Path-connected subsets of Q

Question

It is my understanding that the connected subsets of Q (subset of R containing rational numbers) are all sets that contain one element of Q.

However, I was wondering why this is the case? And further to that, how we can explain that the path-connected subsets of Q are also single elements.

Answer 1

It is, perhaps, easier to show that the only connected components of $\mathbb{Q}$ are the singletons. This can be done via a contradiction:

Assume that $C$ is a connected subset of $\mathbb{Q}$ with two distinct elements, say, $p,q\in C$. We may assume that $p<q$. Then let $x$ be an irrational number with $p<x<q$. Then the two sets $(-\infty,x)\cap C$ and $(x,\infty)\cap C$ form a separation of $C$, and this contradicts the fact that $C$ was connected. Therefore, if $C$ is connected, it can only contain a single element.

This line of reasoning extends to the path connected case: namely, any path connecting $p$ to $q$ must go through the point $x$.

Answer 2

Singletons are always path connected by the constant path $f(t)=c$ for some constant $c$. The irrationals are dense in the reals so every two distinct rational numbers have an irrational between them which impedes the path connecting them.