# Thread: A problem with sets

1. ## A problem with sets

Hi.

problem:

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

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attempt:

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

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

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

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


\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}

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

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

Thank you.

2. Originally Posted by Mollier
Hi.

problem:

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

---------------------------------------------------------------------------------------------------------------

attempt:

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

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

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

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

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

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


\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!

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

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

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

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