Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  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?

share|cite|improve this question
#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
up vote 1 down vote accepted

(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, $L_a\cap\Bbb Q^2=\varnothing$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.