View source for 2006 Romanian NMO Problems/Grade 10/Problem 1

==Problem==
Let <math>M</math> be a set composed of <math>n</math> elements and let <math>\mathcal P (M)</math> be its power set. Find all functions <math>f : \mathcal P (M) \to \{ 0,1,2,\ldots,n \}</math> that have the properties

(a) <math>f(A) \neq 0</math>, for <math>A \neq \phi</math>;

(b) <math>f \left( A \cup B \right) = f \left( A \cap B \right) + f \left( A \Delta B \right)</math>, for all <math>A,B \in \mathcal P (M)</math>, where <math>A \Delta B = \left( A \cup B \right) \backslash \left( A \cap B \right)</math>.

==Solution==
===Solution 1===
I claim that <math>f(A)=|A|</math>, for all <math>A\in \mathcal P(M)</math>. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by proving that any such function <math>f</math> that satisfies both properties must be <math>f(A)=|A|</math>.

We have <math>f(\emptyset) = f(\emptyset) + f(\emptyset)</math>, hence <math>f(\emptyset) = 0</math>. Also, if <math>A \subset B \subset M</math>, then <math>f(B) = f(A) + f(B-A)</math> since <math>B = B \cup A</math>, <math>A = B \cap A</math>, and <math>B-A = B\Delta A</math>.

Let <cmath>A_{0} = \emptyset \subsetneq A_{1} \subsetneq \dotsb \subsetneq A_{n-1} \subsetneq A_{n} = M</cmath> be an increasing sequence of subsets of <math>M</math>, such that <math>A_{i} = A_{i-1} \cup \{a_{i}\}</math> for all <math>i=1,\dotsc,n</math>. Then <math>|A_{i}| = i</math> for all <math>i</math>.

Note that <math>f(a_{i}) \ge 1</math> for any <math>i</math>, from property (a). Hence <cmath>f(A_{0}) = 0 < f(A_{1}) < \dotsb < f(A_{n-1}) < f(A_{n}) \le n \;,</cmath> which implies <math>f(A_{i}) = i</math> for all <math>i</math>.
===Solution 2===
I claim that <math>f(A)=|A|</math>, for all <math>A\in \mathcal P(M)</math>. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by constructing a function <math>f</math> that does satisfy both properties, and then showing that this function is in fact <math>f(A)=|A|</math>.

We have <math>f(\emptyset) = f(\emptyset) + f(\emptyset)</math>, hence <math>f(\emptyset) = 0</math>. Note that if <math>|A|=1</math>, then <math>f(A)\geq 1</math>, from property (a). Let <math>X=\{x\}</math> and <math>Y=\{x,y\}</math> for some <math>x,y\in M</math>. We then have from property <math>B</math> that <math>f(Y)=f(X)+f(Y\backslash X)</math>. Since <math>|X|=|Y\backslash X|=1</math>, <math>f(Y)\geq 2</math>. This shows that for all <math>Y</math> with <math>|Y|=2</math>, <math>f(Y)\geq 2</math>.

 Lemma: <math>f(A)\geq |A|</math> for all <math>A\in \mathcal P(M)</math>.
 Proof: We proceed by induction on <math>|A|</math>. Clearly <math>f(\phi)\geq 0</math> and <math>f(|A|)\geq 1</math> if <math>|A|=1</math> from property (a). This is our base case.
 Now we do the inductive step. Assume that for all <math>i\leq k\leq n</math> for some positive integer <math>k</math>, <math>f(A)\leq |A|</math> if <math>|A|=i</math>.
 Let <math>X</math> and <math>Y</math> be subsets of <math>M</math> such that <math>|X|=1</math>, <math>|Y|=k+1</math> and <math>X\subset Y</math>.
 We then have from property 2 that <math>f(Y)=f(X)+f(Y\backslash X)</math>. From our inductive hypothesis, this sum is at least <math>k+1</math>.
 This completes the inductive step, which completes the proof.

I shall now prove that <math>f(A)=|A|</math> for all <math>A\in \mathcal P(M)</math>.

Note that our lemma implies that <math>f(M)=n</math>. Consider a set <math>A\subset M</math> with cardinality <math>k</math>. We then have that <math>A \cup M=A</math> and <math> A \Delta M=M\backslash A</math>. We then have that <math>f(A)\geq k</math> and <math>f(M\backslash A)\geq n-k</math>, so their sum (which equals <math>f(M)</math> from property (b)) is at least <math>k+n-k=n</math>. This then implies that <math>f(A)=k</math> and <math>f(M\backslash A)=n-k</math> for all <math>A</math>. We defined <math>k</math> to be <math>|A|</math>, so <math>f(A)=|A|</math>. This completes the proof.

==See Also==
*[[2006 Romanian NMO Problems/Grade 9/Problem 4 | Previous problem]]
*[[2006 Romanian NMO Problems/Grade 10/Problem 2 | Next problem]]
*[[2006 Romanian NMO Problems]]
[[Category:Olympiad Combinatorics Problems]]