A problem with sets

Nov 2009
234
12
Norway
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
Apr 2005
20,249
7,909
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 \(\displaystyle f(E\cap F)\subseteq f(E) \cap f(F)\).

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

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 \(\displaystyle y\in f(E)\cap f(F)\).
 
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Reactions: Mollier
Nov 2009
234
12
Norway
That was a crystal clear explanation, thank you very much!