Ah, the actual embedding is the compactification of that construction (which is missing a cantor set, which is also the set of ends the book is talking about)
Technically you can also take the standardly embedded sphere in $\Bbb R^3$ and throw away a cantor subset of the sphere and get an object with a cantor set of ends
It's just not a very interesting embedding though :)
Yeah, if $\overline{U_i}$ were compact then you could consider another sequence of compact sets with $K_n\cup U_i$ in place of $K_n$ for all $n\ge i$ so we don't want to have ends in $U_i$
Consider the one point compactification $L^*$ of $L$. $\{U_{\alpha_k}\}$ as defined is exactly a sequence of connected, decreasing neighborhoods of $\infty$ isn't it.
I'm not sure I'm following then, let's work with $\Bbb R$, its one point compactification is $S^1$ and the nbhds of $\infty$ are $\Bbb R\setminus [-n,n]$, but there is an end $\Bbb R\setminus (-\infty,n]$
Which makes sense since we want to track "how many ways" are there to go to infinity, while the one point compactification only cares about whether you can go to infinity
@BalarkaSen Right, the book describes the topology of $L\cup\text{End}(L)$ on page 112 (how does one make the big epsilon the authors use to denote the set of ends in latex?)
Oh, if $e \in \text{End}(L)$ and $U$ is an open subset of $L$ containing a fundamental system which defines $e$, then the set of $e' \in \text{End}(L)$ such that $e'$ is defined by a fundamental system contained in $U$ is an open nbhd of $e$ in $\text{End}(L)$
Pretty sure those are the basic open neighborhoods of $e$ in $\text{End}(L)$
@AlessandroCodenotti Yep
OK I get it, and the topology on $L \cup \text{End}(L)$ is also very clear.
A right notion of boundary should have a good topology on it
In this case, it's just defining how a sequence of ends converge to other ends.
$\{e_k\}$, a sequence of elements of $\text{End}(L)$, converges to $e \in \text{End}(L)$ if for every neighborhood $U \subset L$ containing a fundamental system that defines $e$ also contains fundamental systems of all but finitely many $e_k$'s.
@BalarkaSen So for the topology not to be discrete you need some end $e$ for which there is no open set $U\subset L$ such that $U$ contains a fundamental system defining $e$, but no fundamental system defining any other end, as in figure 4.2.3
Any open set of $L$ in 4.2.3 that contains a fundamental system defining $e_\infty$ contains the fundamental system defining $e_n$ for some large enough $n$
Ok, I convinced myself that the first step of the proof works, assuming such an exhaustion by compact submanifolds exists, is its existence obvious? Or known for other reasons?
I think an approach to get such an exhaustion is to impose a metric $d$ on $L$, consider $B_n(p)$ to be the ball of radius $n$ in $L$ under the metric $d$ so that $\{B_n(p)\}$ is a countable cover of $L$ and take an atlas $U_\alpha$ of $L$ such that for all $i$ $U_i$ is contained in $B_n(p)$ for some $n$.
Then take $K_n$ to be union of all such $U_\alpha$'s which intersect $B_n(p)$
In the definition of $d(e,e')$ I hope the authors made a typo and meant $2^{-r(e,e')}$, because using $e$ has a basis with all the $e$s already floating around is evil
Then $\mathcal{A}$ bijects to a directed tree whose nodes/vertices correspond to the $U$'s and two vertices are joined by a directed edge if $U \subset V$ (direction is towards "$U \to V$")
This is infinite upwards, towards the direction arrow points, and each branch of this tree (a path starting from anywhere and following the direction given) limits to an end
@BalarkaSen That depends on the space, it could have a finite amount of ends, I think you're thinking about a complete binary tree or something similar
Ok, good point, so we'd have to work that out explicitly. Let's just think about the combinatorial problem for now: We have a directed tree $T$ and $\text{End}(T)$ be the end space of the tree defined by the set of equivalence classes of directed edge-paths $\gamma$ starting from the root (so "rays"), under the equivalence $\gamma_0 \sim \gamma_1$ if there is another ray $\sigma$ which contains infinitely many vertices of $\gamma_0$ and $\gamma_1$ both.
Hm, actually this tree isn't necessary binary, is it? So a path in tree corresponds to an element of $\Bbb N^{\Bbb N}$ (each node still has at most countable many outgoing arrows, otherwise the space wouldn't be second countable)
Ok, I suggest this. Pick $n_k, m_k$ for each natural number $k$ such that $\{r(e_{n_k}, e_{m_k})\}_k$ is a strictly increasing sequence. You can always do this as $\lim_{m, n \to \infty} r(e_n, e_m) \to \infty$, I think.
Well, suppose $U_0=L$ for simplicity (just pick $K_0=\varnothing$), you can construct a tree as earlier whose edges represent inclusion among the components of the $U_k$, this has infinite depth, hence there is an infinite path
You can't pick a random infinite path. You have to pick one which goes through vertices representing either the component containing $e_{n_k}$ or the component containing $e_{m_k}$ at the $k$-stage
Say $\gamma$ is a path in $X$. Length of $\gamma$ makes sense
Namely, just define it as infimum of sum of $d(\gamma(t_i), \gamma(t_{i+1}))$ for a partition $t_i$ of $[0, 1]$, infimum taken over all such partitions
