Let $X=\{(x,y) \in \Bbb R^2 \mid x = 0 \text{ or } x^2+y^2 \in \Bbb Q\} \subset \Bbb R^2$. Is $X$ connected? Is it locally-connected?
$X$ is the $y$-axis union rational points on the unit disc. I think this is not connected as the rational points are "discrete", but I don't know how to formalize this. Can I use the fact that $\Bbb Q$ is not connected?
For the latter part I know that $\Bbb Q$ is also not locally-connected, but I don't think I can argue with just this since there is also the condition $x=0$. What is the way to figure out whether the set is locally connected or not?
Actually, $X$ consists of the $y$-axis united to all the circumferences with center $0$ and rational square of the radius (since it is the squared norm of $(x,y)$ which has to be rational).
This is a good example of a connected but not locally connected space. It is clearly path-connected because you can move from one of the circumferences to another one just by reaching the $y$-axis and going up or down on it. However, it is not locally connected because if you restrict your attention to a small open neighborhood $U=B \cap X$ of a point in $X$ $\backslash \left\{ x=0 \right\}$ (with $B$ being a ball in $\mathbb{R}^2$) then, chosen a suitable $t$ such that $t^2 \not\in \mathbb{Q}$ and $B$ intersects the circumference with radius $t$, the sets $$C_1=\left\{ (x,y) \in U : x^2 + y^2 \leq t^2\right\}$$ $$C_2=\left\{ (x,y) \in U : x^2 + y^2 \geq t^2\right\}$$ form a partition of two closed non-empty disjoint subsets.