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Explanation: How Functions interact with Set Operations
(related to Lemma: Behavior of Functions with Set Operations)
In the above lemma, the rules $2, 5,7$ and $8$ do not allow equality in the general case. The proof given for the lemma is simple but highly technical and it might be interesting to see a counterexample. Such a counterexample is given below.
In the figure, a function $f:A\mapsto A$ is defined and where we set $X:=\{u\}$ and $Y:=\{w\}.$ We calculate the above rules to see that, indeed, no equality can be achieved:
 Rule $2$: Assume equality $f[X\cap Y] = f[X]\cap f[Y]$ for all $X,Y\subseteq A.$ This cannot be true since $f[A\cap B]=f[\emptyset]=\emptyset$ on the one side of the equation and $f[X]\cap f[Y]=\{u\}\cap \{u\}=\{u\}$ on the other.
 Rule $5$: Assume equality $f[A\setminus X]= f[A]\setminus f[X]$ for all $X \subseteq A.$ This cannot be true since $f[A\setminus X]=f[\{w\}]=\{u\}$ on the one side of the equation and $f[A]\setminus f[X]=\{u\}\setminus\{u\}=\emptyset$ on the other.
 Rule $7$: Assume equality $f^{1}[f[X]]=X$ for all $X\subseteq A.$ This cannot be true since $f^{1}[f[X]]=f^{1}[\{u\}]=\{u,w\}=A$ on the one side of the equation and $X\subset A$ (proper subset) on the other.
 Rule $8$: Assume equality $f[f^{1}[X]]=X$ for all $X\subseteq B.$ This cannot be true since if we choose $X=\{w\}$, then $f[f^{1}[X]]=f[\emptyset]=\emptyset$ on the one side of the equation and $X=\{w\}$ on the other.
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References
Bibliography
 Reinhardt F., Soeder H.: "dtvAtlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition
 Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994