4. If the set $A=\left\{\lg x, \lg y, \lg \left(x+\frac{y}{x}\right)\right\}$ has a subset $B=\{0,1\}$, then the maximum element of $A$ could be
A. 1
B. $2 \lg 3$
C. $\lg 11$
D. $\lg 15$
4. C Exhaustive Method, from the problem we have $\left\{\begin{array}{l}\lg x=0 \\ \lg y=1\end{array}\right.$ or $\left\{\begin{array}{l}\lg x=1 \\ \lg y=0\end{array}\right.$ or $\left\{\begin{array}{l}\lg x=0 \\ \lg \left(x+\frac{y}{x}\right)=1\end{array}\right.$ or $\left\{\begin{array}{l}\lg x=1 \\ \lg \left(x+\frac{y}{x}\right)=0\end{array}\right.$ or $\left\{\begin{array}{l}\lg y=1 \\ \lg \left(x+\frac{y}{x}\right)=0\end{array}\right.$ or $\left\{\begin{array}{l}\lg y=0 \\ \lg \left(x+\frac{y}{x}\right)=1\end{array}\right.$. In these 6 cases, the other element in set $A$ is, $\lg \left(x+\frac{y}{x}\right)=\lg 11, \lg 10.1, \lg y=\lg 9, \lg y$ does not exist, $\lg x$ does not exist, $\lg x=\lg (5 \pm 2 \sqrt{6})$, where $\lg 11$ is the largest.