Is the complement of countably many disjoint closed disks path connected?

Question

Let $\{D_n\}_{n=1}^\infty$ be a family of pairwise disjoint closed disks in $\mathbb{R}^2$. Is the complement $$ \mathbb{R}^2 -\bigcup_{n=1}^\infty D_n $$ always path connected?

Here “disk” means a round, geometric disk. (As shown below, the answer is no if we allow arbitrary topological disks.)

Note that the union $\bigcup_{n=1}^\infty D_n$ can be dense in the plane. For example, it's possible to find a collection of disjoint closed disks of positive radius whose union contains $\mathbb{Q}\times\mathbb{Q}$.

It is easy to show that the complement of a countable set of points in the plane is always path connected. In particular, if $S \subset\mathbb{R}^2$ is countable, then there is always path between any two points in $\mathbb{R}^2-S$ consisting of two line segments.

It is not true that the complement of a countable set of disjoint line segments is path connected, as shown in the following figure.

By thickening the line segments slightly, one can find a countable collection of disjoint topological disks whose complement is not path connected.

Answer 1

Contrary to the request of the OP, this is still a sketch. I put it here because it points to some relevant literature, hoping that somebody more expert than me will help to clarify it. The question reminded me of this paper by Fischer and Zastrow: the relevant part is the end of p. 12. See also this answer by George Lowther.

You can assume that $\cup D_n$ is dense. Now choose some nice diffeomorphism $\phi:\mathbb{R}^2\to B_1(0)$. Once you apply it, the images of the disks are no longer disks, but $X:=\overline{B_1(0)}\setminus\bigcup_n \phi(\text{int }D_n)$ is still a planar 1-dimensional Peano continuum without local cutpoints (I'm not sure about the meaning of 1-dimensional..), so it is homeomorphic to the Sierpinski carpet by a theorem of Whyburn (Whyburn, Topological characterization of the Sierpinski curve, Fund. Math. 45 (1958) 320–324).

Our space is homeomorphic to $X\setminus \left(\partial B_1(0)\cup\bigcup_n\phi(\partial D_n)\right)$, but these sets we are removing are characterized as being the only non-separating simple closed curves. So under the homeomorphism with the Sierpinski carpet they map exactly to the boundaries of the holes (plus its exterior boundary) and we are left to show that:

The Sierpinski carpet is still path connected even after removing the boundaries of its holes and its exterior boundary.

But this is easy. Call $S'$ this new awful noncompact carpet. Put $C_1:=[\frac{1}{3},\frac{2}{3}]$, $C_2:=[\frac{1}{9},\frac{2}{9}]\cup [\frac{4}{9},\frac{5}{9}]\cup [\frac{7}{9},\frac{8}{9}]$, and so on. We have $S'=(0,1)^2\setminus\bigcup_{i=1}^\infty C_i\times C_i$.
Put also $G:=(0,1)\setminus\bigcup_{i=1}^\infty C_i$ and observe that $T:=(0,1)\times G\cup G\times (0,1)$ is path connected and $T\subset S'$.
Now fix any $p_0\in T$. Given any $x\in S'$, we wish to connect it to $p_0$ with a path in $S'$. Let $x\in Q_1$ where $Q_1$ is one of the $9$ closed cubes with side $\frac{1}{3}$ (those in which $[0,1]^2$ is divided when constructing the Sierpinski carpet). Choose some $p_1\in T\cap \text{int }Q_1$ and connect $p$ to $p_1$ with a path of length $<3$ (namely a path formed by at most three line segments).
Replace $[0,1]^2$ with $Q_1$ and iterate this step using self-similarity. Iterating infinitely many times we get a sequence $p_n\to x$ and paths in $S'$ connecting $p_n$ to $p_{n+1}$ with length $<\frac{3}{3^n}$, so that concatenating them we get a path from $p_0$ to $x$. $\blacksquare$

Answer 2

I think it is path-connected.

Let $S = \{ D_n \, | \, n \in \mathbb{N} \}$ be the set of closed disjoint disks under consideration. For any integer $m \ge 1$, let $N_m = \{ n \in \mathbb{N} \, | \, \mathrm{radius}(D_n) \in [1/m, 1/m-1) \}$ (with the convention $1/0 = \infty$) and $S_m = \{ D_n \, | \, n \in N_m \}$.

Let $P, Q \in \mathbb{R}^2 \backslash S$ and consider some path $\gamma_0$ between the two, say the straight line segment. For area considerations, there is a finite number of disks in $S_1$ intersecting $\gamma_0$ ; Let $J_1$ denote the reunion of these disks and $K_1$ denote the reunion of $J_1$ and $\gamma_0$. For area and compactness considerations, there is an $\epsilon_1 > 0$ such that the closed $\epsilon_1$-neighborhood $K'_1$ of $K_1$ intersects no other disks of $S_1$. For similar reasons, there is an $\delta_1 \in (0, \epsilon_1/4)$ such that the open $\delta_1$-neighborhood $J'_1$ of $J_1$ only intersects disks of $S$ that are contained in the open $\frac{\epsilon_1}{2}$-neighborhood of $J_1$ (and which have as such rather little radius). Consider the compact and path-connected set $R_1 := K'_1 \backslash J'_1$. One can perturb $\gamma_0$ to a continuous path $\gamma_1$ in $R_1$. Note that this implies that $\gamma_1$ is at a distance $\delta_1 > 0$ of any disk in $S_1$.

It is important to note why $J_1'$ was chosen this way : $\delta_1$ is small enough compared to $\epsilon_1$ to assure that $R_1 \backslash S_2$ is also path-connected. Moreover, points that have been pushed out at the first step don't have to be pushed out a lot in order to avoid disks of $S_2$, since it is sufficient (near those points) to perturb the path inside the intersection of $R_1$ and the $\frac{\epsilon_1}{2}$-neighbordhood of $J_1$. Points that have not been pushed out at the first step might be pushed a lot at the second step, but the same reasoning shows that they won't have to be perturbed a lot from then on.

This suggests that a by careful induction process, we can construct a nested sequence of compact path-connected sets $R_1 \supset R_2 \supset R_3 \supset \dots$ and a sequence of paths $\gamma_m \subset R_m$ joining $P$ and $Q$ such that $d(\gamma_j, S_m) > \delta_m > 0$ whenever $j \ge m$. If the perturbation process is done right (that is, if it is kind of minimal), the sequence $\gamma_m$ is equicontinuous. By the Arzelà-Ascoli theorem, we deduce that the set $\{ \gamma_m \, | \, m \in \mathbb{N}\} \subset C(I, \mathbb{R}^2)$ is relatively compact. Hence, there exists a continuous path $\gamma : I \to \mathbb{R}^2$ (still joining $P$ and $Q$) which is arbitrary well approximated by the set of $\gamma_m$. In fact, since the $R_m$s form a decreasing sequence of nested compact sets, we deduce that $\gamma \subset \cap_{m \in \mathbb{N}} R_m$. By construction, we see that $d(\gamma, S_m) > \delta_m > 0$ and consequently, $\gamma$ intersects none of the disks $D_n$.

Answer 3

So here is my attempt to clean up and clarify the other answer. We pick $A,B \in \Bbb R^2 \setminus S$

For some disks $D_n = B(c_n,r_n)$ we're going to find a "thickening" of $D_n$, $T_n$, such that its border $B_n$ is disjoint from every disk, and find a continuous projecti

on $f_n : T_n \setminus \{ c_n\} \to B_n$. We also want the $T_n$ to be disjoint form each other, to not contain $A$ and $B$, and we want every disk to be included in exactly one of them.

By area considerations, for any compact set $K$ and radius $r$, there is a distance $d(K,r)$ such that for any disk $D$ of radius larger than $r$, the distance from $K$ to $D$ is either $0$ or strictly larger than $d(K,r)$.
Also I denote $B(K,d)$ the set of points that are at distance less than or equal to $d$ from $K$

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We start by enumerating the disks, and repeatedly look at the smallest index $n$ of a disk not contained in a already constructed thickening of a previous disk. We have to build a thickening $T_n$ of $D_n$.

We start with $K_0 = D_n$ and $\phi_0 : K_0 \setminus \{c_n\} \to K_0$ to be the radial projection onto its boundary.

Because there have been finitely many thickenings constructed so far, $K_0$ is at a positive distance from their reunion, and so there is a distance $d_0$ such that $B(K_0,d_0)$ doesn't intersect any thickening, any disk of radius $>1$, nor $A$ or $B$. We will want to have $T_n$ to stay within distance $d_0$ of $K_0$.
Let $e_0 \le \min(d_0/6,d(K_0,d_0/6))$ such that $L_0 = B(K_0,e_0)$ is not tangent to any disk (this is possible because there are countably many disks and uncountably many reals). There is an obvious continuous projection from $L_0 \setminus K_0^\circ$ to its boundary so we notice that it moves every point by at most $e_0$ and compose it with $\phi_0$ to get $\psi_0$.

Now the boundary of $L_0$ may intersect infinitely many disks (of radius less than $d_0/6$), but only a finite number of disks of radius larger than $e_0$.

Let $K_1$ be the reunion of $L_0$ and the disks of radius larger than $e_0$ it intersects. Since there were finitely many disks, $K_1$ is compact and stays within $3d_0/6$ of $K_0$. For every disk added we project the region that sticks out to its boundary, for example by choosing a point of its circumference inside $L_0$ and projecting radially away from that point. Doing so moves every point by at most $2d_0/6$ and we compose it with $\psi_0$ to get $\phi_1$.
The boundary of $K_1$ is made of finitely many arc circles, with no two consecutive arc radius larger than $e_0 \le d_0/6$.

Now we pick $d_1 \le \min (d_0/6, d(K_1, e_0))$ such that the $d_1$-thickening from an arc of $K_1$ doesn't meet the thickening from any other arc, except for consecutive arcs. Then $B(K_1,d_1) \subset B(K_0,d_0)$

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Now the goal is to build a simply connected compact set $K_2$ and a distance $d_2$ such that $K_1 \subset K_2^\circ \subset K_2 \subset B(K_2,d_2) \subset B(K_1,d_1)$ ; $K_2$'s boundary is made of finitely many arc circles, where among two consecutive arc circles, one of them has radius $\le d_1/6$. Together with a projection from $K_2 \setminus K_1$ to the boundary of $K_2$ that moves points by at most some $O(e_0)$.

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Then we choose $e_1 \le \min(d_1/6,d(K_1,d_1/6))$ such that $L_1 = B(K_1,e_1)$ is not tangent to any disk. Moving the boundary of $K_1$ to the boundary of $L_1$ can also be done radially from each center of the disks from each arc, except at the corner cases where two bands overlap or go backwards. We can send all those points in a straight line onto the new corner and we still have a continuous projection that moves every point by at most $e_1$.

Here there may be (possibly infinitely many, possibly zero) disks of radius less than $d_1/6$ that touch the $e_1$-thickening of consecutive arcs. So at each corner we push the area in the corner to the outer border of a hypothetical disk of radius $d_1/6$ tangent to both arcs. Such a disk stays within $d_1$ of $K_1$ (because $e_1+2d_1/6 < d_1$) and so it can't touch non consecutive arcs. We move the points for example by projecting in a direction parallel to the tangent line in the corner. The maximum displacement is bounded by the circumferences of the two smaller two disks out of three, so it's a $O(d_1/6+(e_0+e_1)) = O(e_0)$. Let $M_1$ be the resulting set after pushing out every corner.

Now the boundary of $M_1$ may intersect infinitely many disks (of radius less than $d_1/6$), but only a finite number of disks of radius larger than $e_1$. Among two consecutive arcs of $M_1$, one of them has radius less than $d_1/6$. Let $K_2$ be the union of $M_1$ and the disks of radius larger than $e_1$ it intersects, again along with the possible corners filled up. Filling the corner displaces points by a $O(e_1+d_1/6)$, and then pushing up to the outer border of the new disks by $2e_1$.

$K_2$ is simply connected, compact and stays within $5d_1/6$ of $K_1$ so it is still within $d_1$ of $K_1$. Its border is made of finitely many arcs, with no two consecutive arc radius larger than $d_1/6$

And so on.

At every step, we get $K_i \subset K_{i+1}^\circ \subset K_{i+1} \subset B(K_{i+1},d_{i+1}) \subset B(K_i,d_i)$, and every disk of radius greater than $e_i$ is either completely outside $B(K_{i+1},d_{i+1})$ or completely inside $K_{i+1}$. We take $T_n = \cup K_i$

From $\phi_i$ to $\phi_{i+1}$ we only displace points by at most some $O(d_i)$. Since this sequence converges to $0$ geometrically, the sequence $(\phi_i)$ converges uniformly to a continuous projection $f_n$ Since the image of $f_n$ stays squeezed between $K_{i+1}$ and $B(K_{i+1},d_{i+1})$, $f_n$ pushes out everypoint of $T_n$ (which is not necessarily closed) on its boundary, which doesn't intersect any disk at all.

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After covering every disk of $S$ by a thickening $T$, we compose every projection together to get a continuous map $\Bbb R^2 \setminus C \to \Bbb R^2 \setminus S$ where $C$ is a countable set consisting of the center of the disks we selected previously.

Then we apply this map on any path from $A$ to $B$ that doesn't pass through any point in $C$, and we obtain a path from $A$ to $B$ that stays outisde $S$.

Answer 4

EDIT: This answer has a flaw in itself.

It seems that the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ will not always be path connected, under the assumption that my line of reasoning is correct an that there is no flaw in the proof below.

So let us construct the countable set of pairwise disjoint disks in the following way:

Let $A_{11}$ be the disk with radius $1/2$ centered at the point $(-1,0)$ and $A_{12}$ the disk also with radius $1/2$ centered at the point $(1,0)$. Let the first step be to construct disk $A_{13}$ with radius $1/4$ centered at the point $(0,0)$. Now, construct $A_{14}$ with radius $1/16$ so that the center of $A_{14}$ is midpoint between the centers of $A_{11}$ and $A_{13}$ and construct $A_{15}$ with radius also $1/16$ so that the center of $A_{15}$ is midpoint between the centers of $A_{12}$ and $A_{13}$. Next step is to construct $4$ disks such that every disk has radius $1/64$ and the centers of those $4$ disks are at the midpoints of the line segments that join centers of five already constructed disks (in such a way that the centers of those $4$ disks are between neighbouring disks and when I say neighbouring disks I mean that the two disks are neighbours if there are no other disks between them). If we continue in such a fashion at the $n$-th step we construct $2^{n-1}$ disks, all with the same radius $1/(2^{2n})$. Now we pass to the limit and denote this construction as $\bigcup_{n=1}^\infty A_{1n}$. Let now $A_{21}$ be the circle with radius $1/2$ centered at the point $(3,0)$ and let $A_{22}=A_{12}$. Now we do exactly the same construction of the sequence of disks between $A_{21}$ and $A_{22}$ and denote this construction as $\bigcup_{n=1}^\infty A_{2n}$. Next, let $A_{31}$ be the disk with radius $1/2$ centered at the point $(-3,0)$ and let $A_{32}=A_{11}$. Again we do exactly the same construction between $A_{31}$ and $A_{32}$ and denote it as $\bigcup_{n=1}^\infty A_{3n}$. We proceed in the same way and place centers of the disks $A_{2k,1}$ at the points $(2k+1,0)$ and centers of the disks $A_{2k+1,1}$ at the points $˙(-(2k+1),0)$ and do the same constructions as before to obtain sets $\bigcup_{n=1}^\infty A_{(2k),n}$ and $\bigcup_{n=1}^\infty A_{(2k+1),n}$.

Let ${\bigcup_{k=1}^\infty} {\bigcup_{n=1}^\infty A_{k,n}}=\bigcup_{n=1}^\infty D_n$. Suppose now that $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ is path connected. Choose one point above the $x$-axis and one below $x$-axis. If $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ is path connected then there must be a path connecting those two points. Since one point is below the $x$-axis and the other above the path must cross $x$-axis at some point $x_0$. Now suppose that the point in which the path crosses the $x$-axis is not in some of the disks and is not on the boundary of some disk. That means that the point is at some positive distance from the closest disk, in other words, that there is a $\varepsilon>0$ such that open interval $({x_0}-{\varepsilon}, {x_0}+{\varepsilon})$ has empty intersection with every disc.

That would mean that on the $x$-axis there is disc-free zone of length $2 \varepsilon$ between two neighbouring disks of radius $1/2$ but that is not possible because by construction the distance over the $x$-axis between two neighbouring disks of radius $1/2$ is $1$ and is filled with smaller disks because smaller disks between two discs of radius $1/2$ have total length equal to $1(1/2)+2(1/2^3)+4(1/2^5)+...+2^{n-1}(1/2^{(2n-1)})+...=1$.

Answer 5

I will add another answer, which depends on the assumption that the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ is uncountable and on the assumption that for every point $(x,y)$ in the plane there is a point in the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ that is at the positive distance(I mean strictly positive here and below in the proof where I use the word positive ($>0$)) from the point $(x,y)$. And in this one I am forced to use the axiom of choice (at least it looks to me that I am using it because I am not completely sure how to interpret this process that leads me to the final conclusion and because I do not know enough about that axiom) because I do not see some clever and rigorous enough way to resolve this question without using that axiom.

So let us now choose two points from the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ and call them $A$ and $B$. I will put $A=(0,0)$ and $B=(1,0)$ because no matter what two points we choose we can always rotate and scale the plane so that the problem reduces to the problem between those two points.

Now, define the sets $S(\varepsilon)$ as $S(\varepsilon)=[0, \frac {1}{2} + \varepsilon]$ where $\varepsilon \in [0, \frac{1}{2}]$. Let us choose now some $\varepsilon_0$ from $[0,\frac{1}{2}]$ just to illustrate the process of building a path. This $\varepsilon_0$ corresponds to the set $S(\varepsilon_0)=[0,\frac{1}{2}+\varepsilon_0]$.

We build a path in this way. First we start from the straight line segment that goes from point $A$ to the point $B$. When we have chosen $\varepsilon_0$ it determines the point $(0,\frac{1}{2}+\varepsilon_0)$. Now we choose some point in the plane from the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ that is at the positive distance from the point $(0,\frac{1}{2}+\varepsilon_0)$ (under the assumption that the point $(0,\frac{1}{2}+\varepsilon_0)$ itself is not in the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$, if it is we can leave it in her place or we may not, the proof does not depend on that constructive process), let us denote it as $P_1(\varepsilon_0)$ and let us use that point to build a new path that consists of two line segments, the path goes from point $A$ to $P_1(\varepsilon_0)$ and then from $P_1(\varepsilon_0)$ to $B$. Now choose midpoint of the line segment that goes from $A$ to $(0,\frac{1}{2}+\varepsilon_0)$ and for that point choose some point from the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ that is at the positive distance from that point and do exactly the same for midpoint of the line segment that goes from the point $(0,\frac{1}{2}+\varepsilon_0)$ to the point $B$. With those two points chosen from the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ and denoted as $P_2(\varepsilon_0)$ and $P_3(\varepsilon_0)$ we build a new path that consists of the four line segments, a path that goes from $A$ to $P_2(\varepsilon_0)$ and then from $P_2(\varepsilon_0)$ to $P_1(\varepsilon_0)$ and then from $P_1(\varepsilon_0)$ to $P_3(\varepsilon_0)$ and finally from $P_3(\varepsilon_0)$ to $B$. After continuing in this fashion at the $n$-th step we obtain $2^n$ new midpoints and the curve consists of at most $2^{n+1}$ line segments (at most, because we could in this process leave some points fixed if they already belong to the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$). We now let this process tend to infinity. When doing so we obtain a curve such that the points of that curve that are from the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ are dense on the curve but that is still not enough for curve to be completely a subset of the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$.

What to do next? Well, we choose all (at once, or one by one? haha :D ) $\varepsilon$ from the set $[0, \frac{1}{2}]$ and we do exactly the same process for each of the sets $S(\varepsilon)$ as we did for $S(\varepsilon_0)$. If, in this process of choosing a midpoints of the midpoints for every $\varepsilon$, we stumble upon a points that are already on the curve and also from the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ we could leave them untouched or leave them not, it does not depend on the construction of proof.

So, when doing this process for every $\varepsilon$ from the set $[0, \frac{1}{2}]$ we do some countable process uncountable number of times, to be more precise, we uncountably many times choose countable sets of points and exactly that fact craves for my assumption that the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ is uncountable.

Remark 1: I really do not know if this and processes like this one are allowed in the construction of the proof but to me it was just an idea that came up when I was thinking about this problem. Also, if we allow lines of reasoning like this mine line of reasoning and allow use of countable and uncountable choice then it looks that for every pair of points from the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ we could prove existence of at least countably infinite number of paths that connect those two points.

Remark 2: After re-reading this answer I realized that it also depends on the assumption that if at the $n$-th step we have function $f_n$ whose graph consist only of line segments and of at most $2^{n+1}$ line segments then $\lim_{n\to\infty}f_n$ is continuous.

Remark 3: From the point of view that concerns rigour I am really not satisfied with this proof because it depends on some maybe non-trivial assumptions which I am aware of and maybe on some other which I am not aware of but I post it here as a remainder to all of you who will maybe try to prove the problem in a similar way because somehow I believe that every attempt of proving this problem in a way that is similar to mine will have issues of uncountability of the set in question and issues of strictly positive distance from the set in question and issues of continuity of the function that is the limit of continuous functions defined in this or in some other way.

Answer 6

Here is an outline of an alternate method of proof:

Regard $\mathbb R^2$ as $\mathbb C$ and write $D_n=\bar{B}(z_n,r_n), D^{\circ}_n=B(z_n,r_n)$.

Step 1: $\mathbb C\setminus \{z_1,z_2,\ldots,\}$ is path-connected

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Step 2: There is a continuous surjection $f\colon \mathbb C\setminus \{z_1,z_2,\ldots,\}\to \mathbb C\setminus \bigcup D^{\circ}_n$

Proof of Step 2:

For $w\in \mathbb C\setminus \bigcup D_n$, define $f(w)=w$. Otherwise define $$f(w)= \partial B(z_n,r_n)\cap \{\text{line through }z_n\text{ and }w\}$$ This map is easily seen to be continuous and surjective, and in fact it is a deformation retraction.

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Combining Step 1 and Step 2, we see that $\mathbb C\setminus \bigcup D^{\circ}_n$ is path-connected. Now consider any points $x,y\in \mathbb C\setminus \bigcup D_n$. Connect them via a path in $\mathbb C\setminus \bigcup D^{\circ}_n$. Deform the path such that it remains in $\mathbb C\setminus \bigcup D_n$ to conclude the proof.