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Let $(X,T)$ be a Hausdorff topological space. Suppose $\emptyset\neq A\subset C(X)$, where $C(X)$ is the set of curves in $X$, that is, the set of continuous maps $\alpha:I\to X$, where $I\subset\mathbb{R}$ is an interval (more specifically a non-empty connected set). Every element in $A$ is said to be an admissible path. Suppose $A$ satisfies:

  1. The restriction in any interval of an admissible path is an admissible path
  2. The concatenation of two admissible paths is an admissible path.
  3. Given $x\in X$, there is an admissible path passing through $x$.
  4. Linear reparametrizations of admissibles path become admissibles paths.

Suppose now a map$L:A\to\mathbb{R}\cup\{\infty\}$ (called length) such that:

  1. $L$ is nonnegative.
  2. $L$ is invariant under linear reparametrizations.
  3. $L$ is additive respect to concatenation.
  4. If $\alpha:[a,b]\to X$ is an admissible path. Then the map that measures the length of the restriction of $\alpha$ to the interval $[t,s]$ is continuous for $t$ and $s$.
  5. If $U$ is open and $x\in X$, then $\inf\{L(\alpha): \alpha \text{ an admissible path joining $x$ and $y$}, y\notin U\}>0.$

Under this assumptions, we call $(X,T,A,L)$ a length structure over $X$.

We can define a metric (called intrinsic metric) in $X$ as follows: $$d_L(x,y)=\inf\{L(\alpha):\alpha\text{ an admissible path joining $x$ and $y$}\}.$$ Consider $\inf\emptyset=\infty$. I ommit details, but is not hard to prove that $d_L$ is a metric on $X$.

Now my question: prove that the topology generated by $d_L$ is finer than $T$.

Note: $X$ is simply a topological space. No metric is supposed in $(X,T)$.

I tried it several times, but no results have found.

Thanks for every help.

jdods
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user279325
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1 Answers1

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Let $T_L$ denote the topology defined by $d_L$, we have to show that $T \subseteq T_L$. So let $U\in T$ be given, to show $U \in T_L$, let $x \in U$. Define $$ \epsilon := \frac 12 \inf\{L(\alpha): \alpha \text{ an admissible joining of $x$ and $y$, } y \not\in U\} > 0 $$ Note that $\epsilon > 0$ by (5). Now let $y \in X$ with $d(x,y) < \epsilon$ be given. By definition of $\epsilon$ we have that $y \in U$ (otherwise $\epsilon \le \frac 12d(x,y) < \frac 12\epsilon$). Hence, $B^{d_L}_\epsilon(x) \subseteq U$, so $x$ is an $d_L$-inner point of $U$. Therefore $U$ is $d_L$-open, that is $U \in T_L$.

martini
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