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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:

Visualizing $P$ 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:

  1. $P : [0, 1] \times [0, 1] \to X$ is a homotopy from $p_1$ to $p$.
  2. $P(1 - 2^{2 - k}, t) = p_{1 \to k}(t)$ 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?

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As you indicate, this does not work in general. How do you define the length of a path in a topological space? – Rasmus Jun 29 '12 at 14:19
Oops, did I wrongly assume that path length was defined? If the metric on $X$ is $d$, then is it ok to define the length of a path $p : [0, 1] \to X$ as $\displaystyle\lim_{n \to \infty}\sum_{m = 1}^n d\left(\textstyle p(\frac{m - 1}{n}),p(\frac{m}{n})\right)$? – Herng Yi Jun 29 '12 at 14:22
@HerngYi Please edit your post to indicate that $X$ is a metric space. – user31373 Jun 29 '12 at 17:44
@Kovalev In the part of the question that used the notion of path length, I have added that $X$ had to be a metric space. However, is there a way to guarantee the existence of the limit of the sequence of paths without using the notion of path length? – Herng Yi Jun 30 '12 at 1:03
@Kovalev: If $X$ were a quasiconvex metric space with metric $d$, and we required accordingly that there exists some $C \in \mathbb{R}$ such that $L(p_i) \leq Cd(x_i, x_{i+1})$ for all positive integers $i$, then by the Squeeze Theorem, $0 \leq \lim\limits_{i\to\infty}L(p_i) \leq C\lim\limits_{i\to\infty}d(x_i, x_{i+1}) = 0$. Does this imply the existence of the limit, $p$, of the sequence of paths? – Herng Yi Jun 30 '12 at 1:36
up vote 2 down vote accepted

Here is an example, even a metric example, where the path you want doesn't exist. Take a topologist's sine curve and connect $(0, 0)$ to, say, $\left( \frac{1}{\pi}, 0 \right)$ via a path going below the rest of the curve, so the resulting space is path connected. (I think this is sometimes called the "topologist's circle.") Let $x_k$ be, say, $\left( \frac{1}{\pi k}, 0 \right)$. Then $x_k \to (0, 0)$ but the path you want doesn't exist (the interval is compact but the image of the path you want can't be compact because it has limit points it doesn't contain).

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