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Let $C\subset \mathbb{R}$, $C$ infinite. Suppose that there exist a family of compact sets $\{Q_k\}_{k\in \mathbb{N}}$ such that:

  • $Q_{k+1}\subseteq Q_k$

  • $Q_k\cap C$ is infinite $\forall k.$

By the nested segments intervals theorem, I know that $Q:=\bigcap_{k\in \mathbb{N}}Q_k\neq \emptyset$. I want to know if also is true that $Q\subseteq C$ or at least $Q\cap C\neq \emptyset$. Thanks by your help.

Edit: Assume that $C$ is a perfect set.

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    There are already plenty of answers! If $C$ is a closed subset of $\mathbb{R}$ then the $C\cap Q_k$ are compact, so their intersection is non-empty. As to $Q\subseteq C$, it was mentioned by joriki how easy it is to force this not to hold.2011-09-05

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There's no reason to expect $Q\subseteq C$, since you've made no assumption about the parts of the $Q_k$ outside $C$; for instance, $C=[0,1]$ and $Q=Q_k=[0,2]$ is compatible with your assumptions.

$Q\cap C\neq\emptyset$ also does not follow. Take $C=\mathbb R\setminus\mathbb Q$ and $Q_k=[-\frac1k,\frac1k]$. Then your assumptions are fulfilled, but the only point in $Q$ is $0\notin C$.

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    @Asaf: Thanks :-). I thought they both worked equally well before perfectness got into it; it was mostly coincidence that my answer happened to cover that as well...2011-09-05
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It is not true in general.

Consider $Q_k = [0,\frac{1}{k}]$, and $C=(0,1)$ - the open interval which is indeed perfect.

Each $Q_k$ meets $C$ infinitely often, but the intersection is just $\{0\}$ which is not in $C$ at all. You can take $Q_k = [-1,\frac{1}{k}]$ to have an infinite intersection if you prefer.