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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.

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  • $\begingroup$ Begin by understanding the definition of $f(S)$ and the definition of $A\cap B$. Then apply them one at a time. $\endgroup$
    – Asaf Karagila
    Jan 15, 2015 at 18:57
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – Lord_Farin
    Jan 15, 2015 at 20:37

2 Answers 2

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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.

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$S_{1}\cap S_{2}\subseteq S_{i}$ 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...

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