Take a profinite group $G=\varprojlim G_\alpha$. We know that the inner automorphism group $\text{Inn}(G)$ of $G$ is profinite since $\text{Inn}(G)=G/Z(G)$, and the quotient of a profinite group by a closed normal subgroup (which $Z(G)$ must be as the intersection of centralizers) is profinite. I'm considering another object $\widehat{\text{Inn}}(G)=\varprojlim\text{Inn}(G_\alpha)$. Now I would like to show that $\text{Inn}(G)=\widehat{\text{Inn}}(G)$. A priori there may be things in $\widehat{\text{Inn}}(G)$ that are not in $\text{Inn}(G)$, so I believe that showing this is a worthwhile goal. There is an embedding $\text{Inn}(G)\hookrightarrow\widehat{\text{Inn}}(G)$ and I've managed to show that $\text{Inn}(G)$ is dense in $\widehat{\text{Inn}}(G)$. It remains to show that $\text{Inn}(G)$ is closed in $\widehat{\text{Inn}}(G)$. I believe that one way to go about this is to take a limit point $\varphi$ of $\text{Inn}(G)$ and show that any open subset of $\widehat{\text{Inn}}(G)$ that contains $\varphi$ must intersect $\text{Inn}(G)$. Can someone help me with this argument?
Another possible way involves considering what the basic closed sets look like in a Stone topological space presented as an inverse limit. I know what the basic open sets look like, but what do basic closed sets look like?
Edit: I want to assume all the maps $f:G_\alpha\to G_\beta$ are surjective.
Best Answer
The image of a compact set under a continuous map is compact, and compact subsets of Hausdorff spaces are closed, so the image of $\mathrm{Inn}(G)$ in $\widehat{\mathrm{Inn}}(G)$ is closed.
It follows (assuming you've correctly proved that the canonical map $\mathrm{Inn}(G)\to \widehat{\mathrm{Inn}}(G)$ is injective and has dense image) that this map is an isomorphism of profinite groups.
To answer your second question: What counts as a "basic open" or "basic closed" set in a topological space depends on the basis you've chosen (e.g. if we take the entire topology as a basis, every open set is basic open). Now if we have a Stone space $X$ presented as an inverse limit of finite sets, $X = \varprojlim (X_\alpha)$ and we write $\rho_\alpha\colon X\to X_\alpha$ for the projection maps, then a standard choice of basis is $\{\rho_\alpha^{-1}(Y)\mid Y\subseteq X_\alpha\}$. Note that if $Z = \rho_\alpha^{-1}(Y)$ is basic open, then its complement $X\setminus Z = X\setminus \rho_\alpha^{-1}(Y) = \rho_\alpha^{-1}(X_\alpha\setminus Y)$ is also a basic open set. So this basis $\mathcal{B}$ of open sets is closed under complement, and the corresponding basis of closed sets is again $\mathcal{B}$.