Let $\mathbf{B}$ be a boolean algebra with the universe $B$ and let $F_1$, $F_2$ be distinct ultrafilters of $\mathbf{B}$. Then the set of the form $\{x\in B:F_1\cap F_2\cap \{x,\mathrm{\neg }x\}\ne \mathrm{\varnothing }\}$ is a maximal proper subuniverse of $\mathbf{B}$ which we shall call a basic subuniverse. We prove that every proper subuniverse of $\mathbf{B}$ is an intersection of a family of basic subuniverses. This implies that basic subuniverses are precisely maximal proper subuniverses of a boolean algebra. The same fact proved in another way can be found in [3].