# Clarifying the Defintion of Locally Connected

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.

• 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. – MPW Sep 28 '18 at 18:02
• ah yes, I didn't appreciate the importance of "neighborhood" in this definition. thanks! – yoshi Sep 28 '18 at 18:10
• You're welcome. I will convert my comment to an answer. – MPW Sep 28 '18 at 18:11

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.
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 $$U$$.