5. If the sequence $\left\{a_{n}\right\}: a_{1}, a_{2}, \cdots, a_{5}$ satisfies
$$
0 \leqslant a_{1}<a_{2}<a_{3}<a_{4}<a_{5},
$$

and for any $i, j(1 \leqslant i \leqslant j \leqslant 5)$, $a_{j}-a_{i}$ is in the sequence.
(1) $a_{1}=0$;
(2) $a_{5}=4 a_{2}$;
(3) $\left\{a_{n}\right\}$ is an arithmetic sequence;
(4) The set $A=\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 5\right\}$ contains nine elements.
The number of correct statements is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
5. D.

Since $a_{1}-a_{1}=0 \in \left\{a_{n}\right\}$, we have $a_{1}=0$. Therefore, statement (1) is correct.

Since $0=a_{1}<a_{3}-a_{2}<a_{4}-a_{2}<a_{5}-a_{2}<a_{5}$, and $a_{3}-a_{2}, a_{4}-a_{2}, a_{5}-a_{2} \in \left\{a_{n}\right\}$, we have
$$
\begin{array}{l}
a_{3}-a_{2}=a_{2}, a_{4}-a_{2}=a_{3}, a_{5}-a_{2}=a_{4} \\
\Rightarrow a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=a_{5}-a_{4}.
\end{array}
$$

Thus, $\left\{a_{n}\right\}$ is an arithmetic sequence, and $\left\{a_{n}\right\}$ is
$$
0, d, 2d, 3d, 4d.
$$

Therefore, $a_{5}=4a_{2}$.
Hence, statements (2) and (3) are correct.
The set $A=\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 5\right\}$ contains nine elements.