for all $1 < j < n$. Hence $|S|
\ge \nH-1$.

*(iii) Suppose that $y_1 \in S$ and $x_m \in S$.* If $x_i \notin S$ for some $1 < i < m$, then $y_j \in S$ for every $1 <
j < m$, since $N(y_j) \subset N(x_i)$ and $y_1 \in S \cap (N(x_i)
\setminus N(y_j))$. Then $X \setminus \{x_i\} \subset S$ and thus $|S| \ge \nH-1$ and we are done. Thus assume that $X \subset S$. Then $|Y \cap S| \ge n-1$, otherwise two different vertices from $Y$ would not be fairly dominated by $S$. Hence again $|S| \ge \nH-1$.

In all three cases, we have that $|S| \ge \nH-2$. Since $S$ is an arbitrary FD-set of $H$, it follows that ${{\rm fd}}(H) \ge \nH - 2$. However as shown earlier, ${{\rm fd}}(H) \le \nH - 2$. Consequently, ${{\rm fd}}(H) = \nH -
2$. This completes the proof of Claim I. [[($\Box$)]{}]{}

[Claim II.]{} There exists an infinite family of graphs $F$ of odd order $\nF$ satisfying ${{\rm fd}}(F) = \nF - 2$.

**Proof.** For $n \ge 3$ define the graph $F = F_n$ on $\nF =
2n+1$ vertices as follows: Let $F$ be obtained from the graph $H_n$ of 