# A problem with sets

#### Mollier

Hi.

problem:

Show that if $$\displaystyle f:A\rightarrow B$$ and E,F are subsets of A, then
$$\displaystyle f(E\cup F)=f(E)\cup f(F)$$ and $$\displaystyle f(E\cap F)\subseteq f(E) \cap f(F)$$.

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attempt:

First I try to show that $$\displaystyle f(E\cup F)\subset f(E) \cup f(F)$$ and then that $$\displaystyle f(E) \cup f(F) \subset f(E\cup F)$$.

$$\displaystyle y\in f(E\cup F) \Rightarrow f^{-1}(y)\in E\cup F \Rightarrow f^{-1}(y)\in E \; or \; f^{-1}(y)\in F$$.
$$\displaystyle f(f^{-1}(y))=y\in f(E\cup F)$$ and so $$\displaystyle f(E\cup F)\subset f(E) \cup f(F)$$.

$$\displaystyle y\in f(E)\cup f(F) \Rightarrow f^{-1}(y)\in E or f^{-1}(y)\in F \Rightarrow f^{-1}(y)\in E\cup F$$.
$$\displaystyle f(f^{-1}(y))=y \in f(E\cup F)$$ and so $$\displaystyle f(E) \cup f(F) \subset f(E\cup F)$$.

A friend of mine pointed out that I could show this in the following manner:

\displaystyle \begin{aligned} f(E\cup F) =&\; \{f(x): x\in E \; or \; x\in F\}\\ =&\; \{f(x): x\in E\} \cup \{f(x): x\in F\}\\ =&\; f(E) \cup f(F) \end{aligned}

$$\displaystyle y\in f(E\cap F) \Rightarrow f^{-1}(y)\in E\cap F \Rightarrow f^{-1}(y)\in E \; and \; f^{-1}(y)\in F$$.
Then $$\displaystyle f(f^{-1}(y))\in f(E)\cap f(F) \Rightarrow f(E\cap F)\subset f(E) \cap f(F).$$

I do not know how to continue. If $$\displaystyle y\in f(E) \cap f(F)$$, what then? Should I be doing all of this differently?

Thank you.

#### HallsofIvy

MHF Helper
Hi.

problem:

Show that if $$\displaystyle f:A\rightarrow B$$ and E,F are subsets of A, then
$$\displaystyle f(E\cup F)=f(E)\cup f(F)$$ and $$\displaystyle f(E\cap F)\subseteq f(E) \cap f(F)$$.

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attempt:

First I try to show that $$\displaystyle f(E\cup F)\subset f(E) \cup f(F)$$ and then that $$\displaystyle f(E) \cup f(F) \subset f(E\cup F)$$.

$$\displaystyle y\in f(E\cup F) \Rightarrow f^{-1}(y)\in E\cup F \Rightarrow f^{-1}(y)\in E \; or \; f^{-1}(y)\in F$$.
$$\displaystyle f(f^{-1}(y))=y\in f(E\cup F)$$ and so $$\displaystyle f(E\cup F)\subset f(E) \cup f(F)$$.
I don't see anywhere in the hypotheses that says f is invertible so you should not be writing "$$\displaystyle f^{-1}(y)$$". Write, rather, "if $$\displaystyle y\in f(E\cup F)$$ then there exist x in $$\displaystyle E\cup F$$ such that f(x)= y. Since $$\displaystyle x\in E\cup F$$, either $$\displaystyle x\in E$$ or $$\displaystyle x\in F$$.

Case 1: if $$\displaystyle x\in E$$ then $$\displaystyle y\in f(E)$$.

Case 2: if $$\displaystyle x\in F$$ then $$\displaystyle y\in f(F)$$.

$$\displaystyle y\in f(E)\cup f(F) \Rightarrow f^{-1}(y)\in E or f^{-1}(y)\in F \Rightarrow f^{-1}(y)\in E\cup F$$.
$$\displaystyle f(f^{-1}(y))=y \in f(E\cup F)$$ and so $$\displaystyle f(E) \cup f(F) \subset f(E\cup F)$$.

A friend of mine pointed out that I could show this in the following manner:

\displaystyle \begin{aligned} f(E\cup F) =&\; \{f(x): x\in E \; or \; x\in F\}\\ =&\; \{f(x): x\in E\} \cup \{f(x): x\in F\}\\ =&\; f(E) \cup f(F) \end{aligned}
So your friend was essentially telling you the same thing I just did!

$$\displaystyle y\in f(E\cap F) \Rightarrow f^{-1}(y)\in E\cap F \Rightarrow f^{-1}(y)\in E \; and \; f^{-1}(y)\in F$$.
Then $$\displaystyle f(f^{-1}(y))\in f(E)\cap f(F) \Rightarrow f(E\cap F)\subset f(E) \cap f(F).$$

I do not know how to continue. If $$\displaystyle y\in f(E) \cap f(F)$$, what then? Should I be doing all of this differently?

Thank you.
If $$\displaystyle y\in f(E\cap F)$$ then there exist x in $$\displaystyle E\cap F$$ such that f(x)= E. Since $$\displaystyle x\in E\cap F$$, then $$\displaystyle x\in E$$ so $$\displaystyle y= f(x)\in f(E)$$ and $$\displaystyle x\in F$$ so $$\displaystyle y= f(x)\in f(F)$$. Therefore $$\displaystyle y\in f(E)\cap f(F)$$.

• Mollier

#### Mollier

That was a crystal clear explanation, thank you very much!