Science:Math Exam Resources/Courses/MATH220/December 2009/Question 03 (b)

MATH220 December 2009

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Question 03 (b)
Let $f:A\to {}B$ be a function and let $C\subseteq {}A$. Now let $E\subseteq {}A.$ Prove that if f is injective then $f(C\cap {}E)=f(C)\cap {}f(E)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
This is an equality of sets and so you need to show a double inclusion.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We prove this problem by showing that elements on the left are contained in the right and vice versa. First, suppose that $\displaystyle x\in f(C\cap E)$. Then we can write $\displaystyle x=f(d)$ for an element $\displaystyle d\in C\cap E$, that is, $d\in C$ and $\displaystyle d\in E$. This means that $\displaystyle x\in f(C)$ and $\displaystyle x\in f(E)$ and hence $\displaystyle x\in f(C)\cap f(E)$. Now, suppose that $\displaystyle x\in f(C)\cap f(E)$. Then there exists elements $\displaystyle c\in C$ and $\displaystyle e\in E$ such that $\displaystyle x=f(c)=f(e)$. As $f$ is injective, we have that $\displaystyle c=e$. However, $\displaystyle c\in C$ and $\displaystyle e\in E$ and so this means that $\displaystyle c=e\in C\cap E$. Thus, $\displaystyle x\in f(C\cap E)$ completing the proof.

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