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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.

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  • $\begingroup$ Hey. What did you try so far? $\endgroup$ – Mankind Feb 4 '16 at 15:26
  • $\begingroup$ You must start reducing set notation to logic notation and quantifiers. At least this is the common way. $\endgroup$ – Masacroso Feb 4 '16 at 15:36
  • $\begingroup$ 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. $\endgroup$ – Royun Feb 4 '16 at 15:49
  • $\begingroup$ they look like $\liminf$ and $\limsup$, but they are not. $\endgroup$ – SiXUlm Feb 4 '16 at 19:34
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Hint: Pick $x$ from LHS. $$\exists k_0:x\in A_{k_0,n}\;\forall\; n\implies x\in\cup_{k}A_{k,n}\;\forall n$$

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