# Do locally path connected locally compact metric spaces have an intrinsic metric?

Given a metric space $(X,d)$ we define the path metric $\rho(x,y)$ as the infimum of all paths lengths from $x$ to $y$. The length of a path $\gamma$ means the supremum of all $\sum_i d \big (\gamma(x_i),\gamma(x_{i+1}) \big )$ for all choices of points $0 \le x_1 < x_2 <\ldots < x_n \le 1$.

It is known $\rho$ induces a finer topology than $d$. Sometimes strictly finer. For example consider the harmonic fan: the union of all line segments from points $(1,1/n)$ and $(1,0)$ to $(0,0)$. Then $(X,d)$ is compact but $(X,\rho)$ is noncompact.

The harmonic fan is not locally arc-connected. I wonder what happens if we assume the space is arc-connected.

Suppose the metric space $X$ is arc-connected, locally compact and locally arc-connected. Is the metric on $X$ equivalent to the path metric?

I believe this is equivalent to the following (which I cannot prove either).

Suppose $X$ is above and $x \in X$. Let $U_1 \supset U_2 \supset \ldots$ be a chain of open neighborhoods of $x$ with diameter decreasing to zero. Then the path-diameter of $U_i$ also decreases to zero.

Any ideas?

If it helps we can also assume each point has a basis of closed arc-connected neighborhoods.

• You definition of the path length is not correct. It is the supremum of such sums where $x_i = \gamma(t_i)$ and $t_1,...,t_n$ is a partition of the interval on which $\gamma$ is defined. See for example encyclopediaofmath.org/index.php/Rectifiable_curve. – Paul Frost Aug 2 '18 at 11:49
• How do you treat the case that there exist paths with infinite length? – Paul Frost Aug 2 '18 at 11:57
• Correct. There was an error in the definition of path length. – Daron Aug 2 '18 at 12:17
• We're assuming local arc-connectedness. Arcs in $X$ have finite length. For let $\Gamma \subset X$ be an arc. By compactness the homeomorphism $\Gamma \to [0,1]$ is uniformly continuous and the modulus lets us put a bound on the length of $\Gamma$ in $X$. – Daron Aug 2 '18 at 12:21
• Error not yet corrected. If you have a path $\gamma : [0,1] \to X$ and allow all choices $x_i \in \gamma([0,1])$, the supremum will always be infinite. You have to take the supremum over all sequences $x_1,..., x_n$ having the form $x_i = \gamma(t_i)$ with $0 = t_1 < t_2 < ... < t_n = 1$. – Paul Frost Aug 2 '18 at 12:34

Here are two old results. The first:

Theorem (Bing: A convex metric for a locally-connected continuum, 1928)

Each locally connected metric continuum admits an equivalent convex metrix.

The second:

Theorem (Goebel & Kirk: Topics in Metric Fixed Point Theory p 24)

Each complete and convex metric space has the path metric equivalent to the original metric.

Now suppose $(X,d)$ is as described in your question. The path metric topology is always finer than the original topology. We need to show it is coarser too. So suppose $x_n \to x$ with respect to $d$ and let $K$ be a locally-connected continuum neighborhood of $x$.

Some tail $y_m$ of $x_n$ is contained in the metric subspace $(K,d_K)$. The above theorems say $d_K$ is equivalent to the $d_K$ path metric $\rho_K$. Let $\epsilon>0$ be arbitrary. Some tail $y_N,y_{N+1}, \ldots$ has all $\rho(y_i,x)<\epsilon$. That means there are paths $\gamma_i \subset K$ from $y_i$ to $x$ of $d_K$ length less than $\epsilon$.

But since $d_K$ is the restriction of $K$ these paths $\gamma_i$ are also paths of $d$-length length less than $\epsilon$. Since $\epsilon$ is arbitrary we get $\rho(y_m,x) \to 0$ as required.