If a point has a path-connected neighborhood it also has an open path-connected neighborhood?

Question

If the answer is no: what does it happens if the space in wich lies the point is connected?

PS: sorry for my bad English.

Answer 1

Here's a counterexample. Let $X\subset\mathbb{R}^2$ be the topologist's sine curve $\{0\}\times[-1,1]\cup\{(x,\sin(1/x)):0<x\leq 1\}$ and let $Y$ be an arc from $(0,-1)$ to $(1,\sin 1)$ that does not intersect $X$ except at the endpoints, so $X\cup Y$ is a Warsaw circle. Let $Z$ be an arc that oscillates and approaches a segment of $Y$, so the union of $Z$ and that segment is another topologist's sine curve, and consider the space $A=X\cup Y\cup Z$.

Note that $A$ is connected, since it is the union of the two connected sets $X\cup Y$ and $\overline{Z}$ which have nonempty intersection. The point $(0,0)\in A$ has a path-connected neighborhood in $A$, namely the set $X\cup Y$. But $(0,0)$ has no path-connected open neighborhood in $A$: it is easy to see a path-connected neighborhood of $A$ must contain all of $X\cup Y$, and therefore must intersect $Z$ if it is open, but no path in $A$ can have points in both $Z$ and $X\cup Y$.