# Intersection of $\mathbb{Q}^k$ and a connected set in $\mathbb{R}^k$

1. How to prove that $\mathbb{R}^k$ is connected?

2. Let $C$ be an infinite connected set in $\mathbb{R}^k$. How can I show that $C\bigcap \mathbb{Q}^k$ is nonempty?

-
#2 seems false to me: Take the set $\{(e,x)|x\in R\}$ in $R^2.$ –  only Oct 18 '12 at 3:42
In fact, #2 is even worse: for $k \ge 2$, the complement $\mathbb{R}^k \setminus \mathbb{Q}^k$ is an infinite connected set. (Path connected, even.) –  Nate Eldredge Oct 18 '12 at 4:01

(1) I’ll suggest one of many approaches to proving that $\Bbb R^2$ is connected; it generalizes easily to $\Bbb R^k$. Note that $\Bbb R^2$ is the union of the $x$-axis, which I’ll call $A$, and the straight lines $L_a$ whose equations are $x=a$ for real numbers $a$. Each of these sets is homeomorphic to $\Bbb R$, so each is connected. Moreover, $A\cap L_a\ne\varnothing$ for each $a\in\Bbb R$. Now prove this useful little
Theorem: If $A$ and $C_i$ for $i\in I$ are all connected sets, and $A\cap C_i\ne\varnothing$ for each $i\in I$, then $A\cup\bigcup_{i\in I}C_i$ is connected.
(2) is false: if $a$ is irrational, $L_a\cap\Bbb Q^2=\varnothing$.
1) For any $\,a,b\in\Bbb R^k\,$ , the straighline $\,\{(t-1)a+tb\;:\;t\in\Bbb R\}\,$ is completely contained in $\,R^k\,$, so it is path connected and thus connected.