Infinite intersection non-empty if finite intersections are non-empty proof

Question

If $ A_1 \cap A_2 \cap \dots \cap A_n \neq \emptyset $ for all $n \in \mathbb{N}$

Is it also true that $ A_1 \cap A_2 \cap \dots \neq \emptyset$ ?

I'm guessing no, could I have a hint?

Answer 1

No definitely not. Consider the sequence of sets $(0, 1/n) \subseteq \mathbb{R}$ for $n \in \mathbb{N}$. Then clearly $$\bigcap_{n \in \mathbb{N}}(0,1/n) = \varnothing$$ Since if $x \in \bigcap_{n \in \mathbb{N}}(0,1/n)$, we have $$0 < x < 1/n$$ for any $n \in \mathbb{N}$. Since limits preserve inequalities, we would have $$0 < x \leq \lim_{n \to \infty} 1/n = 0$$ which is impossible.

Answer 2

Try $A_n = \{k\in \Bbb N: k \ge n\}$

Answer 3

Consider the sets $$ A_n = \{x \ |\ 0n$. We have that $$ \begin{align} A_1\cap A_2\cap\cdots\cap A_n =& A_n\\ \neq&\emptyset \end{align} $$ but $$ \begin{align} A_1\cap A_2\cap\cdots =&\lim_{n\rightarrow\infty} A_n\\ =& \{x \ |\ 0

Answer 4

Here's another one: Let $A_n$ be the set of integers divisible by $n$. Then for any finite collection of sets of the form $A_k$, the intersection consists of integers divisible by every subscript -- in other words integers divisible by the least common multiple of the subscripts. So it's nonempty. But the intersection of all of the sets is empty, because no integers is divisible by everything.