[proofplan]
We prove that any connected component of the intersection of two domains of holomorphy is itself a domain of holomorphy. The strategy uses the pseudoconvexity characterisation: since each $\Omega_j$ is a domain of holomorphy, the [Solution of the Levi Problem](/theorems/3416) provides a continuous psh exhaustion $\phi_j$ for each $\Omega_j$. The pointwise maximum $\max(\phi_1, \phi_2)$ is a continuous psh exhaustion of $\Omega_1 \cap \Omega_2$, establishing pseudoconvexity, and the equivalence theorem then gives the domain-of-holomorphy property.
[/proofplan]
[step:Construct psh exhaustions for each $\Omega_j$ via the equivalence theorem]
Since $\Omega_1$ and $\Omega_2$ are domains of holomorphy, the [Solution of the Levi Problem](/theorems/3416) implies that each is pseudoconvex. Therefore there exist continuous plurisubharmonic exhaustion functions
\begin{align*}
\phi_j: \Omega_j \to \mathbb{R}, \quad j = 1, 2,
\end{align*}
with $\{\phi_j < c\} \Subset \Omega_j$ for all $c \in \mathbb{R}$. Concretely, one may take $\phi_j(z) = -\log d(z, \partial\Omega_j)$, which is psh by the equivalence (condition (iv) of the [Solution of the Levi Problem](/theorems/3416)) and is an exhaustion since $\phi_j(z) \to +\infty$ as $z \to \partial\Omega_j$.
[/step]
[step:Verify that $\max(\phi_1, \phi_2)$ is a continuous psh exhaustion of $\Omega_1 \cap \Omega_2$]
Define
\begin{align*}
\psi: \Omega_1 \cap \Omega_2 &\to \mathbb{R} \\
z &\mapsto \max(\phi_1(z), \phi_2(z)).
\end{align*}
Both $\phi_1$ and $\phi_2$ are well-defined on $\Omega_1 \cap \Omega_2$ since this set is contained in each $\Omega_j$.
**Continuity:** Since $\phi_1$ and $\phi_2$ are continuous on $\Omega_1 \cap \Omega_2$, the pointwise maximum $\psi = \max(\phi_1, \phi_2)$ is continuous.
**Plurisubharmonicity:** By the [Stability Properties of PSH Functions](/theorems/3404) (property (1): the pointwise maximum of finitely many psh functions is psh), the function $\psi$ is psh on $\Omega_1 \cap \Omega_2$. To verify directly: $\psi$ is upper semicontinuous (as the maximum of two continuous, hence upper semicontinuous, functions). For any $z \in \Omega_1 \cap \Omega_2$ and $w \in \mathbb{C}^n$, the restriction $\zeta \mapsto \psi(z + \zeta w) = \max(\phi_1(z + \zeta w), \phi_2(z + \zeta w))$ is the pointwise maximum of two subharmonic functions of $\zeta$, hence subharmonic (the sub-mean-value inequality holds for the one achieving the maximum).
**Exhaustion property:** For each $c \in \mathbb{R}$,
\begin{align*}
\{\psi < c\} = \{\max(\phi_1, \phi_2) < c\} = \{\phi_1 < c\} \cap \{\phi_2 < c\}.
\end{align*}
Since $\{\phi_j < c\} \Subset \Omega_j$ for $j = 1, 2$, the set $\{\phi_1 < c\} \cap \{\phi_2 < c\}$ is contained in $\{\phi_1 < c\}$, which is compactly contained in $\Omega_1$, and also contained in $\{\phi_2 < c\}$, which is compactly contained in $\Omega_2$. Therefore $\{\psi < c\}$ is compactly contained in $\Omega_1 \cap \Omega_2$.
This confirms that $\psi$ is a continuous psh exhaustion of $\Omega_1 \cap \Omega_2$.
[guided]
The exhaustion property is the step that requires the most care. Why is $\{\psi < c\} \Subset \Omega_1 \cap \Omega_2$? We need to show that the closure of $\{\psi < c\}$ is a compact subset of $\Omega_1 \cap \Omega_2$.
Since $\{\psi < c\} \subset \{\phi_1 < c\}$ and $\overline{\{\phi_1 < c\}}$ is compact in $\Omega_1$ (by the exhaustion property of $\phi_1$), the set $\overline{\{\psi < c\}} \subset \overline{\{\phi_1 < c\}}$ is bounded and contained in $\Omega_1$. Similarly, $\overline{\{\psi < c\}} \subset \overline{\{\phi_2 < c\}} \subset \Omega_2$. Therefore $\overline{\{\psi < c\}} \subset \Omega_1 \cap \Omega_2$, and since it is bounded and closed in $\Omega_1 \cap \Omega_2$ (and contained in a compact subset of $\mathbb{C}^n$), it is compact.
The maximum operation is natural here: it takes the "more restrictive" of the two exhaustion values, ensuring that $\psi$ blows up whenever either $\phi_1$ or $\phi_2$ blows up -- that is, whenever $z$ approaches the boundary of either $\Omega_1$ or $\Omega_2$.
[/guided]
[/step]
[step:Restrict to a connected component and apply the equivalence theorem]
If $\Omega_1 \cap \Omega_2$ is connected, the function $\psi$ is a continuous psh exhaustion of $\Omega_1 \cap \Omega_2$, so $\Omega_1 \cap \Omega_2$ is pseudoconvex.
If $\Omega_1 \cap \Omega_2$ is disconnected, let $\Omega'$ be any connected component of $\Omega_1 \cap \Omega_2$. Since $\Omega'$ is open in $\mathbb{C}^n$ (connected components of open sets are open), the restriction $\psi|_{\Omega'}$ is a continuous psh exhaustion of $\Omega'$: continuity and plurisubharmonicity are local properties preserved by restriction, and the exhaustion property holds because $\{\psi|_{\Omega'} < c\} = \{\psi < c\} \cap \Omega' \subset \{\psi < c\} \Subset \Omega_1 \cap \Omega_2$, so $\{\psi|_{\Omega'} < c\} \Subset \Omega'$ (a compact subset of $\Omega_1 \cap \Omega_2$ intersected with a closed-in-$\Omega_1 \cap \Omega_2$ set $\Omega'$ gives a compact subset of $\Omega'$).
Therefore $\Omega'$ is pseudoconvex. By the [Solution of the Levi Problem](/theorems/3416), pseudoconvexity is equivalent to the domain-of-holomorphy property, so $\Omega'$ is a domain of holomorphy.
[/step]
