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Jan 7, 2012 at 18:32 comment added Benjamin Steinberg Nice question!!!
Jan 7, 2012 at 18:31 comment added Benjamin Steinberg The equicontinuity argument I give above shows by the Arzela-Ascoli theorem in the uniform setting that in the profinite case the closure is compact and the compact open topology and pointwise convergence coincide. You are correct that the general case is much more subtle.
Jan 7, 2012 at 17:25 comment added Niemi By the way, you are right, the topology of pointwise convergence on $F$ is certainly not discrete. I have corrected that above.
Jan 7, 2012 at 16:44 comment added Niemi I did not doubt the fact that the act functions become continuous in the profinite case. In the general case this does not work as you cannot use projections.
Jan 7, 2012 at 15:22 history edited Benjamin Steinberg CC BY-SA 3.0
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Jan 7, 2012 at 14:39 history edited Benjamin Steinberg CC BY-SA 3.0
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Jan 7, 2012 at 14:32 history edited Benjamin Steinberg CC BY-SA 3.0
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Jan 7, 2012 at 13:17 comment added Benjamin Steinberg Of course these operations are cts in the profinite case since they are for finite groups and by naturality they commute with projective limits. Look in the book of Ribes and Zalesskii where they have a section on raising elements of a profinite group to a power from $\widehat Z$. See also the paper of Almeida in transactions of the AMS on dynamics of implicit operations where a more general setting is cts. I think all this works more generally for compact Hausdorff groups but there are details I have no time to check.
Jan 7, 2012 at 11:42 comment added Niemi 3. In your construction, there is no argument that substantiates that the maps $\nu_{G} \colon G \to G$ are continuous. So, what you have described is just the closure of $F$ in $G^G$. However, the point of my question was that I would like to know the closure in $C(G,G)$ in a more concrete manner than just stating that it is the intersection of the closure of $F$ in $G^G$ intersected with $C(G,G)$.
Jan 7, 2012 at 11:41 comment added Niemi 2. I think you are wrong with your assumption that the compact-open topology and the topology of pointwise convergence coincide. In fact, I seriously doubt that this happens in the profinite case.
Jan 7, 2012 at 11:41 comment added Niemi 1. There is no reason why the closure of $F$ must be compact. Note that $C(G,G)$ does not need to be closed in $G^G$.
Jan 7, 2012 at 11:41 comment added Niemi Thank you for your answer. However, I do have some comments on what you say. For a better overview (and due to the character limits for comments), I will poste them in separate comments.
Jan 6, 2012 at 20:19 history edited Benjamin Steinberg CC BY-SA 3.0
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Jan 6, 2012 at 20:13 history answered Benjamin Steinberg CC BY-SA 3.0