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Let us define $\mathbb{P}=\mathbb{R}\backslash\mathbb{Q}$. Find all path-connected components of $\mathbb{P}^n\cup\mathbb{Q}^n$

I tried solving this question but I'm out of luck. Any help, please?

It is not a duplicate of Connectedness of points with both rational or irrational coordinates in the plane? because they prove connectivity of the set and not path-connectivity.

In the paragraph above there is an explanation of why this is NOT a duplicate of an existing post, please look closely at the questions asked...

Shahar Romem Peled
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  • Welcome to MSE! What approach(es) have you tried? –  Jan 07 '18 at 10:15
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    @Brahadeesh I was able to show that all points in the form $r∗(q_1,...q_n)$ where $r\in\mathbb{R}$ and $q_1,...q_n$ are rational, are path connected – Shahar Romem Peled Jan 07 '18 at 11:49
  • @ShaharRomemPeled You can say slightly more. From the answer to the linked question, you can deduce that for every $\mathbf{q} = (q_1,\dots,q_n) \in \mathbb{Q}^n$ and every $t \in \mathbb{R}$ the points $(q_1 \pm t,\dots, q_n \pm t)$ are path-connected to $\mathbf{q}$ for every choice of sign in each coordinate. –  Jan 07 '18 at 12:15
  • Since any two points in $\mathbb{Q}^n$ can be joined by "diagonal" line segments, it follows that $\mathbb{Q}^n$ is contained in a path-component. –  Jan 07 '18 at 12:19
  • @Brahadeesh yes but what about others points? – Shahar Romem Peled Jan 07 '18 at 18:51
  • I’m not sure, still thinking over it :) I will comment/answer if and when I solve it. –  Jan 08 '18 at 03:53
  • I had updated my answer with a complete solution, but forgotten to ping you. Do check it, please :) –  Dec 19 '18 at 16:46
  • @Brahadeesh Thank you for the complete solution!! Really helped me. – Shahar Romem Peled Jan 11 '19 at 15:01

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The second half of this answer is based on @RobertIsrael's comment here.

Let $\mathbf{q} = (q_1,\dots,q_n) \in \mathbb{Q}^n$. For every $t \in \mathbb{R}$, we have that $$ (q_1 \pm t, \dots, q_n \pm t) \in \mathbb{P}^n \cup \mathbb{Q}^n $$ for every choice of sign in each coordinate. This is because if $t \in \mathbb{Q}$, then the above element lies in $\mathbb{Q}^n$ and if $t \in \mathbb{P}$ then the above element lies in $\mathbb{P^n}$. Any two elements in $\mathbb{Q}^n$ can be connected by means of such straight-line paths. Hence, $\mathbb{Q}^n$ is contained in a path-connected component.

Next, let $\mathbf{r} = (r_1,\dots,r_n) \in \mathbb{P}^n$. There is a sequence $\mathbf{q}_j = (q_{1j},\dots,q_{nj}) \in \mathbb{Q}^n$ such that $\lim_{j \to \infty} \mathbf{q}_j = \mathbf{r}$, that is, $\lim_{j \to \infty} q_{ij} = r_i$ for each $1 \leq i \leq n$. By the previous discussion, there is a path in $\mathbb{P^n} \cup \mathbb{Q}^n$ connecting $\mathbf{q}_j$ to $\mathbf{q}_{j+1}$ for each $j \in \mathbb{N}$. By concatenating these paths, we get a path in $\mathbb{P}^n \cup \mathbb{Q}^n$ connecting $\mathbf{q}_1 \in \mathbb{Q}^n$ with $\mathbf{r} \in \mathbb{P}^n$. This shows that $\mathbb{P}^n$ is contained in the same path-connected component as $\mathbb{Q}^n$.

Thus, $\mathbb{P}^n \cup \mathbb{Q}^n$ is path-connected.