while studying some general topology I came across the following topological space: on $\mathbb{R}$ we define an equivalence relation by $x \sim y$ if and only if $x=y$ or $|x|=|y|$ and $x \notin \mathbb{Q}$, and then we set $X=\mathbb{R}/\sim$ with the quotient topology induced by the euclidean one on $\mathbb{R}$.
I was able to prove that it is $T_1$ but not $T_2$, for instance, but then I asked myself whether $X$ is locally path connected or locally connected.
My guess is that the locally path connected claim is true (and hence it implies the other). Indeed, if $p$ denotes the quotient map, for the class of $x \notin \mathbb{Q}$ one can take the image of the union of two balls $$ U_\epsilon(x):=p(B_\epsilon(x)) \cup p(B_\epsilon(-x)) $$ as $\epsilon$ varies. The $U_\epsilon$ are path connected.
But for $x \in \mathbb{Q}$ the same argument does not apply, because for $x \neq 0$ the open subset $U_1(x)\setminus\{[-x]\}$ does not contain any $U_\epsilon(x)$. I think one has to take at least $U_\epsilon(x)$ minus the image via $p$ of any subset of $\mathbb{Q}$ with $-x$ as the only possible accumulation point, but I wonder whether this is correct: they seem to open in the quotient topology, but are enough to build a neighbourhood system?
