In the GNS-construction for an $C^*$-algebra $\mathcal A$ (see this script on page 30) one starts with a state $\phi:\mathcal A\rightarrow \mathbb C$ (positive linear functional with $\|\phi\|=1$). Take $N_\phi = \{A\in\mathcal A:\phi(A^*A)=0\}$ and show, that $N_\phi$ is a closed left ideal of $\mathcal A$. Then one continues using the completion of $\mathcal A/N_\phi$ with the inner product $\langle A,B\rangle=\phi(A^*B)$ (this completion shall be an Hilbert space).
I understand, that $N_\phi$ must be an ideal. So $\mathcal A/N_\phi$ is still an algebra. But it is necessary that $N_\phi$ is closed? If $N_\phi$ wouldn't be closed, can $\mathcal A/N_\phi$ still be a pre-Hilbert space? Which property of $\mathcal A/N_\phi$ needs the closeness of $N_\phi$?
Closedness of an ideal (or more generally subspace) is useful because it makes the quotient space Hausdorff. A topological vector space is Hausdorff if and only if $\{0\}$ is closed (by translation invariance of the topology, that means the space is $T_1$, and since it's a uniform space, that implies Hausdorff), and that means a quotient space $E/F$ is Hausdorff if and only if $F$ is closed.
Since non-Hausdorff topological vector spaces are often inconvenient, in situations where such would arise, one usually considers the associated Hausdorff TVS $E/\overline{\{0\}}$.
In the given situation, $N_\phi$ is closed because it is the level set of a continuous function ($A \mapsto \phi(A^\ast A)$) to a Hausdorff space ($\mathbb{C}$).