Arcwise connected part of $\mathbb R^2$

Solution 1:

HINT: Not only is $\Bbb R^2\setminus D$ arcwise connected, but you can connect any two points with an arc consisting of at most two straight line segments.

Suppose that $p,q\in\Bbb R^2\setminus D$. There are uncountably many straight lines through $p$, and only countably many of those lines intersect $D$, so there are uncountably many straight lines through $p$ that don’t hit $D$. Similarly, there are uncountably many straight lines through $q$ that don’t hit $D$. Can you finish it from here?

Solution 2:

Any two points in $\mathbb R^2\setminus D$ are connected by uncountably many disjoint arcs of circles, of which only countably many may intersect $D$.

(This is adapted from one of my student's solutions to a related exam problem.)

Solution 3:

Adding an answer to address the questions raised in the bounty message.

Let $p, q$ be two distinct points in $\mathbb{R}^2 \setminus D$. If the straight line segment joining $p$ to $q$ lies in $\mathbb{R}^2 \setminus D$, then you are done: one possible parametrization of the path is $\gamma(t) = (1 - t)p + tq$, $0 \leq t \leq 1$.

So, suppose that the above does not work. Now, note that there are uncountably infinitely many straight lines passing through the point $p$. Since $D$ is only countable, in particular there are uncountably infinitely many straight lines passing through $p$ and not intersecting $D$. Fix any one such line $L_1$. Once more, there are uncountably infinitely many straight lines passing through $q$ that also intersect $L$. In particular, there are uncountably infinitely many such lines that also do not intersect $D$. Fix any one such line $L_2$.

Suppose that $L_1$ and $L_2$ intersect at the point $r$. Then, you can take the path joining $p$ and $q$ to be the one that starts at $p$, traverses along the line $L_1$ to the point $r$, then traverses along the line $L_2$ to the point $q$. One possible parametrization of this path is $$ \gamma(t) = \begin{cases} (1 - 2t)p + 2tr, & 0 \leq t \leq \tfrac{1}{2};\\ (2 - 2t)r + (2t - 1)q, & \tfrac{1}{2} \leq t \leq 1. \end{cases} $$ Note that the pasting lemma shows that $\gamma$ is indeed continuous.

As for how one arrives at these specific parametrizations, one first needs to know the parametric equation of a straight line. For instance, see Parametric form of a line for a quick refresher. Secondly, one needs to adjust the parameter $t$ by scaling and translating in order to get it to lie in the "correct" range for one's purposes. This is another linear change, so it should not be too difficult.

But, just to belabor the point further, here's one way you can mentally figure out the parametrization $\gamma$ in the second case. To join the points $p$ and $r$, we want to vary $t$ from $0$ to $1/2$ such that at $t = 0$ we are at $p$ and at $t = 1/2$ we are at $r$. So, if the parametrization for this part looks like $\gamma(t) = f(t)p + g(t)r$, then we want $f(0) = 1$ and $f(1/2) = 0$, and we also want $g(0) = 0$ and $g(1/2) = 1$. Can we find linear functions $f(t) = \alpha_1 t + \beta_1$ and $g(t) = \alpha_2 t + \beta_2$ that satisfy these conditions? Sure, just substitute those values of $(t, f(t))$ and $(t, g(t))$ that we wrote down for $t = 0, 1/2$ into the respective equations to solve for $\alpha_i, \beta_i$. Repeat the process for the line joining $r$ to $q$, and you are done.