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Munkres writes the following defintion for local connectedness at x:

A space $X$ is said to be locally connected at $x$ if for every neighborhood $U$ of $x$, there is a connected neighborhood of $V$ of $x$ contained in $U$.

I'm confused about why the topologist sine curve is not locally connected. I've read that to see this: simply take an open neighborhood on ${0} \times [0,1]$. The ball will always contain parts of the sine curve. Isn't this consistent with the definition? Indeed there is a connected component contained $U$. (In fact, many)

I think I'm misunderstanding the definition of locally connected, but wiki is not making the situation clearer either.

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    $\begingroup$ The requirement is that there be a connected neighborhood of $x$ in $U$. Of course the component of $U$ containing $x$ is connected, but it isn't a neighborhood of $x$. As you yourself point out, every neighborhood of such an $x$ contains (infinitely) many components, so can't be itself connected. $\endgroup$ – MPW Sep 28 '18 at 18:02
  • $\begingroup$ ah yes, I didn't appreciate the importance of "neighborhood" in this definition. thanks! $\endgroup$ – yoshi Sep 28 '18 at 18:10
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    $\begingroup$ You're welcome. I will convert my comment to an answer. $\endgroup$ – MPW Sep 28 '18 at 18:11
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[Converted from comment]

The requirement is that there be a connected neighborhood of $x$ in $U$. Of course the component of $U$ containing $x$ is connected, but it isn't a neighborhood of $x$. As you yourself point out, every neighborhood of such an $x$ contains (infinitely) many components, so can't be itself connected.

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Let $U$ be a small open ball about, say, $(0,0)$. It will contain some points $(x,\sin(1/x))$ for $x > 0$, but these will not be connected to $(0,0)$ within $U$: any connected subset of $X$ containing $(0,0)$ and $(x, \sin(1/x))$ will have to contain $(t,1)$ for some $t$ and thus is not contained in $U$.

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