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Let $\tilde{\mathbb N}$ be the Abelian semigroup (under addition) given by $\mathbb N\cup\{0,\infty\}$, and let $S_n$ be the Abelian monoid $\tilde{\mathbb N}^{2^n}$ under point-wise addition. Introduce the transition maps $\{\phi_n:S_{n+1}\to S_n\}_{n\in\mathbb N}$ defined by

$$\phi_{n-1}(n_1,n_2,n_3,n_4,\ldots,n_{2^n-1},n_{2^n}) := (n_1+n_2,n_3+n_4,\ldots,n_{2^n-1}+n_{2^n}),$$

i.e. summing all consecutive pairs of integers.

Now consider the sequence $(S_n,\phi_n)_{n\in\mathbb{N}}$. Does the projective limit

$$S=\lim_{\longleftarrow}S_n\subset\prod_{k\in\mathbb N}S_n$$

have a name as an Abelian semigroup, namely is such $S$ an object in a well-studied family of semigroups so that it can be actually identified among them?

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Let ACom be the class of all finite aperiodic commutative monoids (aperiodic means each subgroup is trivial, i.e. the monoid satisfies some identity of the form x^n=x^{n+1}); let pro-ACom be the class of all projective limits of monoids in ACom. The class pro-ACom has free objects on each compact totally disconnected space, in particular on each finite set (considered as a discrete space). Now \tilde\mathbb{N}=S_0 is the free pro-ACom object on a singleton (namely {1}) --- in this context S_0 has to be considered as one-point compactification of the discrete \mathbb{N}_0=S_0\setminus{\infty}. The monoid S in question is the free pro-ACom object on the Cantor set. Inside S the Cantor set of generators can be identified as those elements whose projection to S_0 is 1.

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The object $S$ in question arises from considering the C*-algebra C(X) of continuous functions on the Cantor set as first argument in a bivariant functor $F(C(X), \mathcal K)$, where $\mathcal K$ is the C*-algebra of compact operators. So the answer you provided makes a lot of sense in this context. Thank you very much! –  Phoenix87 Apr 4 '13 at 23:28
I was trying to locate some references for such category of semigroups, in particular some sources about how to construct transition maps between the semigroups in the projective limit given the topological space (the Cantor set in the example in question)? Could you please refer me to some literature? That would be great. Cheers! –  Phoenix87 Apr 5 '13 at 15:48
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Dear Phoenix87, I don't know where to write the answer to your latest question, so I put it here, though I'm afraid this will be interpreted as another answer to the original question. Anyway, two books where profinite monoids and semigroups are developed are: "Finite semigroups and universal algebra" by J. Almeida (in this book the term "profinite" is not used, yet such topics are treated) and "The q-theory of finite semigroups" by J. Rhodes and B. Steinberg.

Maybe I should make a further remark to your example which perhaps supports your understanding.

As I mentioned already, $S_0$ is the free proACom object on a single generator (lets call it $x_1$); more generally, $S_n$ is the free proACom object on a finite space consisting of $2^n$ elements, lets call this $X_n$ (the set consisting of $x_1,x_2,\dots, x_{2^n}$). Now the homomorphism $\phi_n:S_{n+1}\to S_n$ is completely determined by the values it takes on the free generators $X_{n+1}$ of $S_{n+1}$, these values are all in $X_n$ (free generators of $S_n$), according to your definition this mapping is just

$$x_1,x_2\mapsto x_1, x_3,x_4\mapsto x_2, x_5,x_6\mapsto x_3 \dots.$$

So you have an inverse system of finite discrete spaces

$$\dots \to X_{n+1}\to X_n\to X_{n-1}\to \dots\to X_0.$$

From the universal property of the projective limit it follows that the free object on the projective limit of the $X_n$s is just the projective limit of the free objects on the $X_n$s (you may interchange forming the projective limit and the free object). But the projective limit of the $X_n$s is just the Cantor set.

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Dear kar, thank you very much for the suggested reading and for your further explanation, it helped a lot and I think I can more clearly see the link with the Cantor set. I actually have a similar construction with $\mathbb N$ viewed as the limit of $\{1,\ldots,n\}$ with transition maps that act, on the generators, as $x_k\mapsto x_k$ for any $k=1,\ldots,n$ and $x_{n+1}$ is dropped. Hence this should lead to $\tilde{\mathbb N}^{\mathbb N}$ in the limit on the free pro-ACom semigroups. –  Phoenix87 Apr 5 '13 at 18:59
"$x_{n+1}$ is dropped" I guess, should be interpreted as each homomorphism $\phi_n:T_{n+1}\to T_n$ ($T_n$ the free object on $\{x_1,\dots,x_n\}$ as opposed to $S_n$ in the earlier example) maps $x_i$ to $x_i$ for $i\le n$ but $x_{n+1}$ to the identity element in the monoid $T_n$. The resulting concept is known as free profinite object "on an infinite set $X=\{x_1,x_2,\dots\}$ converging to $1$" and is a slightly different one (meaning the universal property in the definition of free object is subject to a slight restriction which mappings defined on $X$ are allowed) and $X$ is not compact. –  kar Apr 5 '13 at 19:26
Yes I actually meant 0 by "dropping" in this particular example. So the limit would not be a pro-ACom object, but I guess this non compact $X$ is just $\mathbb N$ in this other particular example? –  Phoenix87 Apr 5 '13 at 21:10
The limit is still a free pro-ACom object $F$, but not "on a compact, totally disconnected space $X$" but rather on "a set $X$ converging to 1". In the former case, each continuous map $f:X\to S$ ($S$ any (finite) monoid in ACom (equipped with discrete topology)) can be continued uniquely to a continuous homomorphism $\hat f:F\to S$ while in the latter case so does each map $f:X\to S$ which sends all but finitely elements of $X$ to the identity of $S$. So the difference is in the definition of the universal property the free object should satisfy.Yes, in the latter example $X$ is $\mathbb N$. –  kar Apr 5 '13 at 21:42
Ok so the difference is just in the set $X$, and the restrictions on the universal property account for the different nature of such set $X$. Anyway that is exactly what I was expecting, for this former case actually comes from $F(C_0(\mathbb N),\mathcal K)$, where $C_0(\mathbb N)$ denotes the C*-algebra of continuous functions on $\mathcal N$ (which isn't compact) vanishing at $\infty$. So thanks a lot for your help with this matter! –  Phoenix87 Apr 5 '13 at 22:32
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