Which countable set satisfies $\cup_{n=0}^{\infty}(\cap_{k=n}^{\infty} M_k)\neq \cap_{n=0}^{\infty}(\cup_{k=n}^{\infty} M_k)$?

Question

Find a collection $\{M_n\}_{n=0}^{\infty}$, where $M_n$ is countable for every $n$ and such that $\bigcup_{n=0}^{\infty}(\bigcap_{k=n}^{\infty} M_k)\neq \bigcap_{n=0}^{\infty}(\bigcup_{k=n}^{\infty} M_k)$.

I'm not getting idea of what set satisfies the above relation.

Please give suggestions...

Thank you!!

Answer 1

Consider $M_{2n+1} = \{0\}$ and $M_{2n}=\{1\}$.

Then $$\cup_{n=0}^{\infty}(\cap_{k=n}^{\infty} M_k) = \{\}$$ and $$ \cap_{n=0}^{\infty}(\cup_{k=n}^{\infty} M_k) = \{0,1\}.$$

Answer 2

Hint:

An element $x$ is in $\bigcup_{n=0}^\infty \left(\bigcap_{k=n}^\infty M_k\right)$ if there is some $n\in \mathbb{N}$ such that for every $k\geq n$, $x\in M_k$.

An element $x$ is in $\bigcap_{n=0}^\infty \left(\bigcup_{k=n}^\infty M_k\right)$ if for every $n\in \mathbb{N}$, there is some $k\geq n$ such that $x\in M_k$.

So to make these sets different, we just need to find a family of sets $(M_n)_{n=0}^\infty$ and an element $x$ such that $x$ is in infinitely many of the sets $M_n$ (then $x$ is in the second set), but there is no $n$ such that $x$ is in $M_k$ for all $k\geq n$ (then $x$ is not in the first set).

Mouse over below to reveal a possible solution.

Define $$M_n = \begin{cases} \{0\}& \text{if $n$ is even}\\ \varnothing&\text{if $n$ is odd}\end{cases}$$ Then $\bigcup_{n=0}^\infty \left(\bigcap_{k=n}^\infty M_k\right) = \varnothing$, but $\bigcap_{n=0}^\infty \left(\bigcup_{k=n}^\infty M_k\right) = \{0\}$.