I'm looking at the following question, whose answer I disagree with: How to understand completed path algebra?
For context, let $Q = (V, A)$ be a quiver with infinitely many arrows. Then it is often useful to consider the completed path algebra $\widehat{kQ}$: this is the $I$-adic completion of $kQ$, where $I$ is the ideal generated generated by the set of arrows $A$. In other words, $\widehat{kQ}$ is the limit of the diagram $$\cdots \to kQ/I^3 \to kQ/I^2 \to kQ/I.$$ In the linked question, the accepted answer confirms that the underlying $\mathbb{C}$-vector space of $\widehat{\mathbb{C} Q}$ is the product $P = \prod_{p \text{ path}} \mathbb{C} p$. However, I just don't see how this can be true.
Consider the canonical projection $\widehat{\mathbb{C} Q} \twoheadrightarrow \mathbb{C} Q/I^2$. Notice that $\mathbb{C} Q/I^2$ can be identified as the $\mathbb{C}$-vector space spanned by the paths of length strictly less than $2$. Let $q = (q_p)_p \in P$ be the tuple consisting precisely of the arrows (and having $0$-entries elsewhere), i.e. $q_p = p$ whenever $p \in A$ is an arrow, and $q_p = 0$ otherwise. Therefore, the image of $q$ in $kQ/I^2$ should be the sum of the arrows in $Q$. However, if $Q$ has infinitely many arrows, $q$ has infinitely many non-zero terms, so the sum is not defined.
Instead, if we construct the limit $\widehat{\mathbb{C} Q} = \varprojlim \mathbb{C} Q/I^n$ in the canonical way—as a closed subalgebra of the product $\prod_n \mathbb{C} Q/I^n$—shouldn't we find that $$\widehat{\mathbb{C} Q} \cong \prod_{n \geq 0} \left( \bigoplus_{\operatorname{length}(p) = n} \mathbb{C} p \right)$$ consists of series, indexed by $n \geq 0$, of finite $\mathbb{C}$-linear combinations of paths of length $n$?