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Let $\langle\mathbf{R}^n,\mathscr{O}\rangle$ be the $n$-dimensional Euclidean space. Define $\mathbf{Q}\subseteq\mathcal{P}(\mathscr{O})$ to consist of all sets $\mathsf{Q}$ which simultanously satisfy the following four conditions:

  1. $\mathsf{Q}\neq\emptyset$ and $\mathsf{Q}\neq\{\emptyset\}$
  2. $\forall_{U\in\mathsf{Q}}\exists_{V\in\mathsf{Q}}\,\mathrm{Cl}\,{V}\subseteq U$
  3. $\forall_{U,V\in\mathsf{Q}}\,(U\neq V\rightarrow \mathrm{Cl}\,V\subseteq U\vee \mathrm{Cl}\,U\subseteq V)$
  4. $\forall_{A,B\in\mathscr{O}}\,((\forall_{U\in\mathsf{Q}}\,A\cap U\neq\emptyset\neq B\cap U)\rightarrow\mathrm{Cl}\,A\cap \mathrm{Cl}\,B\neq\emptyset)$

So every element of the family $\mathbf{Q}$ is a set in which well-inside relation ($\mathrm{Cl}\,U\subseteq V$) is connected (in the sense of order) and which is decreasing with respect to this relation, but also satisfies the requirement according to which all open sets with non-empty intersections with all elements from the set must have a common adherent point.

Is it true that for every $\mathsf{Q}\in\mathbf{Q}$ there is a point $p\in\mathbf{R}^n$ such that $\bigcap\mathsf{Q}=\{p\}$? I will be grateful for your help.

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  • $\begingroup$ It seems that $\mathsf{Q}=\{\emptyset\}$ has all your properties, but not your goal property. $\endgroup$ Commented Mar 26, 2016 at 18:29
  • $\begingroup$ Apart from that, I am confused by property 4. How can you hope to achieve this? If $\mathsf{Q}$ has a nonempty open set $U$ in it, then we can find two disjoint balls $A$ and $B$ inside $U$, with disjoint closures. So property 4 seems impossible except in the trivial case of my previous comment. Have I misunderstood? $\endgroup$ Commented Mar 26, 2016 at 18:31
  • $\begingroup$ @JoelDavidHamkins Thanks for pointing the singleton of $\emptyset$ - I corrected the question and ruled it out explicitly. As for the second part of your question. The only family with which you could obtain what you write about is $\{\mathbf{R}^n\}$, since this is the only clopen non-empty set in the space, and thus can satisfy 2. But the intention of 4 is to rule this case and guarantee that sets are <<shrinking>> around some precise location in the space. François's answer proves that 4 is a good embodiment of the intention. Is it OK, or have I misunderstood your question? $\endgroup$ Commented Mar 26, 2016 at 19:09
  • $\begingroup$ Ah, I had mis-read the scope of the second quantifier in statement 4, which makes it trivial as I had indicated. I suggest editing the parentheses to indicate that the implication is not under the scope of the quantifier. $\endgroup$ Commented Mar 26, 2016 at 19:18
  • $\begingroup$ @JoelDavidHamkins But the parentheses seem to be unambiguous, aren't they? $\endgroup$ Commented Mar 26, 2016 at 23:31

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Yes. (Assuming $\varnothing$ isn't allowed in your $Q$'s, as Joel pointed out.)

First note that for such $Q$, $\bigcap Q$ cannot contain two points. Indeed, if $a,b \in \bigcap Q$ are distinct, we can pick open neghborhoods $a \in A$, $b \in B$ such that $\operatorname{Cl} A \cap \operatorname{Cl} B = \varnothing$ and these two open sets $A$, $B$ contradict condition 4. So it suffices to rule out the case that $\bigcap Q = \varnothing$.

Suppose, for the sake of contradiction, that $\bigcap Q = \varnothing$. Pick a cofinal sequence $U_0 \supset U_1 \supset \cdots$ of elements of $Q$. Define $V_n = U_n \setminus \operatorname{Cl} U_{n+1}$ and consider $A = \bigcup_{n=0}^\infty V_{4n}$ and $B = \bigcup_{n=0}^\infty V_{4n+2}$. Note that $\operatorname{Cl} A = \bigcup_{n=0}^\infty \operatorname{Cl} V_{4n}$ and $\operatorname{Cl} B = \bigcup_{n=0}^\infty \operatorname{Cl} V_{4n+2}$ because $\bigcap_{n=0}^\infty \operatorname{Cl} U_n = \varnothing$. Thus $\operatorname{Cl} A \cap \operatorname{Cl} B = \varnothing$ and yet $A \cap U \neq \varnothing \neq B \cap U$ for any $U \in Q$, which contradicts condition 4.

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  • $\begingroup$ François, your proof demonstrates that it is enough to assume that the space is Urysohn and has a countable dense subset. Am I missing something? $\endgroup$ Commented Mar 26, 2016 at 22:08
  • $\begingroup$ @MadHatter: That's right. Countable dense subset is a little much, I think the countable chain condition (ccc) is enough. $\endgroup$ Commented Mar 26, 2016 at 22:26
  • $\begingroup$ François, some second thoughts. In order to apply ccc one must know that when $U\neq V$ are from $Q$, then (in case $\operatorname{Cl} U\subseteq V$), $V\setminus U$ contains a non-empty open set. This of course follows from $V\setminus \operatorname{Cl} U$ being non-empty. Yet I fail to see how to get it from Urysohn property. (The same applies to separability). $\endgroup$ Commented Mar 26, 2016 at 23:56
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    $\begingroup$ @MadHatter: The key is that $\operatorname{Cl}(\bigcup_{n=m}^\infty V_n) \subseteq \operatorname{Cl} U_m$. So $\operatorname{Cl} A \subseteq \operatorname{Cl} V_0 \cup \cdots \cup \operatorname{Cl} V_{4n} \cup \operatorname{Cl} U_{4n+4}$ and take the limit as $n \to \infty$. $\endgroup$ Commented Apr 19, 2016 at 23:39
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    $\begingroup$ François, I get the full picture now. By the way, to my taste this is a beautiful proof. Thank you very much for the hint and the help. $\endgroup$ Commented Apr 21, 2016 at 8:35

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