Prove: $m^*({x_0}) = 0 = m^*(\emptyset)$

Prove: $ m^*({x_0}) = 0 = m^*(\emptyset)$ for any $ x_0\in \mathbb{R}$

By definition for $ E\subset \mathbb{R}$

$ $ m^*(E)=inf\{\sum_{n=1}^{\infty}|I_n|:E\subset\cup_{n=1}^{\infty} I_n\}$ $

Where $ I_n$ are open intervals.

let $ I_n=(x_0-\epsilon_n,x_0+\epsilon_n)$ for an arbitrary small $ \epsilon$ there for $ |l_n|=2\epsilon$ and $ m^*(\{x_0\})=0$

As for $ m^*(\emptyset)$ we can say that any cover including $ 0$ will cover the empty set?