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?