> Let $A$ be a complex Banach algebra and $M_A$ be the space of all non-zero multiplicative linear functionals on $A$ equipped with the weak$^*$-topology. Let $\widehat A$ be the image of $A$ under the Gelfand transform. Then a set $E \subseteq M_A$ is said to be a boundary for the Banach algebra $A$ if every $\widehat f \in \widehat A$ admits its maximum on $E$ i.e. for any $\psi \in M_A$ and $f \in A$ we have $$\left \lvert \psi (f) \right \rvert \leq \sup\limits_{\phi \in E} \left \lvert \phi (f) \right \rvert$$
> 
> Shilov showed that intersection of all the closed boundaries of $A$ is again a closed boundary of $A$ which is known as the Shilov boundary of $A.$

But I don't know what is meant by saying the Shilov boundary of the polydisc $\mathbb D^n \subseteq \mathbb C^n$ is $\mathbb T^n$ because $\mathbb D^n$ is not even a closed subset of $\mathbb C^n.$ Could anyone kindly give me some suggestion regarding this?

Thanks in advance.