0
$\begingroup$

Suppose $\mathfrak A$ is a set of closed subsets of $\Bbb R$ such that $∩_{A∈\mathfrak A}A = ∅$. Prove that if $\mathfrak A$ contains a bounded set then $\exists M \in \Bbb Z^+$ s.t $A_1 \cap \cdots A_M=∅$.

I am getting it how to solve I know that intersection of closed sets is closed. Now if it is not true then for any $m \in \Bbb N$ we have $\cap_{i=1}^n A_i\neq ∅$. Now for some $k \in \Bbb N$ if the intersection $\cap_{i=1}^k A_i$ is finite then we will go on increasing $i$ and at least an element has to be in the countably many sets $A_i$. But how can I bring uncountability here?

If this is infinite then I think that we have to use that closed and bounded set which is compact and use some limit property; but how?

Please help...

$\endgroup$
  • 1
    $\begingroup$ $\phi$ is phi and $\emptyset$ is emptyset and never the twain shall meet. $\endgroup$ – Umberto P. Sep 20 '18 at 23:39
  • $\begingroup$ sorry I will correct it. $\endgroup$ – user545623 Sep 20 '18 at 23:59
2
$\begingroup$

Let $B$ be a bounded subset of $\mathcal U$. Since $\cap_{\{A\in \mathcal U\}} A=\emptyset$ wehave $\cup_{\{A\in \mathcal U\}} A^{c} =\mathbb R$. Hence the compact set $B$ is covered by the open sets $A^{c},A \in \mathcal U$. There is a finite subcover, say, $B \in \cup_{i=1}^{n} A_i^{c}$. We then get $B\cap \cap_{i=1}^{n} A_i =\emptyset$. (We can take $M=n+1$).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy