5. Given real numbers $a, b, c$ satisfy
$$
a+b+c=2, ab+bc+ca=0 \text{, and } abc=-1 \text{.}
$$

Then the set of real numbers $a, b, c$ is
$\qquad$
5. $\left\{\frac{1-\sqrt{5}}{2}, 1, \frac{1+\sqrt{5}}{2}\right\}$.

From the problem, we know
$$
a+b=2-c, a b+c(a+b)=0, a b=-\frac{1}{c},
$$

we get
$$
\begin{array}{l}
-\frac{1}{c}+c(2-c)=0 \\
\Rightarrow c^{3}-2 c^{2}+1=0 \\
\Rightarrow(c-1)\left(c^{2}-c-1\right)=0 \\
\Rightarrow c=\frac{1 \pm \sqrt{5}}{2} \text { or } 1 .
\end{array}
$$

By symmetry, the required set is $\left\{\frac{1-\sqrt{5}}{2}, 1, \frac{1+\sqrt{5}}{2}\right\}$.