Proving That Two Sets Are Equal

Let $A = \{ 2x \mid x \text{ is a natural number and $2x<10$} \}$. We need to show that $S\subseteq A$ and $A\subseteq S$.

Let's start by showing that $S\subseteq A$. We pick $n\in S$ and need to show that $n\in A$. Since $n\in S$, this means that $n$ is an even natural number less than 10. Because $n$ is even, we know $n=2x$ for some integer $x$. Because $n$ is a natural number, we know $n>0$ and hence $x>0$ is a natural number. Because $n<10$, we know that $2x<10$. Thus $n=2x$ where $x$ is a natural number and $2x<10$ which means $n\in A$. This shows that every element in $S$ is an element of $A$, in other words we've shown that $S\subseteq A$.

We now need to show that $A\subseteq S$. We let $n\in A$ which means that $n=2x$ where $x$ is a natural number and $2x<10$. Since $n=2x$, we know that $n$ is even. Since $x$ is a natural number, we know $n$ is a natural number. Since $2x<10$, we know that $n<10$. This shows that $n$ is an even natural number less than 10, and hence $n\in S$. This shows that every element in $A$ is also in $S$, which means we've shown $A\subseteq S$.

Since we've show that both $S\subseteq A$ and $A\subseteq S$, we have proven that $S=A$. $\blacksquare$