How to prove $\displaystyle\bigcup^\infty_{k=1}(\bigcap^\infty_{n=1}A_{k,n})\subset\bigcap^\infty_{n=1}(\bigcup^\infty_{k=1}A_{k,n})$

Want to show

$$\displaystyle\bigcup^\infty_{k=1}\left(\bigcap^\infty_{n=1}A_{k,n}\right)\subset\bigcap^\infty_{n=1}\left(\bigcup^\infty_{k=1}A_{k,n}\right)$$

Note the bottoms are $k=1,n=1$ and $n=1,k=1$, rather than $k\geq n, n=1$; $n\geq k,k=1$.

I did it via the same way like $\limsup$ and $\liminf$, but failed. $\displaystyle\bigcap^\infty_{n=k}A_n$ is an increasing sequence of $k$, and $\displaystyle\bigcup^\infty_{n=k}A_n$ is a decreasing one. Yet when $n=1$, they are not true.

• Hey. What did you try so far? – Mankind Feb 4 '16 at 15:26
• You must start reducing set notation to logic notation and quantifiers. At least this is the common way. – Masacroso Feb 4 '16 at 15:36
• Thank u for the remark. I did it via the same way like $\limsup$ and $\liminf$, but failed. $\cap^\infty_{n=k}A_n$ is an increasing sequence of $k$, and $\cup^\infty_{n=k}A_n$ is a decreasing one, yet when $n=1$, they are not true. – Royun Feb 4 '16 at 15:49
• they look like $\liminf$ and $\limsup$, but they are not. – SiXUlm Feb 4 '16 at 19:34

Hint: Pick $x$ from LHS. $$\exists k_0:x\in A_{k_0,n}\;\forall\; n\implies x\in\cup_{k}A_{k,n}\;\forall n$$