证明通常是如何在写作中构建的，一行还是分成多行？

Question

这是以两种不同方式格式化的相同证明。我的问题是，像家庭作业这样的写作的标准是什么？我认为第一个看起来更好，但我不认为这是我通常看到的方式。如果有人能说出哪一个更好，我将不胜感激。

这是第一个输出的乳胶代码。

\subsection*{13.}
\subsubsection*{(a)}
In order to show $f(A_1 \cup A_2) = f(A_1)\cup f(A_2)$, we want to show $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$ and $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$. 
\\
Let $y \in f(A_1 \cup A_2)$. 
\\
Then there exists a $x \in A_1 \cup A_2$ such that $f(x) = y$. 
\\
If $x \in A_1$, then $y \in f(A_1)$. 
\\
As a result, $y \in f(A_1) \cup f(A_2)$. 
\\
If $x \in A_2$, then $y \in f(A_2)$. 
\\
So, $y \in f(A_1) \cup f(A_2)$. 
\\
Therefore, $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$. 
\\
\\
Moreover, we want to show $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$. 
\\
Let $y \in f(A_1) \cup f(A_2)$. 
\\
If $y \in f(A_1)$, then there exists a $x \in A_1$ such that $f(x) = y$. 
\\
Since $x \in A_1$, then $x \in A_1 \cup A_2$. 
\\
If $y \in f(A_2)$, then there exists a $x \in A_2$ such that $f(x) = y$. 
\\
Since $x \in A_2$, then $x \in A_1 \cup A_2$. 
\\
As a result $y \in f(A_1 \cup A_2)$. 
\\
Therefore, $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$. 
\\
Since $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$ and $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$, we can conclude that $f(A_1 \cup A_2) = f(A_1)\cup f(A_2)$.

Answer 1

1. 不要使用\\转到新行，如果需要新行，请使用\newline，如果需要新段落，请留空行。\\仅在表格中使用。
2. 为什么要自己命名\subsection而\subsubsection不是让 LaTeX 替你命名？是为了匹配作业编号还是只是因为你不能修改编号布局？

这就是我要做的。当然，我不了解这个主题，所以你可以改进它，把 displaymath 放在真正需要的地方。

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\subsection*{13.}% Why are you numbering the section and don't let LaTeX do it for you? 
\subsubsection*{(a)}
In order to show \[f(A_1 \cup A_2) = f(A_1)\cup f(A_2)\] we want to show \[f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2) \quad \text{and} \quad f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\,\text{.}\] 

Let $y \in f(A_1 \cup A_2)$. 
Then there exists a $x \in A_1 \cup A_2$ such that $f(x) = y$.
If $x \in A_1$, then $y \in f(A_1)$. 
As a result, \[y \in f(A_1) \cup f(A_2)\,\text{.}\] 
If $x \in A_2$, then $y \in f(A_2)$. 
So, $y \in f(A_1) \cup f(A_2)$. 
Therefore, \[f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)\,\text{.}\] 

Moreover, we want to show \[f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\,\text{.}\] 

Let $y \in f(A_1) \cup f(A_2)$. 
If $y \in f(A_1)$, then there exists a $x \in A_1$ such that $f(x) = y$. 
Since $x \in A_1$, then $x \in A_1 \cup A_2$. 
If $y \in f(A_2)$, then there exists a $x \in A_2$ such that $f(x) = y$. 
Since $x \in A_2$, then $x \in A_1 \cup A_2$. 
As a result \[y \in f(A_1 \cup A_2)\,\text{.}\]

Therefore, \[f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\,\text{.}\]

Since $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$ and $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$, we can conclude that \[f(A_1 \cup A_2) = f(A_1)\cup f(A_2)\,\text{.}\]
\end{document}