-x^{(N) } \|_\infty <\epsilon$. Hence $x_i \to 0$ and so $x \in c_0$.
\end {document}
是否可以对其进行格式化，以使普通文本之间的行距保持不变？
非常感谢所提供的任何帮助！
\documentclass[a4paper]{article}
\usepackage{amsfonts,amsmath,parskip}
\usepackage{ebgaramond} %font
\addtolength\textwidth{2cm} % I have not used this
\begin {document}
\textbf{$c_0$ is a complete metric space.} Suppose $x $ is a limit
point of $c_0$. Then there exists
$(x^{(k)} )_{k \in \mathbb{N}} \in c_0$ and such that $x^{(k)} \to
x$. Let $\epsilon>0$ and pick $N$ such that
$\|x^{(k)}-x\|_\infty < \frac{1}{2} \epsilon$ for $k \ge N$. Since
$x^{(N)} \in c_0$, there is some $N'$ such that
$|x_i^{(N)} | < \frac{1}{2} \epsilon$ for $i \ge N'$. Then
$|x_i| \le |x_i^{(N)} | +|x_i-x_i^{(N)} | \le |x_i^{(N)} | +\|x-x^{(N)
} \|_\infty <\epsilon$. Hence $x_i \to 0$ and so $x \in c_0$.
\end{document}
答案1
问题出在表达式上x_i^{(N)}。你可以通过破坏下标来强制 TeX 降低指数，但这样括号就会与“i”发生冲突。这可以通过稍微降低下标来解决。
\documentclass[a4paper]{article}
\usepackage{amsfonts,amsmath,parskip}
\usepackage{ebgaramond} %font
%\usepackage{ebgaramond-maths}
%\let\epsilon\varepsilon
\addtolength\textwidth{2cm} % I have not used this
\newcommand{\ssub}[1]{_{\smash{\raisebox{-1pt}{$\scriptstyle#1$}}}}
\begin {document}
\emph{$c_0$ is a complete metric space.} Suppose $x $ is a limit
point of $c_0$. Then there exists
$(x^{(k)} )_{k \in \mathbb{N}} \in c_0$ and such that $x^{(k)} \to
x$. Let $\epsilon>0$ and pick $N$ such that
$\|x^{(k)}-x\|_\infty < \frac{1}{2} \epsilon$ for $k \ge N$. Since
$x^{(N)} \in c_0$, there is some $N'$ such that
$|x\ssub{i}^{(N)} | < \frac{1}{2} \epsilon$ for $i \ge N'$. Then
$|x\ssub{i}| \le |x\ssub{i}^{(N)} |
+|x\ssub{i}-x\ssub{i}^{(N)} | \le |x\ssub{i}^{(N)} | +\|x-x^{(N)
} \|_\infty <\epsilon$. Hence $x\ssub{i} \to 0$ and so $x \in c_0$.
\end{document}
我将粗体文本改为斜体：使用粗体 Garamond 是令人厌恶的。
取消注释上述代码中的两行也将使用 Garamond 进行数学运算。
使用不同的符号可能会改善外观。
\documentclass[a4paper]{article}
\usepackage{amsfonts,amsmath,parskip}
\usepackage{ebgaramond} %font
\usepackage{ebgaramond-maths}
\let\epsilon\varepsilon
\addtolength\textwidth{2cm} % I have not used this
\newcommand{\ssub}[1]{_{\smash{\raisebox{-1pt}{$\scriptstyle#1$}}}}
\begin {document}
\emph{$c_0$ is a complete metric space.} Suppose $x$ is a limit
point of $c_0$. Then there exists
$(x_k)_{k \in \mathbb{N}} \in c_0$ and such that $x_k \to x$.
Let $\epsilon>0$ and pick $N$ such that
$\|x_k-x\|_\infty < \epsilon/2$ for $k \ge N$. Since
$x_N \in c_0$, there is some $N'$ such that
$|x_{N,i}| < \epsilon/2$ for $i \ge N'$. Then
$|x_i| \le |x_{N,i}| + |x_i-x_{N,i}| \le |x_{N,i}| +\|x-x_N \|_\infty <\epsilon$.
Hence $x_i \to 0$ and so $x \in c_0$.
\end{document}
答案2
\documentclass[a4paper]{article}
\usepackage{amsfonts,amsmath,parskip}
\addtolength\textwidth{2cm}
\newcommand\test{%
\textbf{$c_0$ is a complete metric space.} Suppose $x $ is a limit
point of $c_0$. Then there exists
$(x^{(k)} )_{k \in \mathbb{N}} \in c_0$ and such that $x^{(k)} \to
x$. Let $\epsilon>0$ and pick zzzz $N_y$ such that
$\|x^{(k)}-x\|_\infty < \frac{1}{2} \epsilon$ for $k \ge N_{0}$. Since
$x^{(N)} \in c_0$, there is some $N'$ such that
$|x_i^{(N^2)} | < \frac{1}{2} \epsilon$ for $i \ge N'$. Then
$|x_i| \le |x_i^{(N)} | +|x_i-x_i^{(N)} | \le |x_i^{(N)} | +\|x-x^{(N)
} \|_\infty <\epsilon$. Hence $x_i \to 0$ and so $x \in c_0$.
}
\begin {document}
1 \test
{\lineskiplimit=-\maxdimen
2 \test
}
{\lineskiplimit=0pt \lineskip=0pt
3 \test
}
{\baselineskip=15pt
4 \test
}
5 \textbf{$c_0$ is a complete metric space.} Suppose $x $ is a limit
point of $c_0$. Then there exists
$(x^{(k)} )_{k \in \mathbb{N}} \in c_0$ and such that $x^{(k)} \to
x$. Let $\epsilon>0$ and pick zzzz $N_y$ such that
\[\|x^{(k)}-x\|_\infty < \frac{1}{2} \epsilon \quad\text{for $k \ge N_{0}$.}\]
Since
$x^{(N)} \in c_0$, there is some $N'$ such that
$|x_i^{(N^2)} | < \frac{1}{2} \epsilon$ for $i \ge N'$. Then
\[|x_i| \le |x_i^{(N)} | +|x_i-x_i^{(N)} | \le |x_i^{(N)} | +\|x-x^{(N)
} \|_\infty <\epsilon\;.\]
Hence $x_i \to 0$ and so $x \in c_0$.
\end{document}
1) 为原始版本（示例略有更改，以显示重叠）
2）使用-\maxdimen此方法优先考虑等距行距一切包括可读性，如果需要，它将使线条重叠而没有任何警告。
3）使用 0pt 的设置\lineskiplimit，因此如果确实需要，它将允许线条接触，但如果线条确实不适合空间，那么它将增加间距而不是过度打印。
4)（如果我真的必须将其设置为内联，我就会这么做）将行距打开到最小值，以允许所需的内联数学适合。
5）（如果可以选择的话我会这么做）使用显示的数学来表示更大的表达式。
答案3
如果你不介意浪费空间，你可以使用
\documentclass[a4paper]{article}
\usepackage{amsfonts,amsmath,parskip}
\usepackage{ebgaramond} %font
\addtolength\textwidth{2cm} % I have not used this
\begin {document}
{\baselineskip=1.2\baselineskip
\textbf{$c_0$ is a complete metric space.} Suppose $x $ is a limit
point of $c_0$. Then there exists
$(x^{(k)} )_{k \in \mathbb{N}} \in c_0$ and such that $x^{(k)} \to
x$. Let $\epsilon>0$ and pick $N$ such that
$\|x^{(k)}-x\|_\infty < \frac{1}{2} \epsilon$ for $k \ge N$. Since
$x^{(N)} \in c_0$, there is some $N'$ such that
$|x_i^{(N)} | < \frac{1}{2} \epsilon$ for $i \ge N'$. Then
$|x_i| \le |x_i^{(N)} | +|x_i-x_i^{(N)} | \le |x_i^{(N)} | +\|x-x^{(N)
} \|_\infty <\epsilon$. Hence $x_i \to 0$ and so $x \in c_0$.
\par}
\end{document}
可能有一种方法可以做到这一点\linespread，但到目前为止我还没有找到方法。