Path connectedness of real subset

Question

Let $A=\{(x,y)\in\mathbb{R}^ 2|x\in\mathbb{Q}\text{ or }y=0\}$.

1. How do I show that $A$ is path connected?
2. How do I show that $A$ is NOT locally path connected?

What I thought:
1. So we need a continous map $f:[0,1]\rightarrow X$ from $(x_1,y_1)$ to $(x_2,y_2)$. I was thinking $$f(t)=t(x_2,y_2)+(1-t)(x_1,y_1).$$ But how can I show that this is in $X$ for all $t\in[0,1]$?
2. I need to find a $(x,y)\in X$ that does not have a path connected neighborhood. Which one could this be?

Answer 1

So $A$ consists of the $x$-axis and all vertical lines at rational distance from the $y$-axis. So you can get from $(0,0)$ to any point in $A$ by going across then up or down.

If you take a small neighbourhood of $(0,1)$ in $A$ it will contain just segments of vertical lines; how can you walk from $(0,1)$ to any other of these lines within this neighbourhood?

Answer 2

Some hints:

1. The map you give will not stay within the set $A$ (which you later call $X$ I think) as between any two rational numbers, there lies an irrational number. Instead, given $(x_1, y_1), (x_2, y_2) \in A$, think about why the line segments $[(x_1, y_1), (x_1, 0)]$, $[(x_1, 0), (x_2, 0)]$ and $[(x_2, 0), (x_2, y_2)]$ all lie in $A$; connecting them yields the desired path from $(x_1, y_1)$ to $(x_2, y_2)$.
2. This method does not work for local path connectedness if an open set e.g. does not intersect the $x$-axis. Think about why it is then no longer possible to make a path.