Let's make a community wiki answer based on the above comments.
Given the subset
$A \subset \Omega$
The relative complement $A^c$ in $\Omega$ is defined as the set of elements of $\Omega$ that are not elements of $A$.
So in the case that $A = \Omega$, the relative complement of $A$ in $\Omega$ is equal to $\emptyset$.
That is, for every set $x$
we have $x \notin A^c$.
In particular we have $\emptyset \notin A^c$.
However, this is probably the "wrong" question to ask, because $\emptyset$ might not be an element of $\Omega$ at all.
Notice that the $\emptyset$ is a subset of every set, so in particular $\emptyset \subset \Omega$.
This is because to check whether $X \subset Y$ you take an arbitrary element of $X$ and check that it is also in $Y$. If $X$ is empty, then there is nothing to check.
The empty set is indeed a set (the set of no elements) and it is a subset of every set, including itself. $$\forall A: \emptyset \subseteq A,\;\text{ including if}\;\; A =\emptyset: \;\emptyset \subseteq \emptyset$$
$$\text{BUT:}\quad\emptyset \notin \emptyset \;\text{ (since the empty set, by definition, has no elements!)}$$
That is, being a subset of a set is NOT the same as being an element of a set: $$\quad\subseteq\;\, \neq \;\,\in: \;\; (\emptyset \subseteq \emptyset), \;\;(\emptyset \notin \emptyset).$$
$\emptyset \;\subseteq \;\{1, 2, 3, 4, 5\},\quad$ whereas $\;\;\emptyset \;\notin \;\{1, 2, 3, 4, 5\},\;$.
$\{3\} \subseteq \{1, 2, 3, 4, 5\},\quad$ whereas $\;\;3 \nsubseteq \{1, 2, 3, 4, 5\}, \text{... but}\; 3 \in \{1, 2, 3, 4, 5\}$.
Best Answer
Yes, you are correct. The universal set $U$ is characterized by $\forall x \colon x \in U$. Taking the complement yields a set $U^{c}$ that is characterized by $\forall x \colon x \not \in U^{c}$. This is equivalent to the statement $\neg \exists x \colon x \in U^{c}$ and hence $U^{c}$ is an (the) empty set. (Depending on your theory, there may not be a unique empty set.)