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2006 Romanian NMO Problems/Grade 10/Problem 1

Revision as of 14:19, 7 May 2012 by 1=2 (talk | contribs) (made some progress, wrote it down. Haven't finished yet, feel free to take a crack at it)

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

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 $f(A)=|A|$.

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 $X$ is a subset of $Y$.
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.

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See Also

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