Prove that if $\mathfrak A$ contains a bounded set then $\exists M \in \Bbb Z^+$ s.t $A_1 \cap \cdots A_M=∅$

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?

• $\phi$ is phi and $\emptyset$ is emptyset and never the twain shall meet. – Umberto P. Sep 20 '18 at 23:39
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 $$M=n+1$$).