Hello I am working on an exercice which aims to show that any complete connected locally connected metric space $E$ is path-connected.
One of the question is showing that some set is open and closed.
I would like to know if my answer is correct, and if so if there's any better way to do it (my proof feels a bit overcomplicated to me).
For $U\subseteq E$ open and connected, $a\in U$ and $\epsilon > 0$ they define $U^*$ as the set of $x\in U$ such that there exists connected open sets $V_0,..,V_n$ satisfying:
- $a\in V_0$ and $x\in V_n$
- $\overline V_i \subset U$ and $\text{diam}(V_i) < \epsilon$ for $i=0,1,2,...,n$
- $V_{i-1} \cap V_i \neq \emptyset$ for $i=1,2,...,n$
Show that $U^*$ is open and closed
(openness)
For $x \in U^*$ we have the associated sequence $V_0,..,V_n$ with $x\in V_n$ and $V_n$ open so $\exists r>0$ such that $B(x,r) \subset V_n$ and naturally any $y\in B(x,r)$ is an element of $U^*$. Hence $U^*$ is open.
(closedness)
We show that the closure of $U^*$ is a subset of $U^*$.
Let $x\in \overline U^*$, $E$ a metric space so there's a sequence $(x_n)$ of elements of $U^*$ that converges towards $x$.
$E$ is locally connected so there is $V$ a connected neighborhood of $x$ and by convergence $\exists n\in\mathbb{N}$ s.t. $x_n \in V$. We have $x_n \in U^*$ so there exists sets $V_0,..,V_n$ satisfying the three conditions.
Note that $V\cap (E\setminus\overline V_n)$ and $V\cap V_n$ form a disjoint open cover of $V$. Since $V$ is connected and $V \cap V_n$ isn't empty because $x_n \in V \cap V_n$ we have that $V\cap (E\setminus\overline V_n) = \emptyset$ hence necessarily $x\in\overline V_n$
Therefore, \begin{gather} x\in\overline V_n \subset U \end{gather}
Now we wish to construct a connected open set $V'$ such that $V_0,..,V_{n-1},V'$ has the three desired properties so we can conclude that $x\in U^*$.
$U$ is open so there exists $0<r<\epsilon - \text{diam}(V_n)$ such that $B(x,r)\subset U$. Now since $E$ is a metric space it is regular so there is a neighborhood $\Gamma$ of $x$ such that $x\in \overline \Gamma \subset B(x,r)$.
Finally since $E$ is locally connected there is an open connected neighborhood $C\subset\Gamma$ of $x$ such that
\begin{gather} x\in C\subset\overline\Gamma\subset B(x,r) \end{gather}
Now consider \begin{gather} V' = C \cup V_n \end{gather}
It's open since it's the union of two open sets.
Note that $V_n$ and $C$ are both connected and $C\cap V_n \neq \emptyset$ since $C$ is a neighborhood of $x\in\overline V_n$. Hence $V'$ is connected.
Concerning the three conditions,
$x\in V'$ since $x\in C$
$\overline V' = \overline C \cup \overline V_n$ and $\overline C \subset \overline \Gamma \subset U$ since $\overline V_n \subset U$ we have $\overline V' \subset U$.
The condition on the radius of the ball implies that $\text{diam}(C) < \epsilon - \text{diam}(V_n)$ so we easily get $\text{diam}(V') < \epsilon$ by triangular inequality.
And obviously $V' \cap V_{n-1} \neq \emptyset$.
Therefore $V_0,...,V_{n-1},V'$ respect all the desired properties and thus $x\in U^*$.
Consequently $U^*$ is closed.