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Let $K\subset \mathbb{R}^n$ be a compact set of non empty interior and denote $\mathcal{P}(\mathbb{N})$ the set of subsets of $\mathbb{N}$. Consider a mapping $\phi:K \rightarrow \mathcal{P}(\mathbb{N})$ that associate to each $x\in K$ an infinite subset of integers. In other words, for all $x\in K$, $\phi(x) \subset \mathbb{N}$ verifies $|\phi(x)| = \infty$. Show that there exists $L \subset K$ verifying $\left|\bigcap_{x \in L} \phi(x)\right| = \infty$ and such that $\text{int}(\text{cl}(L)) \neq \emptyset$ (ie the closure of $L$ is of non empty interior).

Another way of seeing the problem is as follow: for all $x\in K$ we have a subsequence $\phi(x) \in \mathbb{N}^{\mathbb{N}}$ of the identity sequence $u_n = n, \; \forall n$. Show that there exsits $L\subset K$ with $\text{int}(\text{cl}(L)) \neq \emptyset$ such that we can extract the same subsequence from all $\{\phi(x) \; : \; x \in L\}$ (that is $|\bigcap_{x\in L} \phi(x)| = \infty$).

Amine
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1 Answers1

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The result is false as stated.

Let $\mathscr{A}$ be a family of infinite subsets of $\Bbb N$. $\mathscr{A}$ is an almost disjoint family if $A\cap B$ is finite whenever $A,B\in\mathscr{A}$ and $A\ne B$. The answer to this question and the answers to the earlier question linked from it show how to construct almost disjoint families of cardinality $2^\omega=\mathfrak{c}$, and a straightforward Zorn’s lemma argument shows that there is a maximal almost disjoint family of $\mathscr{M}$ of cardinality $\mathfrak{c}$.

Since $K$ has non-empty interior, $|K|=\mathfrak{c}$, and we may take $\varphi:K\to\mathscr{M}$ to be a bijection. Let $L\subseteq K$. If $|L|>1$, then $\bigcap_{x\in L}\varphi(x)$ is finite, so $\bigcap_{x\in L}\varphi(x)$ is infinite iff $L=\{x\}$ for some $x\in K$, in which case $\operatorname{int}\operatorname{cl}L=\varnothing$.

Brian M. Scott
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