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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 $n+1$? – Qiaochu Yuan Aug 4 '11 at 15:26
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 $n$. – Mustafa Gokhan Benli Aug 4 '11 at 15:29
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? – Qiaochu Yuan Aug 4 '11 at 15:30
Yes, I mean every node has exactly 2 children. Maybe a better name is "regular" binary tree. – Mustafa Gokhan Benli Aug 4 '11 at 15:31
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. – Qiaochu Yuan Aug 4 '11 at 15:42
up vote 1 down vote accepted

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. – Mustafa Gokhan Benli Aug 4 '11 at 19:17

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