Show that $X = \{ (x,y) \in\mathbb{R}^2\mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\}$ is path connected.

general-topology connectedness rational-numbers

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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 path-connected (given 2 points in $A$, you can connect them with a path consisting of 3 smaller paths, something like horizontal-vertical-horizontal, 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 path-connected.

2,298

Author by

sarah

Updated on November 25, 2020

Comments

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):
1. 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) )
1. 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 years
  
  My 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 years
  
  I'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 years
  
  I 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 years

what is the theorem you are referring to?