I need to prove that if $S_1$ and $S_2$ are subsets of a set $X$, and if $f: X \to Y$ is an injection, prove that $f(S_1 \cap S_2) = f(S_1) \cap f(S_2)$.
I know I need to show inclusion in both directions but not sure how to begin.
I need to prove that if $S_1$ and $S_2$ are subsets of a set $X$, and if $f: X \to Y$ is an injection, prove that $f(S_1 \cap S_2) = f(S_1) \cap f(S_2)$.
I know I need to show inclusion in both directions but not sure how to begin.
A typical element in $f(S_1\cap S_2)$ is of the form $f(x)$ where $x\in S_1\cap S_2$. In particular, $x\in S_1$ and $x\in S_2$. So $f(x)\in f(S_1)$ since $x\in S_1$ and $f(x)\in f(S_2)$ since $x\in S_2$. Thus $f(x)\in f(S_1)\cap f(S_2)$. Since $f(x)$ was arbitrary in $f(S_1\cap S_2)$, this shows that $f(S_1\cap S_2)\subseteq f(S_1)\cap f(S_2)$.
To show that $f(S_1)\cap f(S_2)\subseteq f(S_1\cap S_2)$, take an element $y\in f(S_1)\cap f(S_2)$. Then there exists an element $x_1\in S_1$ such that $y=f(x_1)$ and there exists an element $x_2\in S_2$ such that $y=f(x_2)$. Since $f$ is injective, $y=f(x_1)=f(x_2)\implies x_1=x_2$. Put $x:=x_1=x_2$. Then $x\in S_1$ and $x\in S_2$ since $x=x_1\in S_1$ and $x=x_2\in S_2$. We have found an $x\in S_1\cap S_2$ such that $y=f(x)$. This means that $y\in f(S_1\cap S_2)$. The second inclusion is thus proved.
$S_{1}\cap S_{2}\subseteq S_{i}$ so that $f\left(S_{1}\cap S_{2}\right)\subseteq f\left(S_{i}\right)$ for $i=1,2$ or equivalently $f\left(S_{1}\cap S_{2}\right)\subseteq f\left(S_{1}\right)\cap f\left(S_{2}\right)$. Note that injectivity of $f$ has not been used for this part.
Let $y\in f\left(S_{1}\right)\cap f\left(S_{2}\right)$. Then $y=f\left(s_{1}\right)=f\left(s_{2}\right)$ for elements $s_{1}\in S_{1}$ and $s_{2}\in S_{2}$. Now use the injectivity of $f$ to find out that...