Science:Math Exam Resources/Courses/MATH220/December 2011/Question 04

MATH220 December 2011

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Question 04
Let ƒ: A → B be a function. If C ⊆ A, define $f(C)=\{f(x)\mid x\in C\}$ Prove that if ƒ is injective, then $f(C\cap D)=f(C)\cap f(D)\quad {\text{for all sets }}C,D\subseteq A$

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
Remember that to show that two sets X and Y are equal, you must show that X ⊆ Y and that Y ⊆ X.

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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 want to show that $f(C\cap D)\subseteq f(C)\cap f(D)$ and that $f(C)\cap f(D)\subseteq f(C\cap D)$ Let us start with the first. Suppose that $y\in f(C\cap D)$; that is, there is some $x\in C\cap D$ such that $f(x)=y$. Since $x\in C$ and $x\in D$, it follows that $y\in f(C)$ and that $y\in f(D)$, i.e. that $y\in f(C)\cap f(D)$ as desired. Note: This proof in no way uses injectivity; this is a general statement about all functions. Now we prove the second statement. Suppose that $y\in f(C)\cap f(D)$. That is, $y\in f(C)$ and $y\in f(D)$. Equivalently, there is some $x_{1}\in C$ such that $f(x_{1})=y$, and some $x_{2}\in D$ such that $f(x_{2})=y$. We claim that $x_{1}=x_{2}$, and hence $x_{1}=x_{2}\in C\cap D$. But this follows immediately from the injectivity of $f$. It follows now, since $f(x_{1})=y$ and $x_{1}\in C\cap D$ that $y\in f(C\cap D)$ as claimed.

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