that ${{\rm fd}}(G') \le
n' - 2$. Every ${{\rm fd}}(G')$-set can be extended to a FD-set in $G$ by adding to it the vertices $u$ and $v$, implying that ${{\rm fd}}(G) \le
{{\rm fd}}(G') + 2 \le n' = n-2$. This establishes the desired upper bound.

We show next that the upper bound is sharp. For this purpose we construct an infinite family of graphs $G$ of order $n \ge 3$ with $\delta(G) \ge 1$ satisfying ${{\rm fd}}(G) = n - 2$. We consider two cases in turn, depending on the parity of $n$.

[Claim I.]{} There exists an infinite family of graphs $H$ of even order $\nH$ satisfying ${{\rm fd}}(H) = \nH - 2$.

**Proof.** For $n \ge 3$ define the graph $H = H_n$ on $\nH =
2n$ vertices as follows: Let $V(H) = X \cup Y$, where $X =
\{x_1,\ldots,x_n\}$ and $Y = \{y_1,\ldots,y_n\}$, and where $x_i$ is adjacent to $y_j$ if and only if $i \ge j$. Further $Y$ is an independent set and, for $i,j > 1$, $x_i$ is adjacent to $x_j$. Thus, $d_H(x_1)  = 1$ and $d_H(x_i) = i +
n - 2$ for $1 < i \le n$, while $d_H(y_i)  = n- i 