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\title{Theorem $2.1$}
\author{Adrienne Stanley}
\begin{document}
\begin{abstract}
We prove theorem $2.1$ using the method of proof by way of contradiction. This theorem states that for any set $A$, that in fact the empty set is a subset of $A$, that is $\emptyset \subset A$.
\end{abstract}
\maketitle
We first start with a discussion of subsets.
\begin{definition}
Let $A$ and $B$ be sets. We say $A$ is a subset of $B$ if every element in $A$ is also an element of $B$ and we write $A \subset B$. This can also be written as
$$ (A \subset B) \leftrightarrow \forall x ( x \in A \to x \in B).$$
\end{definition}
Notice that for sets $A$ and $B$, if $A \not\subset B$, then there exists an element $x$ such that $x \in A$ and $x \notin B$. That is,
$$(A \not\subset B) \leftrightarrow \exists x ( x \in A \wedge x \notin B).$$
\begin{example}
Let $A = \{ 1, 2, 3, 4, 5 \} $, $B = \{ 1, 2 \}$ and $C = \{ 1, 7 \}$. We can see that every element in $B$ is an element of $A$. Further, we can see that $C$ contains an element, namely $7$, which is not in $A$. Thus, $B \subset A$ and $C \not\subset A.$
\end{example}
We now prove theorem $2.1$.
\begin{theorem*}[2.1]
For any set $A$, $\emptyset \subset A$.
\end{theorem*}
\begin{proof}
By way of contradiction, suppose that the theorem fails. Let $A$ be a set such that $\emptyset \not\subset A.$ From the above discussion, we can see that there exists an element $x$ such that $x \in \emptyset$ and $x \notin A$. Let $x$ be such an element. Since the emptyset has no elements, then $x \notin \emptyset$. Thus, we have that $x \in \emptyset$ and $x \notin \emptyset.$ This contradiction proves that the theorem is true.
\end{proof}
\end{document}
