4. The number of elements in the set $\left\{(x, y) \left\lvert\, \lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y\right.\right\}$ is
(A) 0 ;
(B) 1 ;
(C) 2 ;
(D) More than 2 .
4. (B)
$$
\begin{aligned}
& \lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y \\
\Leftrightarrow & x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}=x y, x>0, y>0 .
\end{aligned}
$$

By the AM-GM inequality
$$
\begin{array}{c}
x^{3}+\frac{1}{3} y^{3}+\frac{1}{9} \geqslant 3 \sqrt[3]{\left(x^{3}\right)\left(\frac{1}{3} y^{3}\right)\left(\frac{1}{9}\right)}=x y . \\
\left\{\begin{array}{l}
x^{3}=\frac{1}{9} \\
\frac{1}{3} y^{3}=\frac{1}{9} .
\end{array}\right.
\end{array}
$$

The equality holds, solving these two equations gives
$$
x=\sqrt[3]{\frac{1}{9}}, y=\sqrt[3]{\frac{1}{3}},
$$

Thus, the set of points contains only one point:
$$
\left(\sqrt[3]{\frac{1}{9}}, \sqrt[3]{\frac{1}{3}}\right)
$$