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Prove/disprove: On the $x$ axis there's a set with the cardinality $\mathfrak c$ of points that do not belong to any disk of a set $O$ of disjoint disks of positive radius $\{(x,y)\in \mathbb R|(x-a)^2+(y-b)^2\le r^2 \}:a,b,r\in \mathbb R: r \gt 0$.

Note: the disks can't overlap.

I think it's true, by sketching the disks I always get a gap between them, and even by filing these gaps to infinity they can't cover all the points on an interval because they can't overlap. It's impossible for any shape to not have gaps if they can't overlap.

In a previous part of the exercise I already found the cardinality of $O$ to be (at most) $\aleph_0$ by finding a rational point $(q_1,q_2)$ in each disk of $O: q_1,q_2\in \mathbb Q\times\mathbb Q$. So if we'll take all the numbers that do not belong to any disk, they would be a subset of the transcendental numbers and on the other hand, every disk would have two gaps from both sides of its perimeter, so we'll get $2^{\aleph_0}$.

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shinzou
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    What are $a,b,r$? Given any point $(x,0)$ on the $x$ axis, we can pick $a=x,b=0, r=1$ and find that $(x,0)$ is covered by $O$. Also, clearly the cardinality of $O$, as you define it, id $\mathfrak c$ unless $r=0$. – Hagen von Eitzen Jul 05 '14 at 12:25
  • A "filled" circle is usually called a disk for clarity. – hardmath Jul 05 '14 at 13:10
  • I suspect that you are assuming disjointness of these disks ("if they can't overlap"), which is not explicitly stated in the problem as "quoted" (gray highlighting). – hardmath Jul 05 '14 at 13:35
  • @HagenvonEitzen, I've edited what $a,b,r$ are. About the cardinality, every disk has a unique ordered number in $\mathbb Q$ so it can't be $\mathfrak c$. – shinzou Jul 05 '14 at 13:53
  • @hardmath The points do not belong to any circle, I also asked specifically if overlapping is allowed and it is not, also if overlapping is allowed than there can't be a unique ordered pair in $\mathbb Q^2$ to represent each disk. It's also making the question too easy, just fill the everything with disks and we get $\mathbb R^2$. – shinzou Jul 05 '14 at 13:58
  • Disjointness assumption is a good step forward. However your Comment now raises an issue of "a unique ordered pair in $\mathbb{Q}^2$ to represent each disk". Is that part of the problem or something that arises from your approach to the problem? The natural thing seems to be identifying the individual disks by their three (real) parameters $a,b,r$. – hardmath Jul 05 '14 at 14:10
  • But it's enough to represent each disk with an ordered pair in $\mathbb Q^2$, it was in a previous part of the question and it was marked correct so let's regard that as a given. – shinzou Jul 05 '14 at 14:14
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    The question is impossible to answer as stated, since the phrasing is incorrect, mainly because you use wrong quantifiers. Is your question: "Consider a collection of nonoverlapping discs in the plane. Show that there is a subset of the $x$-axis, of size $\mathfrak c$, and disjoint from the union of these discs."? If this is the question, please consider re-editing it for clarity. Currently, what you say instead is that there is a set of points on the axis that "do not belong to any disk of the set $O$". But $O$ itself is a disc, so this in meaningless. And the use of any is incorrect here. – Andrés E. Caicedo Jul 05 '14 at 17:22
  • @AndresCaicedo isn't $O$ the set of all disks on $\mathbb R^2$ ? – shinzou Jul 05 '14 at 19:14
  • No, the collection of all discs covers the whole plane. – Andrés E. Caicedo Jul 05 '14 at 19:30

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With the restriction that the closed disks do not overlap, we can prove the statement (Added, thanks Hagen) assuming the Continuum Hypothesis. [NB: According to Asaf's Answer to this previous Question, we don't need to assume the Continuum Hypothesis for this particular case. More added to the end of this Answer.]

Consider for the sake of contradiction that it was possible to cover all but at most countably many points of the $x$-axis with the closed disks. Taking those points, together with the at most countably many nonempty intersections of the disks with the $x$-axis, we would have expressed the $x$-axis as a union of (at most) countably many disjoint bounded closed intervals (considering a single point $[s,s]$ as a closed interval here). [NB: I put "at most" in parentheses before countably many because clearly any finite number of disjoint closed intervals would have compact union, and the $x$-axis is not compact.]

However it is not possible to do this. According to a theorem of Sierpiński, if $\mathbb{R}$ or any Baire space is expressed as the countable union of disjoint closed sets, at most one of them is nonempty.

Since the whole $x$-axis cannot be realized as the intersection with one of the closed disks in $O$ or as a single point $[s,s]$, we are done.

Added: Consider the intersection of those open sets which are complements of the closed disks' intersections in the $x$-axis (real line). That intersection of countably many open sets is by definition a $G_\delta$ set. By the argument above it is not countable, and by Asaf's Answer to Is the cardinality of uncountable $G_\delta$ set of $\mathbb{R}$ equals the cardinality of the continuum?, we conclude it has cardinality of the continuum without assuming the Continuum Hypothesis.

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hardmath
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  • What does it mean that one of them is not empty ? – shinzou Jul 05 '14 at 19:20
  • The empty set is closed, as is any topological space in itself. So one way to write $\mathbb{R}$ as a countable union of disjoint closed sets is to say $\mathbb{R} = \mathbb{R} \cup \bigcup_{i=1}^\infty A_i$ where each $A_i = \emptyset$. What we are deducing from the theorem of Sierpiński is that this is the only way to write $\mathbb{R}$ as a countable union of disjoint closed sets (since all but one of them must be empty). – hardmath Jul 05 '14 at 19:28
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    Doesn't this only show that the set is larger than $\aleph_0$? The OP asks for $\mathfrak c$. – Hagen von Eitzen Jul 06 '14 at 08:15
  • Thanks, yes I was assuming the Continuum Hypothesis without explicitly saying so. – hardmath Jul 06 '14 at 12:51