11. (20 points) Given $x_{n}=\{\sqrt{2} n\}\left(n \in \mathbf{N}_{+}\right)$, where $\{x\}=x-[x],[x]$ represents the greatest integer not exceeding the real number $x$. If the sequence $\left\{x_{n}\right\}$ satisfies
$$
\left|(m-n)\left(x_{m}-x_{n}\right)\right| \geqslant \frac{1}{p}\left(p \in \mathbf{N}_{+}\right)
$$

for all unequal positive integers $m, n$, find the set of values for $p$.
11. First prove:
$$
|\sqrt{2} m-n|>\frac{1}{3 m}\left(m, n \in \mathbf{N}_{+} ; m \neq n\right) \text {. }
$$

Assume $|\sqrt{2} m-n| \leqslant \frac{1}{3 m}$.
(1) If $0 \leqslant \sqrt{2} m \div n \leqslant \frac{1}{3 m}$, then
$$
\begin{array}{l}
2 n \leqslant \sqrt{2} m+n \leqslant \frac{1}{3 m}+2 n \\
\Rightarrow 1 \leqslant\left|2 m^{2}-n^{2}\right| \leqslant \frac{1}{9 m^{2}}+\frac{2 n}{3 m} \\
\Rightarrow 9 m^{2} \leqslant 1+6 m n \leqslant 1+6 \sqrt{2} m^{2} \\
\Rightarrow m=1 .
\end{array}
$$

But $|\sqrt{2}-n|>|\sqrt{2}-1|>\frac{1}{3}$, which contradicts the assumption.
(2) If $0 \leqslant n-\sqrt{2} m \leqslant \frac{1}{3 m}$, then
$$
\begin{array}{l}
2 \sqrt{2} m \leqslant n+\sqrt{2} m \leqslant \frac{1}{3 m}+2 \sqrt{2} m \\
\Rightarrow 1 \leqslant 1 n^{2}-2 m^{2} \left\lvert\, \leqslant \frac{1}{9 m^{2}}+\frac{2 \sqrt{2}}{3}\right. \\
\Rightarrow m=1 .
\end{array}
$$

Similarly, this contradicts the assumption.
In conclusion, $|\sqrt{2} m-n|>\frac{1}{3 m}\left(m, n \in \mathbf{N}_{+}, m \neq n\right)$ holds.
$$
\begin{array}{l}
\text { Hence }\left|x_{m}-x_{k}\right| \\
=|\sqrt{2}(m-k)-([m \sqrt{2}]-[k \sqrt{2}])| \\
>\frac{1}{3(m-k)} .
\end{array}
$$

Thus, $\left|(m-k)\left(x_{m}-x_{k}\right)\right|>\frac{1}{3}$. Since $\mid 2 \sqrt{2}\}-\{\sqrt{2}\} \left\lvert\,<\frac{1}{2}\right.$, the minimum value of $p$ is 3. Therefore, the set of values for $p$ is $\{p \mid p \geqslant 3, p \in \mathbf{N}\}$.