Determine $\bigcap_{x \in A} \bigcup_{n \in \Bbb N} (\frac1n, x-\frac{1}{n^2})$ where $A=\{ x \in \Bbb R\mid x > \pi\}.$

Question

Determine $\bigcap_{x \in A} \bigcup_{n \in \Bbb N} (\frac1n, x-\frac{1}{n^2})$ where $A=\{ x \in \Bbb R\mid x > \pi\}.$

Isn’t $\bigcup_{n \in \Bbb N} (\frac1n, x-\frac{1}{n^2}) = (\frac1n, x-\frac{1}{n^2})$?

I’m familiar with that $\bigcup_{n \in \Bbb N}A_n= \{a \mid \exists n \in \Bbb N : a \in A_n \}$.

Now I would need to find $\bigcap_{x \in A} (\frac1n, x-\frac{1}{n^2})$ but this feels odd since I’m not iterating over $n$ anymore and would get something like $(\frac1n, \pi - \frac{1}{n^2})$?

This was a practice question for learning about $\limsup$ and $\liminf$.

Answer 1

$\bigcup_{n \in \Bbb N} (\frac1n, x-\frac{1}{n^2})$ cannot depend on $n$ so it cannot be equal to $ (\frac1n, x-\frac{1}{n^2})$

Verify that $\bigcup_n (\frac 1 n, x-\frac 1 {n^{2}})=(0,x)$ and $\bigcap_{x \in A} (0,x)=(0,\pi]$.