Let $f:S\to T$ be an injective function. If $R\subseteq S$, then denote by $f(R)$ the set $\{f(x):x\in R\}$. Prove that

$$f(A\setminus B)=f(A)\setminus f(B).$$

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Solution

We prove this set equality by extensionality.

• $(\subseteq )$: Let $y\in f(A\setminus B)$. Since $f$ is a function, then there exists $x\in A\setminus B$ such that $f(x)=y$. Since $x\in A\setminus B$, then $x\in A$ and $x\notin B$ by definition of set subtraction. We further note that $y=f(x)\in f(A)$ since $x\in A$. Assume for contradiction that $y=f(x)\in f(B)$. Then, there exists $x^{\prime }\in B$ such that $f(x^{\prime })=y$. Since $f$ is injective and $f(x)=f(x^{\prime })$, then $x=x^{\prime }$, which leads to the contradiction that $x\in B$. Thus $y=f(x)\notin f(B)$ by contradiction. Therefore, as $y\in f(A)$ and $y\notin f(B)$, we infer that $y\in f(A)\setminus f(B)$.

• $(\supseteq )$: Let $y\in f(A)\setminus f(B)$. Then $y\in f(A)$ and $y\notin f(B)$ by definition of set subtraction. Since $f$ is a function and $y\in f(A)$, let $x\in A$ be such that $f(x)=y$. Assume for contradiction that $x\in B$. Then, $y=f(x)\in f(B)$ which is a contradiction to the assumption that $y\notin f(B)$. Thus $x\notin B$ by contradiction such that $x\in A\setminus B$. Therefore, $y=f(x)\in f(A\setminus B)$.