A profinite set is a compact, Hausdorff, totally disconnected topological space, another description of a profinite set is that it is a space $X$ that can be written as a (categorical) limit $X = \lim X_i$ where $X_i$ are finite discrete topological spaces.
The topological space $\mathbb{N} \cup \{ \infty \}$ is profinite, how can I writte him as $\mathbb{N} \cup \{ \infty \} = \lim X_i$ explicitly?
Best Answer
To not let the comment by Idéophage go to waste, I'll add my own write-up:
Let $X_n = \{1,2,3,\ldots, n\}$ for all $n \in \Bbb N$ in the discrete topology. The connecting morphisms $f_n: X_n \to X_{n-1}$ ($n \ge 2$) are
$$f_n(i) = \begin{cases} i & i < n \\ n-1 & i=n\end{cases}$$
Now some inverse limits 101:
The (inverse) limit $X_\infty$ consists of all "connected" sequences in $P:= \prod_{n \in \Bbb N} X_n$, i.e.
$$X_\infty = \{ (x_n)_n \mid \forall n: x_n \le n \text{ and } f_n(x_n)=x_{n-1}\}$$
which gets the subspace topology from the product topology on $P$ (and as $P$ is compact, Hausdorff and zero-dimensional, and $X_\infty$ is a closed subset of it, it has those same properties too).
There are just two types of points in $X_\infty$: either we have some $n$ so that $x_n =n$ and $x_{n+1} = n$ (that $n$ is unique and we'll have $x_k=n$ for all $k > n$ onwards, a "stagnant" point, which I will denote by $\mathbf{n} \in X_\infty$) or the point $x_n=n$ for all $n$ which I will denote by $\mathbf{\infty} \in X_\infty$. There are no other points as when we have $x \in X_\infty$ and $x_m < m$ for some $m$ then all $k <m$ the value is determined by the linkedness condition and for all $k>m$ too as a point $<m$ in $X_m$ only has one pre-image under the connecting map. So either we stagnate at a value or we keep increasing.
In terms of open sets, define $U(k,l) := \{x \in X_\infty\mid x_k =l\}$ for $k \in \Bbb N$ and $l \le k$ which are open (as $\pi_k^{-1}[\{l\}] \cap X_\infty$) in $X_\infty$ and by the previous considerations $$\{\mathbf{n}\} = U(n,n)\cap U(n+1,n)$$ so these are isolated points in $X_\infty$ and $U(n,n)$ is a neighbourhood of $\mathbf{\infty}$ that contains $\{\mathbf{k}\mid k \ge n\}$. It's now quite clear that $X_\infty \simeq \Bbb{N} \cup \{\infty\}$ (a convergent sequence, aka as the one-point compactifcation of $\Bbb N$ or $\omega+1$ as an ordinal space).