Show that $A:=\{(x,y) \in \Bbb R^2 \mid x \in \Bbb Q \iff y \in \Bbb Q\} \subset \Bbb R^2$ is locally path-connected.
I've been trying to think about this problem for a while now and I don't even understand the set $\{(x,y) \in \Bbb R^2 \mid x \in \Bbb Q \iff y \in \Bbb Q\}$. The closes I can think of this is just $\Bbb Q^2$, but I don't understand this pathological condition $x \in \Bbb Q \iff y \in \Bbb Q$.
To show it's locally path-connected I think I have to show that for every $x \in A$ there is a neighborhood basis containing only locally path-connected sets?
This definition of locally path-connectedness seems a bit overwhelming. It's like three different conditions. Is there some characterization of this that could be useful here?
$A=\mathbb{Q}^2\cup\mathbb{I}^2$ is locally path-connected, i.e. any $(a,b)\in A$ admits a basis $(V_i)_{i \in I}$ of neighbourhoods in $A$ such that each $V_i$ is path-connected: take $V_r=U_r\cap A$ for any positive $r$, $U_r$ being the square $\{(x,y)\in {\mathbb R}^2: |x-a|+|y-b|<r\}.$ Any two points of $V_r$ are joined inside $V_r$ by a line segment of slope $1$ or $-1$, or by a union of two such segments.