6. Set $A=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$, calculate the sums of any two elements in $A$ to form set $B=$ $\{3,4,5,6,7,8,9,10,11,13\}$. Then $A=$ $\qquad$ .
6. $\{1,2,3,5,8\}$.

Assume $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=3, \\
x_{4}+x_{5}=13 .
\end{array}
$$

Since in the set of binary elements of $A$, each element appears 4 times, the sum of the elements of set $B$ is
$$
\begin{array}{l}
4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right) \\
=3+4+5+6+7+8+9+10+11+13 .
\end{array}
$$

Thus, $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=19$.
From (3) - (1) - () we get
$$
x_{3}=19-3-13=3 \text {. }
$$

Since $x_{1}+x_{3}$ is only greater than $x_{1}+x_{2}$, hence
$$
x_{1}+x_{3}=4 \text {. }
$$

Thus, $x_{1}=1$.
Since $x_{3}+x_{5}$ is only less than $x_{4}+x_{5}$, hence $x_{3}+x_{5}=11$.
Thus, $x_{5}=8$.
Substituting (3) and (6) into (1) and (2) we get
$$
x_{2}=2, x_{4}=5 \text {. }
$$

Therefore, $A=\{1,2,3,5,8\}$.