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I was trying to come up with a counterexample for something and kept failing and failing, so I was hoping someone could help me out. The setting is compact, connected, metric spaces (continua) so everything is nice. Here are the definitions I'm using:

$X$ is locally connected at $x \in X$ if it has a local basis of connected open sets. $X$ is semi-locally connected at $x \in X$ if it has a local neighborhood basis $V_n$ such that $X \setminus V_n$ has only finitely many components for every $n$.

It seems preposterous to assume that if $X$ is both locally connected and semi-locally connected at some point $x$, then we can find a local basis of connected open neighborhoods whose complements each have only finitely many components (i.e. both the 'lc' and 'slc' properties can be satisfied simultaneously with the same local basis).

But I really struggled to come up with a counterexample. The space below was my initial try; the purple fans force the open neighborhoods around the left-most point to contain entire fans (except their right-bottom end points), and then I tried to force those neighborhoods to cut off infinitely many pieces with the green fans. But of course, we can just 'add on' the green fan to the last purple fan we take and get an open, connected neighborhood, the collection of which will satisfy the lc and slc properties simultaneously.

enter image description here

So then I started trying to prove it's true after I couldn't come up with any modifications. The idea is to show that you can always add on little bits (here, like the green fans at the end of the chain of purple fans) that will also converge to $x$. I want to use the following fact proved by Jones:

If $X$ is semi-locally connected at $x$ then for every $y \in X \setminus \lbrace x \rbrace$ there is a compact, connected neighborhood of $y$ not containing $x$.

So to try to prove that in fact you can get this 'double basis' at $x$, suppose that $X$ is a continuum that is lc and slc at $x$. Let $U_n$ be a local basis of connected open neighborhoods of $x$ with $U_{n+1} \subset U_n$. By contradiction, assume $X$ has no such 'double basis'. Then $X \setminus U_n$ has infinitely many components $C_n^k$ for some subsequence of $(U_n)$, which we can just assume is all of $U_n$ after discarding some.

Since $X \setminus U_n$ is closed, each $C_n^k$ is closed and thus the complement of all but finitely many is open. Denote an arbitrary such union of $U_n$ and all but finitely many components of its complement by $W_n$. It's connected by the Boundary Bumping Theorem (which, says that components of complements intersect boundaries in continua), and thus would constitute a 'double basis' assuming that $W_n \rightarrow \lbrace x \rbrace$ (in the Hausdorff sense) for some choice of $W_n$. So because of our contradiction assumption, assume otherwise.

Note that because $U_{n+1} \subset U_n$, by maximality of components every $C_n^j$ is contained in some $C_{n+1}^k$. If all the components of $X \setminus U_n$ are contained in only finitely many components of $X \setminus U_{n+1}$ then we discard those components from $W_{n+1}$ (each closed in $X$), possible since $X \setminus U_{n+1}$ has infinitely components. But then $W_{n+1} \subset U_n$ and thus $W_{n+1} \rightarrow x$, which is against assumption, since $W_{n+1}$ will still be a connected, open neighborhood of $x$.

Thus we can assume there is a component $E_n$ of $W_n \setminus U_n$ not containing any component of $W_j \setminus U_j$ for $j < n$ and such that $E_n$ doesn't converge to $x$, since $W_n$ doesn't and $U_n$ does. Since each intersects $\partial{U_n}$ and $U_n \rightarrow x$, we have that if $E_n' \rightarrow K$ is a convergent subsequence of $(E_n)$ it contains some other point $y \neq x$. Fix such a subsequence and such a $y$.

Let $G$ be a continuum neighborhood of $y$. Thus it intersects every $E_n'$ for $n$ large enough. Now we want to show that this forces $G$ to contain $x$, contradicting the above-cited result of Jones. But for $n$ large these $E_n'$ are disjoint outside of tiny neighborhoods of $x$, so it's impossible.

This is somewhere between a proof and a sketch, and because the result is so non-intuitive I was wondering if anyone knows a counterexample or sees a problem with the proof before I try to write it down precisely.

Thanks for any help!

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  • $\begingroup$ It turned out to be true. $\endgroup$ – John Samples Jun 23 '18 at 4:16
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I don't know connectedness as good as compactness, I am not sure up to details, and I could miss subtleties like Boundary Bumping Theorem which I did not study, but, as for me, the arguments looked good to this point

Since each intersects $\partial{U_n}$ and $U_n \rightarrow x$ we have that if $E_n' \rightarrow K$ is a convergent subsequence of $(E_n)$ it contains some other point $y \neq x$.

Since $K$ is a neighborhood of $y$ it intersects $E_n'$ for $n$ large enough. Now we want to show that this forces $K$ to contain $x$, contradicting the above-cited result of Jones. But for $n$ large these $E_n'$ are disjoint outside of tiny neighborhoods of $x$, so it's impossible.

So you defined $K$ as a limit of $E’_n$. Then is not clear why $K$ is a neighborhood of $y$. Moreover, in order to obtain a contradiction with the result of Jones, you have to show not that some compact (?) $K$ containing $y$ contains $x$, but that each compact connected neighborhood of $y$ contains $x$.

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  • $\begingroup$ Oh thanks for spotting that; indeed it makes no sense. I was fidgeting with that bit, I probably used $K$ twice and then got mixed up. I changed it (here the second usage of $K$ becomes $G$). $\endgroup$ – John Samples Oct 1 '17 at 2:01
  • $\begingroup$ @JohnSamples Now it looks good. But I'm too tired now and don't see the last step: why if for $n$ large these $E_n'$ are disjoint outside of tiny neighborhoods of $x$ then $G$ contains $x$? $\endgroup$ – Alex Ravsky Oct 1 '17 at 2:09
  • $\begingroup$ For $G$ to be connected it has to contain a connected set intersecting each of $G \cap E_n'$, and that will only happen if $x \in G$. $\endgroup$ – John Samples Oct 1 '17 at 2:57
  • $\begingroup$ I think it's not actually that obvious, so let's use an induction argument: Since $E_2'$ is disjoint with $E_1'$ outside of $U_2$, since $X$ is connected there is some point $u_2$ where one of their closures intersects the other set. It will be contained in $\overline{U_2}$. Proceeding inductively you get $u_n \in U_n$ that will converge to $x$. Then use the closedness of everything. $\endgroup$ – John Samples Oct 1 '17 at 3:14

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