@MichaelGreinecker That definition works better, but can still be an empty set. For instance, in the case of the C$^*$-algebra $c_0(X)$ for an uncountable set $X$.

Of course, sometimes a curve is a map from $\mathbb{R}$ into a manifold, and other times it is a Riemann surface, and still other times it is a reduced 1-dimensional separated scheme of finite type over a field.

@July Consider $\mathbb{R}^0$.

Functional analysis generally uses "positive" to mean "nonnegative" because the notion of strictly positive does not generalize well to infinite dimensional spaces (the interior of the positive cone of an ordered Banach space can be empty). Functional analysts also call "nonexpansive" mappings "contractions". It makes sense that the more frequently used concept gets the shorter name.

@მამუკაჯიბლაძე It appears that filtered colimits of commutative unital C*-algebras in the category of unital C*-algebras are commutative (the joint image of the diagram is a dense commutative *-subalgebra, so the whole algebra must be commutative). I didn't do it this way only because I already knew how to prove it in the special case of $C(2^\omega)$, and had never thought of proving the general case. Of course, "filtered" is necessary.

The comment that prompted my earlier comment has been deleted, making it look silly. Someone said that idempotence of sequential closure was necessary.

One does not need idempotence of sequential closure in a topological space $X$ to obtain a sequential topology. One simply takes the complements of sequentially closed sets to be the open sets. This produces a topology with no conditions on $X$'s original topology.

To get an absolutely convex example, you just take the absolutely convex hull (this will then be compact and must inherit the lack of embeddability in locally convex spaces). By Świrszcz's result, this is not an Eilenberg-Moore algebra of the monad of signed Radon measures of total variation $\leq 1$, i.e. there exists a signed Radon measure on it without a barycentre.

You can't get very far with finite or countable absolutely convex combinations. One can make a space with three points $\{x,0,-x\}$ where non-trivial absolutely convex combinations are mapped to 0. Being finite, this is compact in its discrete topology. Regarding local convexity, the necessity of this assumption was shown by the work of J. W. Roberts, who constructed a compact convex set in a topological vector space with no extreme points at all. Here is a more accessible reference: link.springer.com/article/10.1007/BF01578905

Therefore $\chi_C$ is the image of $\chi_{C'}$ under the map from $C(2^n) \rightarrow C(2^\omega)$. We can then define $g(\chi_C) = f_n(\chi_{C'})$. As $2^\omega$ is zero-dimensional, we can extend $g$ from indicator functions of clopens to a map $C(2^\omega) \rightarrow B$. Any other map $C(2^\omega) \rightarrow B$ making the diagram commute would take the same values on indicator functions of clopen sets, and since they have dense span, this proves the uniqueness.

@YemonChoi Your first comment is the correct interpretation, and $2^\omega$ is $\{0,1\}^{\mathbb{N}}$. Then $C(2^\omega)$ is separable because $2^\omega$ is a compact metric space. Gelfand duality is not quite enough to prove that $C(2^\omega)$ is the colimit in noncommutative C*-algebras. Given maps $f_n : C(2^n) \rightarrow B$ for $B$ a noncommutative unital C*-algebra, and $C \subseteq 2^\omega$ a clopen subset of $2^\omega$, there exists an $n$ and a subset $C' \subseteq 2^n$ such that $C$ is the preimage of $C'$ under one of the projection maps.

Section 5 of: ncatlab.org/nlab/show/convergence+space is also relevant.

@Adam The question says "countable", not "infinite", and as the article you linked to explicitly states in the second paragraph of the introduction, it is a theorem of ZF that there is no free complete Boolean algebra on any set containing a countable subset. The complete Boolean algebras in that article are free on Dedekind-finite sets, not $\mathbb{N}$.