Suppose that ($X,d$) is a compact metric space. Prove that $X$ is locally connected if and only if for each $\epsilon>0$, there is a finite cover of X by compact connected sets of diameter less than $\epsilon$.

For ($\rightarrow$), I have already find it in this site.

For ($\leftarrow$), I somehow show that $\forall x \in X$, $\forall G \in \tau$ s.t. $x \in G,$ $\exists V \in \mathfrak{N}(x)$ such that $x \in V \subseteq G$ and $V$ is connected set. By the way, I really don't know how to choose such $V$ to be open instead of just a neighborhood.

Every hint is appreciated.

Thank you


HINT: It’s probably easiest to prove the contrapositive: if $X$ is not locally connected, then there is an $\epsilon>0$ such that $X$ has no finite cover by compact, connected sets of diameter less than $\epsilon$.

If $X$ is not locally connected, then $X$ is not weakly locally connected: there are a point $p\in X$ and an open nbhd $U$ of $p$ such that if $V$ is a connected nbhd of $p$, not necessarily open, then $V\nsubseteq U$. (You can find a short proof that every weakly locally connected space is locally connected here. Choose $\epsilon>0$ such that $B(p,\epsilon)\subseteq U$; then no connected nbhd of $p$ is contained in $B(p,\epsilon)$.

Now suppose that $X$ has a finite cover $\mathscr{C}$ by compact, connected sets of diameter less than $\epsilon$. Let $\mathscr{C}_0=\{C\in\mathscr{C}:p\notin C\}$, and let $K=\bigcup\mathscr{C}_0$; $\mathscr{C}_0$ is finite, so $K$ is closed. Let $W=X\setminus K$; clearly $W$ is an open nbhd of $p$.

  • Show that $W\subseteq\bigcup(\mathscr{C}\setminus\mathscr{C}_0)\subseteq B(p,\epsilon)$.
  • Show that $\bigcup(\mathscr{C}\setminus\mathscr{C}_0)$ is connected.
  • Conclude that $\bigcup(\mathscr{C}\setminus\mathscr{C}_0)$ is a connected nbhd of $p$ contained in $B(p,\epsilon)$, contradicting the choice of $p$ and $\epsilon$.
  • $\begingroup$ I attempted the constructive route and got as far as $\mathscr{C}_0=\{C\in\mathscr{C}:p\notin C\}$ but somehow couldn't seem to get an open connected set out in the end. Maybe contrapositive is the way to go. $\endgroup$ – BigbearZzz Oct 29 '16 at 17:12
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    $\begingroup$ @BigbearZzz: It’s often easier to start with a single bad thing and show that it can’t exist than it is to show that everything is good. $\endgroup$ – Brian M. Scott Oct 29 '16 at 17:26

For the sake of completeness and those who will need it I add the following references are from Ryszard Engelking's «General Topology» (2nd ed., Heldermann, Berlin, 1989).

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Thus (c) directly implies the implication ($\rightarrow$). Implication ($\leftarrow$) hold because every open cover of a compact metric space $X$ has a Lebesgue number.

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