When is $\cap_{i=1}^{\infty} A_i$ non-empty

$A_1\supset A_2\supset\cdots A_n\supset A_{n+1}\supset\cdots$ be an infinite sequence of non-empty subsets of $\mathbb R^3$.Which one of the following ensures that their intersection $\cap_{i=1}^{\infty}A_i$ is non-empty $?$

$A.$ Each $A_i$ uncountable.

$B.$ Each $A_i$ open.

$C.$ Each $A_i$ connected.

$D.$ Each $A_i$ compact.

If I take $$A_n=\left(0,{1\over n}\right)^3$$ then it is both open and connected and their arbitrary intersection being $\emptyset$ it contradicts both $A.$ and $B.$ Also, If all $A_i$ are compact then they satisfy finite intersection property and the arbitrary intersection is non-empty as well.

Only I don't know what happens when all $A_i$ are connected . Help me with that. Thanks.

• Isn't $(0,1/n)^3$ connected? Do you really mean you aren't sure about $D$? – Thomas Andrews Jan 27 '16 at 18:35
• @ThomasAndrews : Right . So, $D$ is the only option left. – user80631 Jan 27 '16 at 18:37
• What definition of compactness are you using? – Thomas Andrews Jan 27 '16 at 18:37
• (0,1/n)^3 eliminates A,B,C so only D is left but it could be that none of them ensure non-emptiness. Is it possible for Ai to be compact and their intersections empty? If not, why not? – fleablood Jan 27 '16 at 18:42

Hint for (D): Given a descending sequence $A_n$ of compact subsets of $\mathbb R^3$, let $U_n=\mathbb R^3\setminus A_n$. If $\bigcap_{i} A_i=\emptyset$, then $\bigcup U_i=\mathbb R^{3}$, so it covers, in particular $A_1$.
But that means that $\bigcup_{i=1}^{n} U_i$ must cover $A_1$, for some $n$. Is that possible?
The choose $a_i\in A_i$. This sequence has to have a limit point $a\in A_1$, in particular. Then use that the $A_i$ are closed to show that $a\in A_i$ for all $i$.
Note that just closed doesn't work - you can take $A_i=[i,+\infty)^3$ to find a sequence of close spaces.