I am trying to determine if the set $A = \{(x,y):x,y \in \mathbb{R} \setminus \mathbb{Q} \} $ is path connected in $\mathbb{R}^{2}$. For a countable set we could use the countability-uncountability argument used for example here.
Asked
Active
Viewed 54 times
1
-
I suspect, it is not. I suggest that you try showing that any path in $\mathbb{R}^2$ must have at least one point with a component in $\mathbb{Q}$. Perhaps by showing that a short enough segment of a path will look like a line and then using compactness of $\mathbb{Q}$ in $\mathbb{R}$. – Mosquite Mar 21 '17 at 18:50
1 Answers
2
This cannot be path-connected, as any path between $(1/e,1/\pi)$ and $(e,\pi)$ has to intersect the unit square. But no point on the unit square falls into the set that you've demarcated. In fact, the set you've mentioned has no path-connected subsets of cardinality greater than $1$.
Arkya
- 628
Stella Biderman
- 31,475