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I would like to explicitly understand the topological space $\hat{\mathbb{N}}$, which is the profinite completion of the space of natural numbers.

First of all, the definition. There is an obvious functor $$ \mathsf{FinSet} \to \mathsf{Top} $$ $\mathsf{Top}$ has all cofiltered limits and the functor only takes values in cocompact objects, hence we obtain an induced fully faithful functor $$ \mathsf{Pro}(\mathsf{FinSet}) \to \mathsf{Top}, \\ \underset{i \in I}{\text{"lim"}} S_i \mapsto \underset{i \in I}{\operatorname{lim}} S_i $$ This is essentially the inclusion of the totally disconnected compact Hausdorff spaces. We may also view it as a kind of "realization" functor.

On the other hand, there's also an obvious functor in the other direction (ignore sizes issues) $$ \mathsf{Top} \to \mathsf{Pro}(\mathsf{FinSet}), \\ X \mapsto \underset{X \to S}{\text{"lim"}} S $$ this is a cofiltered diagram indexed by continuous maps $X \to S$ for finite sets $S$.

This defines an adjunction: $$ \mathsf{Pro}(\mathsf{FinSet}) \leftrightarrows \mathsf{Top} $$ I will call the monad associated to this adjunction $\widehat{(-)}$. To be explicit: This is the functor $$ \widehat{(-)}: \mathsf{Top} \to \mathsf{Top}, \\ X \mapsto \hat{X} = \underset{X \to S}{\operatorname{lim}} S $$

Now, it's easy to describe the underlying set of $\hat{X}$: It's just $$ \hat{X} = \left\{ (a_f)_{f: X \to S} \in \prod\limits_{f: X \to S} S \;\middle|\; \forall g: S \to S' : a_{gf} = g(a_f) \right\} $$

In particular, we can easily describe the unit map $X \to \hat{X}$: $$ X \to \hat{X}, \\ x \mapsto (f(x))_{f: X \to S} $$

Also, the unit map $X \to \hat{X}$ actually factors through the Stone-Cech compactification $X \to \beta X \to \hat{X}$. If $X$ carries the discrete topology, we can describe the Stone-Cech compactification as the space of ultrafilters on $X$, and so we can easily describe the latter map $\beta X \to \hat{X}$: $$ \beta X \to \hat{X}, \\ \mathcal{U} \mapsto \left( \underset{\mathcal{U}}{\operatorname{lim}} f(x) \right)_{f: X \to S} $$ where we may note that $f$ only takes finitely many values, so that there is exactly one value which $f$ takes $\mathcal{U}$-often.

Here's my question:

Is the natural map $\beta \mathbb{N} \to \hat{\mathbb{N}}$ a homeomorphism? If not, is there a more explicit description of $\hat{\mathbb{N}}$?

Smiley1000
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1 Answers1

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The natural map $\beta\mathbb{N}\rightarrow\hat{\mathbb{N}}$ is a homeomorphism. Indeed, the Stone-Čech compactification $\beta\mathbb{N}$ is compact Hausdorff by design and it is also extremally disconnected (see Corollary 6.2.29 in Engelking, General Topology, or here), hence totally disconnected and thus profinite. Then, the universal property of the profinite completion also gives you a map $\widehat{\mathbb{N}}\rightarrow\beta\mathbb{N}$. These maps are uniquely determined by their compatibility with the canonical maps $\mathbb{N}\rightarrow\beta\mathbb{N}$ resp. $\mathbb{N}\rightarrow\hat{\mathbb{N}}$, so their composite in either order is the respective identity by another application of the respective universal property. Thus, they are inverse homeomorphisms. [This argument generalizes to any discrete space instead of $\mathbb{N}$, or even any extremally disconnected Tychonoff space $X$.]

Thorgott
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    Or even more generally, $X$ can be replaced by any strongly zero-dimensional Tychonoff space $X$ (in the sense that any finite cover by cozero sets has a finite clopen refinement; the word cozero can be replaced by open when $X$ is normal). This follows from the known formula $\dim X = \dim \beta X$ and equivalence of totally disconnected and strongly zero-dimensional for compact Hausdorff spaces. In other words, we require that $\dim X = 0$ where $\dim$ is Lebesgue covering dimension. – Jakobian May 01 '25 at 11:33
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    This also gives the most precise result, since it shows $\beta X$ is totally disconnected $\iff X$ is strongly zero-dimensional Tychonoff space – Jakobian May 01 '25 at 11:43