1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $\epsilon$.
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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).

enter image description here enter image description here

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

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.