Your guess is correct and you can see it a little bit more formally as follows. Let $\mathcal{A} = (Q, A, \cdot, q_0, F)$ be a DFA. The Nerode congruence $\sim$ on $Q$ is defined as follows:
$$
p \sim q \text{ if and only if, for every word $u \in A^*$, }\ p \cdot u \in F \iff q \cdot u \in F
$$
The set of states of the minimal automaton of $\mathcal{A}$ is $Q/{\sim}$.
Now if $F$ is the empty set, all the states of $Q$ are $\sim$-equivalent and thus $Q/{\sim}$ has only one element, say $Q/{\sim} = \{1\}$. You have no choice for the transitions and thus $1 \cdot a = 1$ for each letter $a$. Finally $1$ is the initial state, but there is no final state.