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This simple question came to my mind the other day:

Question: Can we fit uncountably many nonempty open sets in $\mathbb{R}^n$ such that each point of $\mathbb{R}^n$ is contained in at most finitely many of the sets?

I spent hours trying to find a clever way of getting such a collection by drawing subsets of $\mathbb{R}^2$, but every attempt failed. Hence my intuition strongly tells me that the answer is 'no', but then I also spent hours trying to disprove the claim without success.

Maybe there is a clever way of using that in a complete metric space a nonempty countable closed subset contains at least one isolated point.

Patrick
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3 Answers3

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No. Let $\mathcal F$ be a family of nonempty open subsets of $\mathbb R^n$ such that each point of $\mathbb R^n$ is in at most finitely many sets of $\mathcal F$. Now consider the set $$ P = \{ (q,A) \in \mathbb Q^n\times \mathcal F \mid q\in A \} $$

On one hand $P$ must be countable, because it has finitely many members for each $q$, and a countable union of finite sets is countable.

On the other hand $P$ has at least as many elements as $\mathcal F$, because each nonempty open set contains a rational point.

Thus $\mathcal F$ is at most countable.

This argument does not depend on $\mathbb R^n$ being metric or complete; it only requires that it is a separable space.

(As bof argues in comments below, this doesn't require any form of the axiom of choice either, assuming the space is second countable).

hmakholm left over Monica
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  • I think you would need AC if the "point-finite" assumption were weakened to "point-countable", but I don't think you need it for the question as stated. – bof May 18 '15 at 22:49
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    Let $\mathcal O$ be the collection of all open sets in $\mathbb R^n.$ Since $\mathbb R^n$ has a countable base, we can effectively (without choice) define an injection from $\mathcal O$ to $\mathcal P(\mathbb N),$ whence we can define a linear ordering of $\mathcal O.$ It follows that the union of countably many finite collections of open sets is countable. – bof May 18 '15 at 22:56
  • @bof: Hmm, that sounds convincing. (What was wrong with your answer, by the way?) – hmakholm left over Monica May 18 '15 at 22:58
  • Nothing wrong with my answer. It just seemed redundant after I saw your answer, and it seemed kind of dumb of me to use a countable base for the topology when only a countable dense set was needed. – bof May 18 '15 at 23:02
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    @bof: It sounded a lot more elegant to me, even if it may be just phrasing rather than deep structure. – hmakholm left over Monica May 18 '15 at 23:18
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I formulate Henning's excellent answer in its greatest possible generality:

Definition 0. Let $X$ denote a set and $\mathcal{O}$ denote a family of subsets of $X$. Then $D \subseteq X$ is called $\mathcal{O}$-dense iff for every $A \in \mathcal{O} \setminus \{\emptyset\}$, the intersection $D \cap A$ is non-empty. The density of $\mathcal{O}$ is the least cardinality of an $\mathcal{O}$-dense subset. We write $\mathrm{den}(\mathcal{O})$ for the density of $\mathcal{O}$.

Note that:

  • $\mathrm{den}(\mathcal{O}) \leq |X|$
  • $\mathrm{den}(\mathrm{Open}(\mathbb{R}^n)) = \aleph_0$

Definition 1. Let $X$ denote a set and $\mathcal{F}$ denote a family of subsets of $X$. The repiticity of $\mathcal{F}$ is the least cardinal $\kappa$ such that for all $x \in X$, we have $$|\{A \in \mathcal{F} \mid x \in A\}| \leq \kappa.$$ We write $\mathrm{rep}(\mathcal{F})$ for the repiticity of $\mathcal{F}$.

Note that:

  • $\mathrm{rep}(\mathcal{F}) \leq |\mathcal{F}|$

Theorem. Let $X$ denote a set and suppose that $\mathcal{O}$ and $\mathcal{F}$ are families of subsets of $X$ such that $\mathcal{F} \subseteq \mathcal{O} \setminus \{\emptyset\}$. Then: $$|\mathcal{F}| \leq \mathrm{rep}(\mathcal{F}) \cdot \mathrm{den}(\mathcal{O})$$

The proof is now at the end of this answer.

We can now answer your question rather simply.

You ask: can we fit uncountably many nonempty open sets in $\mathbb{R}^n$ such that each point of $\mathbb{R}^n$ is contained in at most finitely many of the sets? If we can answer "no" to the variant of this question in which the phrase "at most finitely many" is replaced by "at most countably many," then we can answer "no" to the original question. But this is equivalent to: does there exist uncountable $\mathcal{F} \subseteq \mathrm{Open}(\mathbb{R}^n) \setminus \{\emptyset\}$ such that $\mathrm{rep}(\mathcal{F}) \leq \aleph_0$? Using our theorem, we see that there does not. For suppose $\mathcal{F} \subseteq \mathrm{Open}(\mathbb{R}^n) \setminus \{\emptyset\}$ satisfies $\mathrm{rep}(\mathcal{F}) \leq \aleph_0$. Then:

$$|\mathcal{F}| \leq \mathrm{rep}(\mathcal{F}) \cdot \mathrm{den}(\mathrm{Open}(\mathbb{R}^n)) \leq \aleph_0 \cdot \aleph_0 = \aleph_0$$

So $\mathcal{F}$ is countable.

Proof. Write $D$ for an $\mathcal{O}$-dense subset of $X$, and assume $|D| = \mathrm{den}(\mathcal{O})$. Define:

$$\mathcal{G} = \{ (d,A) \in D \times \mathcal F \mid d\in A \} $$

There is a projection $\pi_1 : \mathcal{G} \rightarrow \mathcal{F}$ given by $\pi_1(d,A) = A$. This is surjective (use that $\mathcal{F} \subseteq \mathcal{O} \setminus \{\emptyset\}$ and the $\mathcal{O}$-density of $D$). Hence $|\mathcal{F}| \leq |\mathcal{G}|.$

Hence:

$$|\mathcal{F}| \leq |\mathcal{G}| = |\{ (d,A) \in D \times \mathcal F \mid d\in A \}| = \left|\bigoplus_{d:D}\{A \in \mathcal{F} \mid d\in A \}\right| $$

$$= \sum_{d:D} \left|\{A \in \mathcal{F} \mid d\in A \}\right| \leq \sum_{d:D} \mathrm{rep}(\mathcal{F}) = \mathrm{rep}(\mathcal{F}) \cdot |D| = \mathrm{rep}(\mathcal{F}) \cdot \mathrm{den}(\mathcal{O})$$

goblin GONE
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just a rephrasing of Henning's idea

if $\{r_n\}$ is any countable dense subset of $\Bbb{R}^n$ we may form: $$ \mathcal{V}=\bigcup_{n\lt \omega}\{U\in \mathcal{F}|r_n \in U\} $$ clearly $\mathcal{V}$ is at most countable, but it exhausts all the members of $\mathcal{F}$ which contain any $r_n$.

hence, since $\mathcal{F}$ is uncountable we may (AC) choose a $V \in \mathcal{F} \setminus \mathcal{V}$ which contains no $r_n$, contradicting the assumption of density

David Holden
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