Given a path connected topological space $X$, consider a sequence $x_1, x_2, x_3, \dotsc$ of points in $X$ converging at some $x \in X$. For each $x_i$ and $x_{i+1}$, there exists some path $p_i$ in $X$ from $x_i$ to $x_{i+1}$. I define the following for all positive integers $i$ and $j$ such that $i + 1 < j$:
$ p_{i \to i+1} = p_i \quad\text{and}\quad p_{i \to j} = p_i \cdot p_{i+1 \to j}, $
where $\cdot$ denotes path composition; for any paths $f, g : [0, 1] \to X$ such that $f(1) = g(0)$, $(f \cdot g) : [0, 1] \to X$ is a path such that
$ (f \cdot g)(t) = \begin{cases} f(2t) & 0 \leq t \leq {\textstyle\frac{1}{2}}\\ g(2t - 1) & {\textstyle\frac{1}{2}} < t \leq 1. \end{cases} $
It can be seen that $p_{i \to j}$ is a path from $x_i$ to $x_j$. Consider the sequence $p_{1 \to 2}, p_{1 \to 3}, p_{1 \to 4}, \dotsc$. Visually, the successive terms of the sequence show a path stretching a bit of its end to touch the next $x_k$. Under what conditions can I say that this sequence of paths converges to a path $p$ in $X$ with the following?
$ p(1 - 2^{-k}) = x_{k+1} \quad\text{and}\quad p(1) = x. $
For instance, does it suffice to require that $X$ has a metric $d$, and $\lim\limits_{i \to \infty} L(p_i) = 0$, where $L(q)$ denotes the length of a path $q$? However, this would impose a metric on $X$, which is a magic bullet I'd rather not use.
In fact, does the notion of path length exist? Are there further conditions I must pose on $X$ so that the following limit exists for some path $q : [0, 1] \to X$?
$ L(q) = \lim_{n\to\infty}\sum_{m=1}^n d\left(\textstyle q(\frac{m-1}{n}),q(\frac{m}{n})\right). $
(The following "appendix" may not be fully relevant but it seems interesting)
Consider the function $P : [0, 1]^2 \to X$, where
$ P(t,s) = \begin{cases} p_k(2 - 2^k(1 - t)) & t \leq s < 1, k = \lceil-\log_2(1 - t)\rceil\\ p_k\left(2 - 2^k\left(1 - \frac{t + s}{2}\right)\right) & s \leq t < 1, k = \left\lceil-\log_2\left(1 - \frac{t + s}{2}\right)\right\rceil\\ x & t = s = 1 \end{cases} $
$P$ can be visualized in the graph below:
P on the $ts$-plane">
Solid lines contain points in the $ts$-plane which are mapped by $P$ to the same point in $X$. If the sequence $p_{1 \to 2}, p_{1 \to 3}, p_{1 \to 4}, \dotsc$ indeed converges to some path $p$ in $X$, then we have the following:
- $P : [0, 1] \times [0, 1] \to X$ is a homotopy from $p_1$ to $p$.
- $P(1 - 2^{2 - k}, t) = p_{1 \to k}(t)$ for any integer $k \geq 2$ and $t \in [0, 1]$.
However, does the converse hold? If $P$ is continuous, then does the "limit path" $p$ exist?