Show that $X = \{ (x,y) \in\mathbb{R}^2\mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\}$ is path connected.
Solution 1
HINT: You can get from any point of $X$ to any other point of $X$ by a path consisting of at most three line segments, all of which are either horizontal or vertical.
Solution 2
Let $A=\left(\mathbb{R}\times\mathbb{Q}\right)\cup\left(\left\{0\right\}\times\mathbb{R}\right)$ and $B=\left(\mathbb{Q}\times\mathbb{R}\right)\cup\left(\mathbb{R}\times\left\{0\right\}\right)$. Show that $A$ and $B$ are pathconnected (given 2 points in $A$, you can connect them with a path consisting of 3 smaller paths, something like horizontalverticalhorizontal, and similarly for $B$). Now, $X=A\cup B$ and $A\cap B$ is not empty (for example $(0,0)\in A\cap B$)). From an easy theorem you probably know, these imply that $X$ is pathconnected.
sarah
Updated on November 25, 2020Comments

sarah about 2 years
How do I show that $X = \left\{ (x,y) \in \mathbb{R}^2 \mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\right\}$ is path connected?
Note that $X$ is a topological space with subspace topology $\tau = \left\{ U \cap X \mid U\text{ is open in }\mathbb{R}^2\right\}$.
I only know two things (with respect to path connectedness):
 A topological space $X$ is path connected iff for every $p,q \in X$, there is a continuous function $f: [a,b] \to X$ with $f(a)=p$ and $f(b) = q$.
Continuity $f$ is defined as for every open set in $X$, its preimage is open in $[a,b]$ (with subspace topology induced by $\mathbb{R}$ (with standard topology) )
 Suppose $X$ and $Y$ are two topological spaces and $f: X \to Y$ is a continuous surjection. If $X$ is path connected, then $Y$ is path connected.

Stefan Hamcke about 9 yearsMy answer for math.stackexchange.com/questions/516916/… can be adjusted to apply it to your problem if you choose $\alpha$ and $\beta$ rational. Or you can look at Brian's answer there and replace irrational by rational and vice versa whenever necessary.

Michael Hardy about 9 yearsI've done some editing on the title. For now I'm not going to clean up the rest of it. But for stackexchange, I'd never have been able to imagine that people's notions of good ways to use TeX or MathJax were so bizarre.

Michael Hardy about 9 yearsI vaguely seem to recall answering nearly this same question here about a year ago.

Brian M. Scott about 9 years@Michael: I had the same memory, but a quick Google search failed to turn up an exact match.

sarah about 9 yearswhat is the theorem you are referring to?