Well-defined way to quotient by a relation involving an infinite sum Take a unital (*-)algebra $\mathcal A$ generated by a finite set of generators, $e_n$, and relations. We can require the generators to be (self-adjoint) projections. Some of the relations are of the form $\sum_{n\in S} e_n = 1$ for various finite sets $S$. This is okay so far.
Now I'd like to define an analogous algebra when there are countably many generators, and the sets $S$ involved in the sum-to-one relations are no longer finite.
Inherent in defining an object by generators and relations is selecting an axiomatic meaning for all symbols used (you have to know, for example, whether you are defining a group or a unital algebra over the complex numbers, etc.). My problem is with the infinite sum in a relation. It seems that the possible interpretations are to use infinite sums formally, or to impose some kind of topology.
What kind of mathematical object could work here?
(The eventual goal of defining this object is to consider representations in the bounded operators $B(\mathcal{H})$ on some Hilbert space $\mathcal H$, and to think about traces.)
 A: There is a new preprint on arxiv that addresses your question: Infinite sum relations on universal $C^\ast$-algebras
For the relations from your question, their construction goes as follows: They start with the free unital $\ast$-algebra $\mathcal{A}$ generated by the projections $e_n$, just as in your OP. A unital $\ast$-representation of $\mathcal{A}$ on a Hilbert space $H$ is called a representation of the defining relations if $\sum_{n\in S}e_n=1$ in the strong operator topology for all defining sets $S$.
They then define the following seminorm on $\mathcal{A}$:
$$
\lVert x\rVert=\sup_{\pi}\lVert \pi(x)\lVert,
$$
where the infimum is taken over all admissible representations $\pi$. Since every element of $\mathcal{A}$ is a finite linear combination of finite products of elements $e_n$, this supremum is indeed finite for every $x\in \mathcal{A}$ (in general, they require a property called admissibility to guarantee that this supremum is finite).
The associated $C^\ast$-algebra $A$ is then defined as the completion of $\mathcal{A}/\{x\in \mathcal{A}: \lVert x\rVert=0\}$. It is universal in the following sense: Whenever there is a unital $C^\ast$-subalgebra $B$ of $B(H)$ for some Hilbert space $H$ with projections $f_n\in B$ such that $\sum_{n\in S}f_n=1$ strongly for all $S$, then there exists a unique unital $\ast$-homomorphism $\phi\colon A\to B$ such that $\phi(e_n)=f_n$.
Also, $A$ has a faithful representation on a Hilbert space $H_u$ in which the defining relations hold in the strong operator topology. However, in their general construction it may happen that the $C^\ast$-algebra has faithful representations that don't satisfy the defining relations in the strong operator topology. In other words, the validity of $s$-$\sum_{n\in S}\pi(e_n)=1$ depends on the representation $\pi$ (in contract to universal $C^\ast$-algebras that involve only "finite" representation).
In the simple example when you only have one relation $\sum_{n=1}^\infty e_n=1$, then $A$ is isomorphic to $c_0+\mathbb{C}1$. Indeed, if projections $p_n$ on a Hilbert space $H$ satisfy $\sum_n p_n=1$ strongly, then they are mutually orthogonal. Thus $A$ is a commutative $C^\ast$-algebra, the linear combinations of the projections $e_n$ and $1$ are dense in $A$, and it suffices to compute its character space.
If $\chi\colon A\to \mathbb{C}$ is a character, then $\chi(e_n)$ is a projection, that is, $\chi(e_n)\in \{0,1\}$. Moreover, since $\sum_{n=1}^N e_n\leq 1$ for all $N\in \mathbb{N}$, for every character $\chi$ there is at most one $n\in \mathbb{N}$ such that $\chi(e_n)=1$. The universal property of $A$ implies for every $n\in \mathbb{N}$ the existence of a character $\chi_n$ such that $\chi_n(e_n)=1$, and it is not hard to see that the weak$^\ast$-topology on $\{\chi_n\mid n\in\mathbb{N}\}$ is discrete. As this space is not compact, the character space of $A$ must be homeomorphic to the one-point compactification of $\mathbb{N}$.
