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This is from an old paper of FB Jones.

Unless I am mistaken, the space is a countable union of closed nowhere dense sets (it is dense in $\mathbb R ^3$ and is a countable union of arcs). Is this correct?

My main question is, why is the example locally connected? I'm having a hard time picturing a connected open neighborhood.

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Unless I am mistaken, the space is a countable union of closed nowhere dense sets (it is dense in $\mathbb R ^3$ and is a countable union of arcs). Is this correct?

Yes, that is correct.

My main question is, why is the example locally connected? I'm having a hard time picturing a connected open neighborhood.

Let the space be called $X$. I claim that if $U\subset\mathbb{R}^3$ is open and convex, then $U\cap X$ is connected. Indeed, let $x,y\in U\cap X$; I will show that any clopen subset $V$ of $U\cap X$ which contains $x$ must also contain $y$. Since $V$ is closed, it suffices to show $V$ intersects any $U_n\subseteq U$ which contains $y$. So suppose $y\in U_n\subseteq U$. Since $V$ is open in $X$, there is some $m$ such that $V$ contains $U_m\cap X$. By construction, $X$ then contains the line segment from some point of $U_m$ to some point of $U_n$. Since $U$ is convex, that entire line segment is also contained in $U$. Since $V$ is clopen in $U\cap X$ and the line segment is connected, $V$ must also contain the entire line segment. Thus $V$ contains a point of $U_n$, as desired.

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  • $\begingroup$ ah, I see. I just couldn't tell how the "straight" part of the construction was useful $\endgroup$ – Forever Mozart Jun 23 '17 at 0:08

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