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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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    I guess you mean the complete binary rooted tree. What exactly does $\phi_n$ do to level $n+1$?2011-08-04
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    What do you mean by complete? If we idetify vertices with 0,1 words and a word $w$ has 2 children $w0$ and $w1$, $\phi_n$ identifies the vertices $w0,w1$ on level $n+1$ with $w$ which is on level $n$.2011-08-04
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    I mean probably what you mean, since you used "the": every node has two children except in the last level. (In general a binary tree is a tree in which every node has _at most_ two children.) Anyway, I'm not sure what you mean by "structure." The inverse limit ought to be the infinite complete binary rooted tree, right?2011-08-04
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    Yes, I mean every node has exactly 2 children. Maybe a better name is "regular" binary tree.2011-08-04
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    The inverse limit will be compact (I guess) so cannot be complete binary tree. But I am not sure.2011-08-04
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    The inverse limit in what category? I'm working in the category of graphs. I guess you're working in the topological category?2011-08-04
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    I am not exactly sure what kind of topology is involved. If we put discrete topology on each $T_n$ then $\phi_n$ will be continuous and the inverse limit will be a compact topological space.2011-08-04
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    The _discrete_ topology? Are you sure you don't want the obvious topology as a subspace of, say, $\mathbb{R}^2$?2011-08-04
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    I was identifying $T_n$ with 0,1 words of length at most $n$ and putting discrete topology on it.2011-08-04
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    So you don't actually want to consider the edges at all? It seems to me you need to be a _lot_ more precise in your wording of this question. It already seems like it admits (at least) three possible interpretations.2011-08-04
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    Qiaochu, I know I was not very precise but 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