Show that $A$ is not path-connected.

Question

I have the following problem:

In $\mathbb{R}^2$ with the usual topology consider the set $A= \{(x,y)\in\mathbb{R}^2:x\in\mathbb{R}\setminus\mathbb{Q},y\geq 0\}\cup\{(x,y)\in\mathbb{R}^2:x\in\mathbb{Q},y<0\}$. Show that the set $A$ is not path connected.

I have an attempt but I believe its wrong.

My Attempt: By contradiction. Suppose that A is path connected, the for $(q_1,-1)$ and $(q_2,-1)$ with $q_1,q_2\in\mathbb{Q}$, exists a path $f$ continuous and satisfies that $f(0)=(q_1,-1)$ and $f(1)=(q_2,-1)$. Since $f$ is continuous, $h:=\Pi_1\circ f$ is continuous where $\Pi_1$ is the proyection on $x$.

We have that $h(0)=(q_1,0)$ and $h(1)=(q_2,0)$ but $h(0),h(1)\notin A$ and then, $h$ is not continuos, and so, $f$ is not continuos. Therefore, A is not path connected.

I hope you can help me to fix it if or give me an idea of how can I prove this, please.

Thanks!!!

Answer 1

Following your notations, just denote $h(t)=(h_1(t),h_2(t))$. Then $h_1(0)=q_1< q_2=h_1(1)$ and $h_2(0)=h_2(1)=-1$. The idea is to consider the first time $t_0$ when $h(t_0)$ is on the upper half plane.

Step 1. Let $$ E = \{t\in [0,1]: h_2(t)\geq 0\} = \{t\in [0,1]: h_1(t)\notin\mathbb{Q}\}. $$ Show that $E\neq\emptyset$ by the fact $q_1\neq q_2$ and the Intermediate Value Theorem.

Step 2. Take $t_0=\inf E$. Show that $t_0\in E$ and hence $t_0\in (0,1)$.

Step 3. Show that $h_1(t)\in\mathbb{Q}$ for all $t\in [0,t_0)$ by the definition of $t_0$.

Step 4. We have shown in Step 2 that $h_1(t_0)\notin\mathbb{Q}$. This together with Step 3 give a contradiction by taking an irrational number between $h_1(0)$ and $h_1(t_0)$ and using the Intermediate Value Theorem.

Answer 2

Let $h:[0,1]\to\mathbb{R}$ be the projection onto the first coordinate of the path $f$ that is connecting $(q_1,-1)$ and $(q_2,-1)$ with $q_1,q_2\in\mathbb{Q}$.

Consider the set $V=h^{-1}(\{q_1\})$. It is closed, since $h$ is continuous. Let $t_0\in V$. If in every neighborhood $U$ of $t_0$ we had points $t\in U$ where $h$ takes other values different from $q_1$, since $h$ is continuous, $h$ would take on $U$ all the values in the interval from $q_1$ to $h(t)$. Therefore, it would take irrational values. At those points $f$ is in the upper half plane. So, $f(t_0)$ is in the open lower half plane and in every neighborhood of $t_0$ there are points where $f$ is in the upper half plane. That cannot be, since $f$ is continuous. Therefore, $h$ is locally constant at $t_0$. Hence $V$ is open.

Since $V$ is non-empty, closed and open, and $[0,1]$ is connected, then $V=[0,1]$. Therefore, $h([0,1])=\{q_1\}$. This contradicts that $h(1)=q_1\neq q_1$.