# Problem A. 464. (November 2008)

**A. 464.** Let *H* a set and denote by *P*(*H*) the family of all subsets of *H*. Suppose that *f* and *g *are *P*(*H*)*P*(*H*) maps such that *f*(*X*)*f*(*Y*)*H* and *g*(*X*)*g*(*Y*)*H *whenever *X**Y**H*. Prove that there exist sets *A*,*B**H* for which *f*(*A*)=*H*\*B* and *g*(*B*)=*H*\*A*.

(5 pont)

**Deadline expired on December 15, 2008.**

**Solution. **For every *X**H*, let *h*(*X*)=*H*\*g*(*H*\*f*(*X*)). Our goal is to construct a set *A**H* for which *h*(*A*)=*A*,

The function the function *h* s nondecreasing, i.e. *h*(*X*)*h*(*Y*) whenever *X**Y**H*.

Call a set *X**H* "nice" if *X**h*(*X*) and let *A* be the union of all nice sets:

We will show that *h*(*A*)=*A*. Let *A*'=*h*(*A*).

For every nice set *X* we have *X**A* and thus *X**h*(*X*)*h*(*A*). Then

Therefore, the set *A* is nice, *A**h*(*A*).

Since *h* is nondecreasing, the set *A*' is nice as well; *A*'=*h*(*A*)*h*(*A*'). On the othe hand, *A* is the union of all nice sets, so *A*'=*h*(*A*) is a subset of *A*.

Hence, *A**h*(*A*) and *h*(*A*)*A*. Therefore *A*=*h*(*A*).

### Statistics:

9 students sent a solution. 5 points: Backhausz Tibor, Bodor Bertalan, Éles András, Tomon István, Tossenberger Anna. 2 points: 2 students. 0 point: 2 students.

Problems in Mathematics of KöMaL, November 2008