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Let $T_n$ be the binary rooted tree of $n$ levels. Let $ \phi_n : T_{n+1} \rightarrow T_n$ be the quotient map collapsing level $n+1$.

What kind of structure does the inverse limit have?

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    Qiaochu, I know I was no$t$ very precise bu$t$ my intention was to get a feeling about what the resulting object is in various cases.2011-08-04

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This is bigger than the infinite binary tree $T_\omega$. $T_\omega$ is contained in the inverse limit as the set of chains $(\ldots,x_n,x_{n-1},\ldots,x_1)$ where $x_n$ is constant for large enough $n$. However there are chains in the inverse limit which are not eventually constant, corresponding to paths in the tree $T_\omega$ starting at the base and going out to $\infty$. Each such path corresponds to a point in the cantor set, so it looks to me like the inverse limit is the compactification of $T_\omega$ by adding a cantor set at $\infty$.

Edit: If you are just considering the nodes of the tree and not the edges, this analysis still holds. (In fact it's slightly easier.) You are still adding a cantor set at infinity to the set of vertices of an infinite binary tree.

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    Thanks for the answer. If $T$ is the binary tree and $\partial T$ is its boundary (i.e infinite sequences) one can put a topology on the disjoint union $T \cup \partial T$ which makes it compact. I think this is what you are describing. I found this in the book of V Nekrashevych about self-similar groups.2011-08-04