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2006 Romanian NMO Problems/Grade 10/Problem 1

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Problem

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

(a) $f(A) \neq 0$, for $A \neq \phi$;

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

Solution

Solution 1

I claim that $f(A)=|A|$, for all $A\in \mathcal P(M)$. 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 $f$ that satisfies both properties must be $f(A)=|A|$.

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

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

Note that $f(a_{i}) \ge 1$ for any $i$, from property (a). Hence \[f(A_{0}) = 0 < f(A_{1}) < \dotsb < f(A_{n-1}) < f(A_{n}) \le n \;,\] which implies $f(A_{i}) = i$ for all $i$.

Solution 2

I claim that $f(A)=|A|$, for all $A\in \mathcal P(M)$. 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 $f$ that does satisfy both properties, and then showing that this function is in fact $f(A)=|A|$.

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

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

I shall now prove that $f(A)=|A|$ for all $A\in \mathcal P(M)$.

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

See Also

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