Intersection of Two Boundaries in a Commutative Banach Algebra Given a Commutative Banach Algebra say $\mathcal{A}$, call it Maximal Ideal Space $\mathcal{M}_{\mathcal{A}}$. Let $\mathcal{B}_{\lambda}$ be a set where every $\phi \in \mathcal{M}_{\mathcal{A}}$ obtain its maximum. We call $\mathcal{B}_{\lambda}$ a boundary of $\mathcal{A}$. My question is given a general Commutative Banach Algebra, is the intersection of two boundaries always a boundary?
I came across a term called Shilov Boundary and in the book called Commutative Normed Rings written by Gelfand (Chapter 11), I found the proof for the unique existence of minimal boundary, which is the Shilov Boundary. However, no explicit construction of Shilov Boundary is given.
Thanks for Martin's correction, I realize in some $C(X)$ (e.g: in the example given by him) the intersection of two boundary could be empty. We need to go into the function system (or function algebra) of $C(X)$, which is a closed subalgebra what separate points and contain constant functions. Let $\mathcal{U}$ be a function system in $C(X)$. In this case I can view boundaries of $\mathcal{U}$ as subsets of $X$. Given two boundaries $\mathcal{B}_{\lambda_1}$ and $\mathcal{B}_{\lambda_2}$, their intersection will not be empty (by definition of boundary and Urysohn's Lemma). Hence $\bigcap_{\lambda \in \Lambda}\,\mathcal{M}_{\lambda}$ will not be empty. If finite intersection of boundaries is also a boundary, then for eacah $f \in C(X)$, there will be a net indexed by $\mathcal{F} = \{\bigcap_{F \subseteq \Lambda}\,\vert\, \vert\,F\,\vert < \infty\}$ (say $\{x_F\}_{F \in \mathcal{F}}$) such that $f(x_F) = \|f\|_{\infty}$. This net will converge to a point in $\bigcap_{\lambda \in \Lambda}\,\mathcal{M}_{\lambda}$ and this also proves the unique existence of Shilov Boundary. 
Note:
The use of net is originally from Commutative Normed Rings Chapter 11 and this question is inspired by Problem 2.27 from Banach Algebra Technique in Operator Theory written by Douglas
 A: The answer is "no", in general. 
Note that if you take all of $C(X)$, then the only boundary is $X$, precisely because of Urysohn. The Silov boundary is an interesting object when you consider proper subalgebras (or even subspaces) of $C(X)$; more properly, one can show that the Silov boundary exists for function systems (that is, subspaces of $C(X)$ that contain $1$ and the conjugates of its elements, and that separate points). Here are a few super basic examples (note that any closed set that contains the Silov boundary is a boundary): 


*

*On $C[0,1]$, let $\mathcal F=\operatorname{span}\{1,x\}$. Then the Silov boundary is $\{0,1\}$. 

*On $C(\mathbb T)$, let $\mathcal F=\operatorname{span}\{1,z,\bar z\}$. Then the only boundary of $\mathcal F$ is $\mathbb T$. 

*On $C(\overline{\mathbb D})$, let $\mathcal F=\operatorname{span}\{1,z,\bar z\}$. The the Silov boundary of $\mathcal F$ is $\mathbb T$. 
The proof that I know of the existence of the Silov boundary for function systems inside $C(X)$ is not complicated, but not trivial either. Regarding your argument, you say that the intersection of boundaries is a boundary; this is trivially true in your case because the only boundary is $X$. In the case of a function system inside $C(X)$, the only proof I know that the intersection of boundaries is a boundary comes from first showing that a minimal boundary exists (i.e., the Silov boundary exists). 
The requirement that the function system separates points is essential. Otherwise, and this answers your question, consider for instance
$$
\mathcal F=\{f\in C[0,1]:\ f(t)=f(1-t),\ t\in[0,1/2]\}|. 
$$
This is a Banach algebra (a C$^*$-algebra, actually), but not a function system because it does not separate points. In this case both $[0,1/2]$ and $[1/2,1]$ are boundaries, but their intersection is clearly not a boundary. In fact, with a simple tweak one can get disjoint boundaries: for instance the subalgebra $\{f(3-t)=f(t)\}\subset C([0,1]\cup[2,3])$ has $[0,1]$ and $[2,3]$ as boundaries. 
