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Suppose (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 $\epsilon$

I've got the forward direction, but I would like someone to check my backward implication

My attempt

Suppose for each $\epsilon >0$, there is a finite cover of X by compact connected sets of diameter less than $\epsilon$. Let $x \in X$ and V be a neighborhood of x. There exists an $\epsilon >0$ such that $B_d(x;\epsilon)\subseteq V$. By assumption there exists a finite cover of X by compact connected sets of diameter less than $\frac{\epsilon}{2}$, say $\{G_1,...,G_n\}$. Since $\cup_{i=1}^nG_i=X$, there exists $k\in\{1,...,n\}$ such that $x\in G_k$. Note that for any $y\in G_k$ we have that $d(x,y)\leq \frac{\epsilon}{2}<\epsilon$ so $y\in B_d(x;\epsilon)$. Hence $G_k$ is a connected neighborhood of x that is contained in V so X is locally connected

Is this alright? It seems like I didn't use the fact that each set in the cover is compact.

In the link below the proof was done using contraposition and weakly locally connectedness which is not covered in class so I'd rather do it the constructive way if it's possible.

Locally connected set in compact metric space

Thank you.

Edited Proof

Suppose for each $\epsilon >0$, there is a finite cover of X by compact connected sets of diameter less than $\epsilon$. Let $x \in X$ and V be a neighborhood of x. There exists an $\epsilon >0$ such that $B_d(x;\epsilon)\subseteq V$. By assumption there exists a finite cover of X by compact connected sets of diameter less than $\frac{\epsilon}{2}$, say $\mathscr{G}=\{G_1,...,G_n\}$.

Let $\mathscr{G_0}=\{G \in \mathscr{G}:x \in G\}$. Let $K=\cup( \mathscr{G} \backslash \mathscr{G_0})$ and $N=\cup \mathscr{G_0}$.

Since $\mathscr{G_0}$ is a family of connected sets with common point x, their union N is also connected. Now if every $G \in \mathscr{G}$ contains x, then K is empty and therefore closed. If on the other hand there exists some index say $n_1,...,n_m$ for which $G_{n_i}$ does not contain x, then K is the finite union of these sets each of which is a compact subset of a metric space thereby closed in X so K is closed. It follows by construction that $x \in K^c$ and that $K^c$ is open. Now let $y \in K^c$. Since $\mathscr{G}$ is a cover of X, $y \in G_i$ for some $i \in \{1,...,n\}$. Since $y \notin K$ it has to be in $G$ for some $G \in \mathscr{G_0}$. Therefore $y \in N$ and $K^c \subseteq N$. Hence N is a connected neighborhood of x so X is locally connected.

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  • $\begingroup$ Notice that the statement you used as title is false. $\endgroup$ – Mariano Suárez-Álvarez Nov 11 '16 at 3:47
  • $\begingroup$ Argh, @MarianoSuárez-Álvarez, you beat me to it! $\endgroup$ – Neal Nov 11 '16 at 3:49
  • $\begingroup$ Sorry about that. When I tried to type in epsilon it showed me an error so I didn't put that in. $\endgroup$ – Sai Nov 11 '16 at 3:51
  • $\begingroup$ There is no need for the title to contain all the details: it simply has to be descriptive enough —something like "a characterization of local connectedness for compact metric spaces", say— and correct. $\endgroup$ – Mariano Suárez-Álvarez Nov 11 '16 at 4:23
  • $\begingroup$ @Mariano :( I'll be more careful next time. $\endgroup$ – Sai Nov 11 '16 at 4:28
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That argument doesn’t quite work, because there’s no reason to think that $G_k$ is a nbhd of $x$. Indeed, the interior of $G_k$ might even be empty, in which case $G_k$ isn’t a nbhd of any point. However, the sets $G_1,\ldots,G_n$ can still be used to get what you want.

HINT: Let $\mathscr{G}=\{G_1,\ldots,G_n\}$, and let $\mathscr{G}_0=\{G\in\mathscr{G}:x\in G\}$. Let $N=\bigcup\mathscr{G}_0$, and let $K=\bigcup(\mathscr{G}\setminus\mathscr{G}_0)$.

  • Show that $N$ is connected.
  • Show that $K$ is closed.
  • Show that $x\in X\setminus K\subseteq N$, and conclude that $N$ is a connected nbhd of $x$.
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  • $\begingroup$ Thank you. I've edited my proof according to your hint. Is this alright? $\endgroup$ – Sai Nov 11 '16 at 3:46
  • $\begingroup$ I forgot that connectedness has nothing to do with being open $\endgroup$ – Sai Nov 11 '16 at 3:48
  • $\begingroup$ @Sai: Yes, that looks much better now. $\endgroup$ – Brian M. Scott Nov 11 '16 at 18:47

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