# Kind of linear property on a function $f: X \to Y$

A function $f: X \to Y$ is called linear iff

$f(x_1+x_2) = f(x_1) + f(x_2)$

It is what I learned.

Is there any concept like 'linear' defined on the functions whose input and output are sets.

A function $f: X \to Y$ is called 'ABCD' iff $f(X1 \cup X2) = f(X1) \cup f(X2)$

Is there any concept corresponding to 'ABCD' here?

// Thanks, lulu I clarify here:

f is defined on some fixed set and that X_i denotes subsets of that set.

• May be "additive"? – Alessandro Blasetti Nov 7 '16 at 14:06
• Well, if we take the lattice of sets with intersection and union, the property becomes join-semilattice homomorphism. – lisyarus Nov 7 '16 at 14:21
• This isn't clear. What are $X,Y$? I guess they are collections of sets closed under union? Are they power sets of fixed sets? – lulu Nov 7 '16 at 14:23
• Well, the point should be clarified. In the posted solution below, the writer is assuming that you have $f$ is defined on some fixed set and that $X_i$ denotes subsets of that set. If that's what you intended then indeed all functions satisfy your rule. If, to the contrary, you meant $X$ is some collection of sets (closed under union) and $f$ is defined on that collection, then not all functions satisfy your rule. – lulu Nov 7 '16 at 14:49

1. If $f(x_1+x_2)=f(x_1)+f(x_2)$ then $f$ is called additive and not linear !

If $X$ and $Y$ are vector spaces over a field $F$ and we have additional

$f( \alpha x)= \alpha f(x)$ for all $\alpha \in F$ and all $x \in X$,

then f is called linear.

1. For each function $f:X \to Y$ we have

$f(X_1 \cup X_2)=f(X_1) \cup f(X_2)$ for all subsets $X_1,X_2$ of $X$

• Thank you for the explanation. So the second statement is always true for all kinds of such functions? Doesn't this theorem have any name? (since it's too trivial?...) – syko Nov 7 '16 at 14:35
• 2. seems false to me. Suppose $X$ and $Y$ are sets of nonempty, finite subsets of $\mathbb R$, and $f(X)$ is a singleton containing $max(X)$ – then $f(A\cup B)=\{max(max(A), max(B))\}$ while $f(A)\cup f(B)=\{max(A), max(B)\}$ which are not the same. What's more, how about $f$ which value is not a set? The RHS $f(A)\cup f(B)$ might not even be defined... – CiaPan Nov 7 '16 at 14:38
• What you have written seems very confused to me ! – Fred Nov 7 '16 at 14:47
• The confusion lies in the OP's intent. If $f$ is just defined on some collection of sets, then the desired property need not hold. For example, say $X$ is the power set on $\{1,2\}$ and for $S\in X$ define $f(S)=\emptyset$ if $S\neq \emptyset$ and $f(\emptyset)=\{1\}$. – lulu Nov 7 '16 at 14:52