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I understand the concept of one coordinate moving while the rest don't change, however I can't make up the exact mapping that would prove this. Can anyone give me the concrete mapping?

Acki
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  • Read this : https://math.stackexchange.com/questions/882234/show-that-s-mathbb-r2-setminus-a-ba-b-in-mathbb-q-is-path-connected?noredirect=1&lq=1 Could qualify as a duplicate as well? Please reply if this answers your question, and any doubts you have on the answers there. – Sarvesh Ravichandran Iyer Jan 19 '21 at 09:16
  • I can't find a mapping defined on that answer either :/ – Acki Jan 19 '21 at 09:19
  • Look at Nishant's answer : the coordinatewise thing is explained. If you want to be exact and precise about the mapping , you will have to break into cases where each particular coordinate is irrational, and so on. Providing the explicit path is then just a case-by-case analysis, which I would not consider necessary to expound on unless you are in an exam or submitting an assignment. – Sarvesh Ravichandran Iyer Jan 19 '21 at 09:20

2 Answers2

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Take $(a_1,a_2,a_3),(b_1,b_2,b_3)\in\Bbb R^3\setminus\Bbb Q^3$. Assume, for instance, that $a_1,b_3\notin\Bbb Q$. Then consider$$\begin{array}{rccc}\gamma\colon&[0,2]&\longrightarrow&\Bbb R^3\\&t&\mapsto&\begin{cases}\bigl(a_1,tb_2+(1-t)a_2,tb_3+(1-t)a_3\bigr)&\text{ if }t\in[0,1]\\\bigl((t-1)b_1+(2-t)a_1,b_2,b_3\bigr)&\text{ if }t\in[1,2].\end{cases}\end{array}$$Since $a_1,b_3\notin\Bbb Q$, $(\forall t\in[0,2]):\gamma(t)\in\Bbb R^3\setminus\Bbb Q^3$.

José Carlos Santos
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Let $(a,b,c)$, $(d,e,f)$ be two points in $\mathbb{R}^3\setminus\mathbb{Q}^3$. Then at least one of $a,b,c$ is not rational. By symmetry, we may assume WLOG that it is $a$.

If $d$ is not rational, we can follow the following path which is entirely contained in $\mathbb{R}^3\setminus\mathbb{Q}^3$:

$$(a,b,c) \to (a,\sqrt2,c)\to(d,\sqrt2,c)\to(d,e,f)$$ where $P\to Q$ signifies a straight line path from $P$ to $Q$.

If $e$ is not rational, we follow: $$(a,b,c) \to (a,e,c) \to (d,e,f)$$

If $f$ is not rational, we follow: $$(a,b,c) \to (a,b,f) \to (d,e,f)$$

Anonymath
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