Supremum, max, min, inf of $S=\bigcup_{n \in \mathbb N}\left[\frac{1}{2^n},8-\frac{1}{n}\right)$

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$S=\bigcup_{n \in \mathbb N}\left[\frac{1}{2^n},8-\frac{1}{n}\right)$

I wrote out the first few intervals and see that this converges to the open interval $(0,8)$. So given that I'm trying to find the union of these intervals, I believe the union of intervals is $\left[\frac{1}{2},7\right]$

If this is true, then

$\text{infS}=\frac{1}{2}$

$\text{minS}=\frac{1}{2}$

$\text{supS}=7$

$\text{maxS}=7$

Did I got about this the correct way?