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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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    If there are no further conditions on $C$, we are in trouble. Let $C$ be the set of all $1/n$, where $n$ ranges over the natural numbers, and for any natural number $k$, let $Q_k=[0,1/k]$.2011-09-05
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    @André: thank you by the observation. I'll to impose some condition in $C$.2011-09-05
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    @leo: joriki's answer was already including the perfect case (the irrationals are a perfect space), and I have edited mine to include the perfect case.2011-09-05
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    Thanks, by your answers, what do you think with the edit in mind?2011-09-05
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    ahhh, ok, thanks @Asaf.2011-09-05
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    Closed is easily enough for the second wish. And closed (or perfect) cannot possibly be enough for $Q\subseteq C$.2011-09-05
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    @André: I think I get it. Thanks2011-09-05
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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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    Your answer is much better than mine, and can be engineered such that the resulting $Q$ will be any countable compact subspace of $\mathbb Q$ (and I don't mean that in an artificial way, but rather at no point no $Q_k$ has isolated points...)2011-09-05
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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.