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I'm aware how to prove that the intersection of nested sequence of compact non-empty sets is compact and non-empty. but I want to generalize this question to transfer the hypothesis of having nested sequence of compact non-empty sets with having a collection of closed subsets which have finite intersection property. so, the new question is:

suppose that $M$ is covering compact and $\mathcal{C}$ is a collection of closed subset of $M$ such that every intersection of finitely many members of $\mathcal{C}$ is non-empty. Prove that the grand intersection $\bigcap _{A_n \in \mathcal{C}} A_n$ is non-empty.

In first case assume that the elements of nested sequence are $B_1 \supset B_2 \supset B_3 ... \supset B_n ...$ by compactness of each element of nested sequence we can say that the intersection is compact. then because each $B_n$ is non-empty, so for each $n \in \mathbb{N}$ we can choose a member $b_n \in B_n$ and make the sequence $(b_n)$. $(b_n) \in B_1$ Since the each $B_n$ is contained in $B_1$. Since $B_1$ is compact, $(b_n)$ has a convergent subsequence in $B_1$ named $ (b_{n_k})$ which converges to a point $p \in B_1$. the limit $p$ also lies in the set $B_2$ since except possibly for the first term, the subsequence $(b_{n_k})$ lies in $B_2$ and $B_2$ is a closed set. The same is true for each $B_n$, so $ p \in \bigcap{B_N}$ and $\bigcap B_n$ is non-empty.

But in my question, we can just say the $\bigcap A_n$ is compact since each $A_n$ is compcat. But what about being non-empty?

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  • $\begingroup$ Are you making any assumptions on your space? Hausdorffness, perhaps? $\endgroup$ Apr 26, 2015 at 8:34
  • $\begingroup$ @hard-boiledwonderland No, the question doesn't have any assumption on space but I would be happy even if I can prove it for a metric a space! $\endgroup$
    – F.K
    Apr 26, 2015 at 15:58

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For each $C\in\mathscr{C}$ let $U_C=M\setminus C$, and let $\mathscr{U}=\{U_C:C\in\mathscr{C}\}$. Suppose that $\bigcap\mathscr{C}=\varnothing$; then

$$\bigcup\mathscr{U}=\bigcup_{C\in\mathscr{C}}(M\setminus C)=M\setminus\bigcap\mathscr{C}=M\setminus\varnothing=M\;,$$

so $\mathscr{U}$ is an open cover of $M$. Now use the compactness of $M$ to get a contradiction to the hypothesis that $\mathscr{C}$ has the finite intersection property.

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  • $\begingroup$ I can't get the end point! I got $\mathscr{U}$ is an open cover of $M$. But I can't get your contradiction! $\endgroup$
    – F.K
    Apr 27, 2015 at 4:16
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    $\begingroup$ @F.K: $M$ is compact, so there is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ that covers $M$. Say $\mathscr{U}_0=\{U_{C_1},\ldots,U_{C_n}\}$; then $U_{C_1}\cup\ldots\cup U_{C_n}=M$. Now what can you say about $C_1\cap\ldots\cap C_n$? $\endgroup$ Apr 27, 2015 at 4:21
  • $\begingroup$ I'm not sure that this intersection is necessarily empty! $\endgroup$
    – F.K
    Apr 27, 2015 at 16:29
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    $\begingroup$ @F.K: It is: just use De Morgan’s laws. $\endgroup$ Apr 27, 2015 at 18:44
  • $\begingroup$ yeah I got it. Thank U so much :) $\endgroup$
    – F.K
    Apr 28, 2015 at 7:06

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