Convex metric on a continuum.
I am having troubles with this problem.
Translation:
A metric $d$ on a continuum is called convex if for any two points $p$ and $q$ of $X$ there is a point $x\in X$ such that $d(p,x)=\frac12d(p,q)=d(q,x)$. Prove that if the metric on $X$ is convex, then the balls are connected, and $X$ is locally connected.
Once the balls shown that the balls are connected, the following implication comes for free.
My idea is to suppose that there are $x_0\in X$ and $\epsilon_0>0$ such that $B(x_0,\epsilon_0)=\{x\in X:d(x,x_0)<\epsilon_0\}$ is not connected. Then there are $U$ and $V$, open, disjoint, and non-empty, such that $B(x_0,\epsilon_0)=U\cup V$.
The next part is complicated to write, but the idea is to construct a Cauchy sequence in $B(x_0,\epsilon_0)$, and since we’re in a compactum it will converge to some element of $X$, which will in fact be in the boundary of $B(x_0,\epsilon_0)$.
To do this we suppose without loss of generality that $x_0\in U$. Let $x_1\in V$. Since $d$ is convex there is an $x_2\in X$ such that $d(x_0,x_2)=\frac12d(x_0,x_1)=d(x_1,x_2)$. Then $x_2\in U\cup V$ (since $d(x_0,x_2)<d(x_0,x_1)<\epsilon_0$).
If $x_2\in U$, then we look at $x_1\in V$, $x_2\in U$, and there is an $x_3\in X$ such that $d(x_1,x_3)=\frac12d(x_1,x_2)=d(x_2,x_3)$. (Again we see that $x_3\in B(x_0,\epsilon_0)$ and moreover $d(x_0,x_3)<\frac12\epsilon_0$.)
If $x_2\in V$, then we look at $x_0\in U$, $x_2\in V$, and there is an $x_3\in X$ such that $d(x_0,x_3)=\frac12d(x_0,x_2)=d(x_2,x_3)$. (Again we see that $x_3\in B(x_0,\epsilon_0)$ and moreover $d(x_0,x_3)<\frac12\epsilon_0$.)
I hope that the construction of the sequence $(x_n)$ is understood. According to me one can prove that we will have satisfied the following two conditions:
i) $d(x_n,x_{n+1})<\frac1{2^n}\epsilon_0$ for all $n$ (so it will be Cauchy)
ii) $d(x_0,x_n)<\frac12\epsilon_0$ for all $n$
Then on the one hand we have $(x_n)\to x$ for some $x\in X$. Then $x\in\operatorname{bdry}B(x_0,\epsilon_0)$. But on the other hand $d(x_0,x)\le d(x_0,x_n)+d(x_n,x)<\frac12\epsilon_0+\frac12\epsilon_0=\epsilon_0$ for each $n\ge k$ for some $k\in \mathbb{N}$. Then $x\in B(x_0,\epsilon_0)$, which is a contradiction. Therefore the open balls are connected, and therefore $X$ is locally connected.
I don’t actually know whether this proof is okay or if there is another easier way. I hope you can help. Thank you.
"Una métrica $d$ en un continuo se llama convexa si para cualesquiera 2 puntos p y q de X, existe un punto $x\in X$ tal que $d(p,x)=\frac{1}{2}d(p,q) =d(q,x)$. Prueba que si la métrica para X es convexa, entonces las bolas son conexas y entonces X es localmente conexa."
Una vez teniendo lo de que las bolas son conexas, la siguiente implicación viene de regalo. Ahora, mi idea es suponer que existe $x_{0} \in X$ y $\epsilon _{0} >0$ talque $B(x_{0},\epsilon _{0})=\left\{ x\in X : d(x,x_{0})<\epsilon_{0} \right\}$ no es conexa. Entonces existen U y V abiertos, ajenos, no vacíos tales que $B(x_{0},\epsilon _{0})=U\cup V$. La siguiente parte es complicada de escribir, pero la idea es construir una sucesión de Cauchy en $B(x_{0},\epsilon _{0})$ y como estamos en un compacto ésta va a converger a algún elemento de X, que de hecho estará en la frontera de $B(x_{0},\epsilon _{0})$. Para hacer esto, supongamos, sin pérdida de generalidad, que $x_{0}\in U$. Sea $x_{1}\in V$, como d es convexa existe $x_2\in X$ tal que $d(x_{0},x_{2})=\frac{1}{2}d(x_{0},x_{1}) =d(x_{1},x_{2})$, entonces (haciendo cuentitas $x_{2}\in B(x_{0},\epsilon _{0})$) entonces $x_{2}\in U\cup V$.
Si $x_{2}\in U$ entonces nos fijamos en $x_{1}\in V,x_{2}\in U$ entonces, existe $x_{3}\in X$ talque $d(x_{1},x_{3})=\frac{1}{2}d(x_{1},x_{2}) =d(x_{2},x_{3})$. (de nuevo haciendo cuentas podemos llegar a que $x_{3}\in B(x_{0},\epsilon _{0})$ pero además $d(x_{0}, x_{3})< \frac{1}{2} \epsilon_{0}$.
Si $x_{2}\in V$ entonces nos fijamos en $x_{0}\in U,x_{2}\in V$ entonces, existe $x_{3}\in X$ talque $d(x_{0},x_{3})=\frac{1}{2}d(x_{0},x_{2}) =d(x_{2},x_{3})$. (de nuevo haciendo cuentas podemos llegar a que $x_{3}\in B(x_{0},\epsilon _{0})$ pero además $d(x_{0}, x_{3})< \frac{1}{2} \epsilon_{0}$.
Espero, que se entienda cómo va a ser la construcción de la sucesión $(x_{n})$ Según yo, se puede probar que así construida vamos a tener las siguientes dos condiciones:
i)$d(x_{n},x_{n+1})< \frac{1}{2^{n}}\epsilon_{0} \forall n$ (por lo tanto va a ser de Cauchy)
ii) $d(x_{0},x_{n})< \frac{1}{2}\epsilon_{0} \forall n$
Entonces, por un lado tenemos que $(x_{n}) \longrightarrow x$ para algún $x\in X$. Entonces $x\in Fr(B(x_{0},\epsilon _{0}))$. Pero por otro lado $d(x_{0}, x)\le d(x_{0}, x_{n})+d(x_{n},x)< \frac{1}{2}\epsilon_{0} + \frac{1}{2}\epsilon_{0}= \epsilon_{0}$ $\forall n\ge k$ para alguna $k\in \mathbb{N}$. Entonces $x\in B(x_{0},\epsilon _{0})$ lo cual es una contradicción. Por lo tanto las bolas abiertas son conexas y por lo tanto X es localmente conexo.
La verdad no se si esta demostración esté bien o si hay otra forma más sencilla. Espero me puedan ayudar. Gracias.
Solution 1:
Your proof is basically correct, the following is a cleaner way to argue.
Definition. A metric space $X$ is called geodesic if for every pair of points $p, q\in X$ there exists an isometric (i.e. distance-preserving) map $f=f_{p,q}: [0,D]\to X$ so that $f(0)=p, f(D)=q$, where $D=d(p,q)$. Since such $f$ is 1-Lipschitz, it is also continuous.
Lemma 1. if $X$ is a geodesic metric space then metric balls in $X$ are path-connected.
Proof. If $q\in B(p,R)$ (ball of radius $R$ centered at $p$), then the image of the geodesic $f_{p,q}$ connecting $p$ to $q$ is clearly contained in $B(p,R)$. qed
Lemma 2. If $X$ is a complete convex metric space, then $X$ is a geodesic metric space.
Proof. Take a pair of points $p, q\in X$ within distance $D$. Consider the interval $I=[0,D]$ and the subset $E\subset I$ consisting of numbers of the form $rD$, where $r$ is a dyadic rational number (a rational number whose denominator is a power of $2$). In other words, $$ E=\bigcup_{n\ge 0} E_n, $$ where $E_n$ consists of numbers of the form $\frac{m}{2^n}D$, where $m, 2^n$ are coprime.
Define a map $f$ from $E$ to $X$ inductively by:
$f(0)=p, f(D)=q$.
If $f$ is defined on $E_i, i\le n$, we extend $f$ to $E_{n+1}$ by the rule: If $t, s$ are consecutive elements of $$ \bigcup_{k\le n} E_k $$ then $$ f(\frac{t+s}{2}) $$ is a midpoint of $f(t)$ and $f(s)$. This midpoint need not be unique, but that's OK.
Then it is not hard to verify (using just the triangle inequality) by induction on $n$ that $f: E\to X$ is an isometric map. Hence, $f$ is 1-Lipschitz; thus it is uniformly continuous and, therefore, extends to a continuous map $f: I\to X$ (here we use completeness of $X$). By continuity, the resulting map $f: I\to X$ is again isometric. qed
Combination of Lemmas 1 and 2 imply that metric balls in a convex complete metric space are path-connected. Local connectivity immediately follows.