3. Let $F$ be the set of all functions $f: P(S) \rightarrow R$ such that for all $X, Y \subseteq P(S)$, we have $f(X \cap Y)=\min (f(X), f(Y))$, where $S$ is a finite set and $P(S)$ is the power set of $S$. Find: $\max _{f \in F} \mid \operatorname{Im}$ $(f) \mid$, where $\operatorname{Im}(f)$ denotes the image set of $f$.
3. Solution: $\max _{f \in F}|\operatorname{Im}(f)|=n+1$. For each $n \in$ $\mathbf{N}^{*}$, let $S_{n}=\{1,2, \cdots, n\}$. Define $f$ as follows: for every $A \subseteq S_{n}$, if $n \notin A$, then let $f(A)=\frac{1}{n+1}$;
if $n \in A$ and $n-1 \notin A$, then let $f(A)=\frac{1}{n} ; \cdots$
In general, if $n, n-1, \cdots, k \in A$, and $k-1 \notin$ $A$, then let $f(A)=\frac{1}{k}$, then $f$ is a mapping from $P(S) \rightarrow R$, and for $X, Y \in P(S)$, suppose $n, n-1, \cdots, k \in(X \cap$ $Y)$, and $k-1 \notin(X \cap Y),(k=1,2, \cdots, n+1)$, then $n, n-1, \cdots, k \in X, Y$ and $k-1$ does not belong to both $X$, $Y$ simultaneously, thus, by the definition of $f$: $f(X), f(Y) \geqslant \frac{1}{k}$ at least one of them equals $\frac{1}{k}$, and $f(X \cap Y)=\frac{1}{k}$, hence $f(X \cap Y)=\min (f(X), f(Y))$, now we have $|\operatorname{Im}(f)|$ $=n+1$, i.e., $\max _{f \in F}|\operatorname{Im}(f)|=n+1$.