Example

Let $\Lambda$ be a finite set and $V\subset \mathbf{C}^{\Lambda}$ a compact convex subset. Then V is polynomially convex. In fact, let W be the polynomially convex envelope of V. Let us prove that $W\subset V$, which will imply the assertion. Let $x\in \mathbf{C}^{\Lambda}-$V. There exists a real hyperplane H in $\mathbf{C}^{\Lambda}$ which strictly separates $x$ and V (EVT, II, p. 41, Prop. 4). Let $f_{\mathbf{R}}$ be an $\mathbf{R}$-linear form on $\mathbf{C}^{\Lambda}$ and $\alpha \in \mathbf{R}$ such that H is the set of $y\in \mathbf{C}^{\Lambda}$ satisfying $f_{\mathbf{R}}(y) =\alpha$. Let $f$ be a linear form on $\mathbf{C}^{\Lambda}$ such that $f_{\mathbf{R}}=\mathscr{R}(f)$. We therefore have

$\mathscr{R}(f(x))>$ sup$_{y\in V}\mathscr{R}(f(y))$.

For every $t\in \mathbf{R}$ and $y\in V$, let us set $f_t(y) =t+f(y)$. We have $|f_t|-\mathscr{R}(f_t)\rightarrow 0$ in $\mathscr{C}(\mathbf{C}^{\Lambda},\mathbf{R})$ endowed with the topology of compact convergence when $t\rightarrow +\infty$. For $t$ sufficiently large, it follows that $|f_t(x)|>$ sup$_{y\in V}|f_t(y)|$ since V is compact. Thus $x\in \mathbf{C}^{\Lambda}-$ W since $f_t$ is a polynomial function.