Complete Circle Packing

@Michael Mendrin @Brian Charlesworth @Arjen Vreugdenhil Alright, I think this constitutes a proper proof. Thoughts? Can I definitively update the answer to no?

Do we intend to cover the side? The side is in the boundary, not the interior

"@Arjen Vreugdenhil – Because the rational points are a set of measure 0, and you are giving up control over the radius, I can cover the rational points by discs of total area < 1 (actually close to 0), which shows that this approach will not work."

Showing that a cover fails is not a proof that any cover would fail.

No matter how small of a circle you make, it'll still leave a gap in the corner.

Not sure if this is helpful, but here's something that I thought :

We consider order the set of all collections of (disjoint) balls contained in the unit square by set inclusion. Then, if $\{B_n\}_{n \in \mathbb{N}}$ maximal collection (which exist by Zorn's lemma) then $(\cup_{n \in \mathbb{N}}B_n)^c$ has empty interior. Here, complements are with respect to the unit square.

"The interval cannot be expressed as the disjoint union of countably many closed intervals (possibly of 0 length)."

This statement represents a significant change to the problem. There is nothing in the original statement of the problem that requires the discs to be disjoint.
It is trivial to represent the interior of (0,1) as a countable union of closed intervals, whose interiors are disjoint.

The comment said "this approach will not work"' but the comment did not show that. It showed that the approach might not work. I'm hard-pressed to see how to fix it, but proving either side of this proposition seems difficult.
I suspect that no proof in either direction will be found. The framing of the question is to take the countable union of disjoint closed sets, but we don't know anything about the countable union of disjoint closed sets (except that the measure of the union is the sum of the measures of the individual sets). While it is easy to create a collection of closed sets with total measure < 1, that fact doesn't imply that it isn't also possible to create a collection of disjoint closed sets with total measure = 1. Of course having total measure 1 would only be a necessary and not sufficient condition for the union to be able to cover the entire interior of the unit square.

Let's suppose that we can place infinitely many closed discs of positive radius in a unit square where the interiors of the discs do not intersect, such that any point given in the interior of the square lies on a disc, then we can take(choose) (assuming axiom of choice) one countable number of points (a sequence of different $a_n$ formed by the centers of some different circles of radius $r_{n} > 0$ ) inside of some of these closed circles. Then, due to Weierstrass theorem, this sequence must have a subsequence $a_{n_k}$ converging to a point $A$ in the clousure of the square, but closed discs of positive radius do not intersect, i. e, the radius $r_{n_k}$ of these circles containing $a_{n_k}$ is decreasing to a limit 0. This leads to a contradiction: Sketch of proof: Take a closed disk of radius $\epsilon > 0$ around $A$ , that is, $\overline{D(A, \epsilon)}$ , ... Following this process, $\displaystyle \lim_{k_1, k_2 \to \infty} d(a_{n_{k_1}}, a_{n_{k_2}}) = 0 \ge \lim_{k_1, k_2 \to \infty} r_{n_{k_1}} + r_{n_{k_2}}$ , and this leads to circles with limit of radius being equal to 0... For visualizing my idea think about three mutual tangent circles, then to cover the space between these circles with circles, in some moment we should use circles of radius 0.

Topologycal proof.-

Let's call $A$ be the interior of the unitary square and $\text{ Cl (A) = A } \cup \{ \text{ sides of this square } \}$ , and let's suppose that $A$ can be covered by $B = \cup \{\text{ countable infinitum closed circles such that... }\}$ then, $A \subset B \subset Cl(A) \Rightarrow B$ is a connected set. And, $\text{Cl (A) } \subseteq \text{Cl (B) } \subseteq \text{Cl (Cl (A)) = Cl(A) } \Rightarrow \text{Cl (A) = Cl (B) }$ and $\text{Int (A) = A} \subseteq \text{Int(B)}$ . Let's take now $C = \cup \{\text{ countable infinitum open circles with the same radius as the circles in B }\}$ . Then, $C$ is contained strictly in $A$ ,because $C$ is an open set and if $A$ was equal to $C$ , $A$ wouldn't be a connected set, however $C \subset A \subset B \subset \text{Cl (C) } \Rightarrow$ $\text{ Cl(B)} = \text{Cl (C)}$ . Conclusion, $C, A$ are open sets, with $C$ contained strictly in $A$ with $\text{ Cl(A)} = \text{Cl (C)} \Rightarrow$ $\text{ int(Cl(A)) = A} = \text{Int(Cl (C)) = Int(C} \cup \text{Boundary(C)}) = C$ (Contradiction)

Question.- If $C$ is an open set, then , $\text{Int(Cl(C))} = C$ ? No, it's not true, $\text{Int(Cl(U))} \neq U$ being $U$ an open set, nevertheless $C$ is the union of disjoint open circles with positive radii, then it's clear that $\text{Int(Cl(C))} \supset C$ because $\text{Int(Cl(C))}$ is the greatest open set contained in $\text{Cl(C))}$ . Let's suppose that $\text{Int(Cl(C))} \neq C$ then would be a point $a \in \text{Int(Cl(C))} = A \subset B$ such that $a \notin C$ , then due to $\text{Int(Cl(C))}$ is an open set, for all neighborhood $N \subset A$ around $a \in A$ , we have two possibilities:

a) $N \cap C = \emptyset$ . This scenario is impossible because if it was true $B$ wouldn't cover $A$ and we are supposing that $B$ covers $A$ .

b) $N \cap C \neq \emptyset$ . This leads also a contradiction because then $a \in \text{Cl(C))}$ and then $\text{Cl(C))} = \text{Int(Cl(C))}$ and this would imply that $\mathbb{R}^2$ is not a connected set.

Therefore, $\text{Int(Cl(C))} = C$ .

You can get closer and closer to the corner, as the radius of the circle approaches 0, but you can never get there.

I'm not posting anything rigorous because I'm not that adept at this. I'm just stating why I think the answer should be NO.

I'm assuming it's not possible to cover the sides with countably infinite circles. Now, let's think there's a circle (I'm calling it the first circle) in the middle of the square. We've to assemble other circles around it. With every step there will be space left behind. So, we will be zooming into the perimeter of the first circle and trying to cover up the non-occupied space (for an infinite amount of time).

As we're zooming (infinitely) into the perimeter of the first circle, it eventually becomes (or, tends to) a straight line. But, we've already assumed that we cannot cover the sides of the unit square which is essentially a straight line. Hence, we can either prove we can cover every single space within the square (including the corners and sides) or we cannot.

Let's suppose that we can place (infinitely many) closed discs (of positive radius) in a unit square, where the interiors of the discs do not intersect, such that any point given in the interior of the square lies on a disc. Due to your observation 1: there are countably many circles... Let's call them, $\{B_n\}_{n \in \mathbb{N}}$ , these closed bounded balls covering the interior of the square, they are compact sets. Due to observation 2, the side of the square can't be covered by these circles. Take $x$ in the side of the square such that $x$ doesn't belong to $\displaystyle \cup_{n \in \mathbb{N}} B_n$ . For example , take $x$ being a corner of the square. For each, $n \in \mathbb{N}$ let's define $d(x, - ) : B_n \longrightarrow \mathbb{R}$ , $\forall y \in B_n$ , $d(x,y) = \text{ euclidean distance from x to y}$ . Being $B_n$ (n is a fixed natural number) a compact set and $| d(x,z) - d(x, y) | \leq d(z,y)$ (triangular inequality), $d(x, - )$ is a continuous function of a compact set in $\mathbb{R}$ which it is bounded, then due to Weierstrass- theorem, this function has a maximum and a minimum. Obviously, this minimum is greater than $0$ , because if the minimum of this function was $0$ , $x$ would belong to $B_n$ . (Contradiction). $\text{ minimum d(x, B}_n) = d(x , y_n)$ being $y_n$ the point in the boundary of $B_n$ joining $x$ with the center of $B_n$ .

Nevertheless, $d(x, \cup B_n) = \text{ infimum } \{ d(x, y), \text{ such that y} \in \cup B_n \} = 0$ , i.e, $\exists r_n \in \cup B_n$ such that $\displaystyle \lim_{n \to \infty} d(x, r_n) = 0 = d(x, \lim_{n \to \infty} r_n)$ , because the function $d(x, - )$ is a continuous function ,and this implies $\displaystyle \lim_{n \to \infty} r_n = x$ .

On the other hand, $d(x, \text{Int(B}_n)) = d(x, y_n), \space \forall n \in \mathbb{N}, \text{ such that n is a fixed natural number}$ althought $y_n \notin \text{Int(B}_n) \Rightarrow$ $d(x, \cup \text{Int(B}_n)) = \text{ infimum } \{ d(x, y), \text{ such that y} \in \cup \text{Int(B}_n))\} = 0$ , i.e, WLOG $\exists (a_n) \in \cup \text{Int(B}_n)$ such that $\displaystyle \lim_{n \to \infty} d(x, a_n) = 0 = d(x, \lim_{n \to \infty} a_n)$ , because the function $d(x, - )$ is a continuous function ,and this implies $\displaystyle \lim_{n \to \infty} a_n = x$ Let's call $D = \cup B_n$ , $A = \text{interior the unitary square}$ and $\text{Cl(A) = A} \cup \text{ sides of the square }$ then we have $A$ is a connected set and $A \subset D \subset \text{Cl(A)} \Rightarrow D$ is a connected set. The continuous image of a connected set is a connected set. Hence $d(x, -) ((D)) = d(x, \cup B_n) = \cup d(x, B_n)$ is a bounded interval (non empty) in $\mathbb{R}$ containig at least two different points...

proof:

the convergence $a_n \to x$ prove that $x \in \text{Cl(} \cup \text{Int(B}_n))$ , and $x \in \text{Cl(} \cup \text{Int(B}_n)) \supset \cup \text{(Cl(Int(}B_n))) = \cup B_n$ and $x \notin \cup B_n$ . I'm going to prove :

$\text{Cl(} \cup \text{Int(B}_n)) = \cup B_n$ .

One inclusion is proved,if $a \in \text{Cl(} \cup \text{Int(B}_n))$ , then for all "very, very small" open neighborhood $N$ around $a$ , $N \cap (\cup \text{Int(B}_n)) = \cup (N \cap \text{Int(B}_n)) \neq \emptyset$ , $\cup \text{Int(B}_n))$ is the union disjoint of open sets, this implies that $a \in \cup B_m$ . (See below note). Therefore, we have proved the above equality, and this implies that $x \in \cup B_n$ and $x \notin \cup B_n$ (contradiction).

Note.- If $A \subset B$ , then $\text{Cl(A)} \subset \text{Cl(B)}$ .

$N \cap (\cup \text{Int(B}_n)) = \cup (N \cap \text{Int(B}_n)) \subset \cup B_n$

Question.- $\cup \text{Int(B}_n)) \subset \text{Int(} \cup B_n) = \cup \text{Int(B}_n)) \text { ? }$ . No. What is the interior of a countable number of closed, bounded balls in $\mathbb{R}^2$ with positive radius and intersection among two of them being at most a point?

$\mathbb{R}^2$ is a Baire space, and all points of intersection ( $d_n$ ) of each two closed balls of $\cup B_n$ have empty interior, ie, the countable union ( $\cup d_n$ ) of all these intersecting points of the closed balls has an empty interior. This says us: $\cup \text{Int(B}_n)) = \cup \text{Int(B}_n) \cup \text{Int(}\cup d_n) \subset \text{Int(} \cup B_n \cup d_n) = \text{Int(} \cup B_n)$ .

"Guide for another possible proof"".-

There can only be 1 closed "disjoint" ball with a radius of 1/2 in the square, and there is space in the interior of square not covered

There can only be 4 closed "disjoint" balls with a radius of 1/4 in the square, and there is space in the interior of square not covered.

There can only be 16 closed "disjoint" balls with a radius of 1/8, and there is space in the interior of square not covered.

There can only be 64 closed "disjoint" balls with a radius of 1/16, and there is space in the interior of square not covered.

There can only be 256 closed "disjoint" balls with a radius of 1/32, and there is space in the interior of square not covered.

There can only be $4^{n - 1}$ closed "disjoint" balls with a radius of $\frac{1}{2^n}$ , and there is space in the interiror of square not covered. As $n \to \infty$ , if the space in the interior of the square could be covered with a infinitum countable number of closed "disjoint" balls with positive radius, and the sequence $\frac{1}{2^n}$ tend to $0$ , and $4^{|\mathbb{N}|}$ is an uncountable number...

"Guide for another possible proof.-" G

Every sequence bounded in $\mathbb{R}$ has a convergent subsequence. Consider to be $r_n$ the radius of the closed and bounded ball $B_n$ . This sequence $r_n$ is bounded in $\mathbb{R}$ , WLOG, let $r_n$ be the subsequence converging. Obviously $r_n$ has to coverge to $0$ . Then, for example, for $\epsilon_k = \frac{1}{10000^{1000^{10000^{k}}}}, \quad k \in \mathbb{Z}^{+}$ , there exist an infinitum countable number of radius $r_{n_{k}}$ smaller than $\epsilon_k$ and only a finitum number of radius $r_{n_{k}}$ bigger than $\epsilon_k$ . Then we can choose radii, WLOG, $r_n$ , being $r_1$ the greatest radius, $r_2 < r_1$ the next greatest radius,i.e, we can stablish a well- order about the radii,i. e, $\ldots < r_n < r_{n - 1} < \ldots < r_2 < r_1$ . Now we choose the centers $c_n$ of these closed balls of radius $r_n$ . The sequence $c_n$ is a bounded sequence and this implies, there exists, WLOG, a subsequence $c_n$ converging. Where can this subsequence $c_n$ converge if the radii are positive?...

another proof.-

Since $\cup B_n)$ only can have a countable number of points in a side of square, and the side of square has an uncountable number of points and it's a point of adherence(clousure) of the interior of square, then is a point of the clousure of $\cup B_n)$ . But around of this point belonging to the side of square which it is not a point of $\cup B_n)$ lead us this union can't be a countable union...

"Another possible proof.-" $\mathbb{R}^2$ is a Baire space.

Lemma.- Every open set in a Baire space is a Baire space.

Theorem.- $X$ is a Baire space if and only if given any numberable family $\{U_n\}$ of open sets in $X$ , each of which is dense in $X$ , its intersection is also a dense set in X.

Lemma.- If $C_1 \supset C_2 \supset \ldots$ is a sequence of closed non-empty sets, in a complete metric space X, (this implies that X is a Baire space), and $\text{diameter (C}_n) \to 0$ then $\cap C_n \neq \emptyset$ .

Then, using G this leads us a other contradiction, do you know how?..