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I had this question on my final:

Let $(X,d)$ be a metric space. Prove that the set of points where $X$ is locally connected is the countable intersection of open subsets of $X$.

I wasn't quite sure how to approach the problem; I know that the components of open subsets of $X$ are open if $X$ itself is locally connected, but I wasn't sure where to go with the information (if anywhere at all). Any hints? If possible, I'd prefer hints to begin with so I can work out a solution. After I have solved it, anyone can return to edit their solution to be more complete so as not to detract from the value of MSE as a reference site.

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I didn't know the definition of "locally connected at $x$", but Wikipedia says it means: for every open set $V$ containing $x$ there exists a connected, open set $U$ with $x \in U \subset V$. Since we're in a metric space, we can surely replace the sets $V$ with open balls of small radius. So I'd try thinking about a countable intersection over $n\ge1$ of some sets obtained by taking the $V$'s to be open balls of radius $1/n$. –  user108903 Dec 21 '12 at 18:11
    
I can see how this statement holds for every component, but how could this hold across components? It should be a union of countable intersections. –  gnometorule Dec 21 '12 at 18:40
    
@user108903: It seems like this definition implies the set of locally connected points would have to be open (which I'm not arguing, it's actually what I claimed on the final for the reason you just said). If it's open, it would have to be a $G_\delta$-set as we can just intersect itself countably many times.... Am I making a completely unjustified argument? –  Clayton Dec 21 '12 at 19:27
    
@gnometorule: Do you perhaps have a counterexample? Just asking... –  Clayton Dec 21 '12 at 19:41
    
I'm pretty confident I have an answer, but it still relies on show the set of points where $X$ is locally connected is open. Can anyone say it is the case that the set is not open? I'll post a solution below for comments regarding it... –  Clayton Dec 21 '12 at 20:14
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Note first that contrary to the claim in comments, this set (call it $LC(X)$) need not be open.

Let $X \subset \mathbb{R}^2$ be the union of all the lines through the origin that have rational slope, with the Euclidean metric. $X$ is locally connected at the origin 0; indeed any open ball in $X$ centered at 0 is path connected. However, for any other point $y \in X$, any neighborhood of $y$ which is so small that it does not contain 0 is not connected (it contains a point on some other line), so $X$ is not locally connected at any other point. Thus $LC(X) = \{0\}$, which is certainly not open in $X$.

To show in general that $LC(X)$ is $G_\delta$, here's a hint: for any $R > 0$, let $LC_R$ be the set of all $x \in X$ such that there exists an $r_x < R$ and a connected open set $U_x$ with $x \in U_x \subset B(x,r_x)$. Show that $LC_R$ is open and that $LC(X)$ is the intersection of countably many of the $LC_R$. (There might be a simpler way but this is the first thing that occurred to me.)

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Thank you! It's so clear and I missed this! I appreciate it. When you figure it out, please post and I'll accept your solution. Right now I have to leave, so I will be gone a few hours. –  Clayton Dec 21 '12 at 22:29
    
@Clayton: I think I got it, and added a hint. –  Nate Eldredge Dec 21 '12 at 22:38
    
I’ve not yet tried hard to make this work, but showing that $LC_R$ is open may not be too easy: I’ve an example showing that $U_x$ need not be a subset of $LC_R$. –  Brian M. Scott Dec 22 '12 at 5:47
    
@BrianM.Scott: Indeed. But we do have $U_x \cap B(x, R-r_x) \subset LC_R$. If $y \in U_x \cap B(x, R-r_x)$, take $r_y = d(x,y) + r_x$; then $r_y < R$ and $y \in U_x \subset B(y, r_y)$, so we can take $U_y = U_x$. –  Nate Eldredge Dec 23 '12 at 4:55
    
Looks good to me. –  Brian M. Scott Dec 23 '12 at 14:04
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Let $LC(X)$ denote the set of points where $X$ is locally connected. If $LC(X)=\varnothing$, it clearly is a $G_\delta$-set as $\varnothing=\bigcap\varnothing$ and $\varnothing$ is open.

Suppose $LC(X)$ is nonempty. Then there is a point $x\in LC(X)$ such that for every open set containing $U$ such that $x\in U$, there is a connected open set $V\subseteq U$ such that $x\in V$. Note then, that every point in $V$ will be in $LC(X)$ since if $y\in V$, then any open set $U'$ with $y\in U'$ intersects $V$ and $V\cap U'\neq\varnothing$, thus it is an open set containing $y$, in which case we can take an open ball around $y$ small enough so that it is contained in the intersection and is connected. This says that we found an open set $V\subseteq LC(X)$ around an arbitrary point $x$, hence $LC(X)$ is open. Hence, $$LC(X)=\bigcap \{LC(X)\}.$$ Does the argument sound complete? It seems to me that it is, but I'm not going to claim without any kind of uncertainty that it is flawless.

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I've yet to read through this, but I think you want to write $LC(X)=\bigcap \{LC(X)\}$. ;) –  tomasz Dec 21 '12 at 20:26
    
Yes, thank you. I'll edit that. –  Clayton Dec 21 '12 at 20:27
    
That looks right to me. The original question was likely either something else or an attempt to trick the examinees into overly elaborate thinking. :) It's also true in any topological space. –  tomasz Dec 21 '12 at 20:31
    
Thanks, I'll wait to see if I garner any more responses. It seems right to me, but I want to be sure one way or the other. –  Clayton Dec 21 '12 at 20:35
    
I don't think this is right. You can construct an open ball around $y$ which is contained in $V \cap U'$ but there is no reason why this ball has to be connected. Indeed, I don't think $LC(X)$ need be open at all, and I'll post a counterexample as an answer. –  Nate Eldredge Dec 21 '12 at 22:16
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