Is $\bigcup\limits_{n=1}^{\infty}[a+\frac{1}{n},\infty)=\bigcup\limits_{n=1}^{\infty}(a+\frac{1}{n},\infty)$? 
Is $\bigcup\limits_{n=1}^{\infty}[a+\frac{1}{n},\infty)=\bigcup\limits_{n=1}^{\infty}(a+\frac{1}{n},\infty)$

There must be a difference, or not ? which is equal to $(a,\infty)$ ? RHS must be open, since countable union of open sets is open, but what is LHS then ?
 A: They are equal. There is no difference. Both sets are equal to $(a, \infty)$. The only point of contention would be $a$ and it does not belong to any of the two sets. 
Supose $x \in \bigcup\limits_{n=1}^{\infty}[a+\frac{1}{n},\infty)$. Then $x \in [a + \frac 1 m, \infty)$ for some $m \in \Bbb N$. 
As long as you can picture that $[a + \frac 1 m, \infty) \subseteq (a+\frac{1}{m + 1},\infty)$ ( EDIT: this is because $\frac{1}{m} \gt \frac{1}{m + 1} \implies x \ge a + \frac{1}{m} \gt a + \frac{1}{m + 1}$) we have that $x \in (a+\frac{1}{m + 1},\infty) $ and hence $x \in \bigcup\limits_{n=1}^{\infty}(a+\frac{1}{n},\infty)$
Therefore $$\bigcup\limits_{n=1}^{\infty}[a+\frac{1}{n},\infty) \subseteq \bigcup\limits_{n=1}^{\infty}(a+\frac{1}{n},\infty)$$
Similarly since $ (a+\frac{1}{m'},\infty) \subseteq  [a+\frac{1}{m'},\infty)$ for $m' \in \Bbb N$ it can be shown that 
$$\bigcup\limits_{n=1}^{\infty}[a+\frac{1}{n},\infty) \supseteq \bigcup\limits_{n=1}^{\infty}(a+\frac{1}{n},\infty)$$ 
whence the equality follows. 
A: Yes, they're the same
The LHS is open as well, and you can prove it by element membership.
It's clear that the LHS has to be a superset of the RHS (since it's a union of supersets of the RHS's sets). It remains to show that anything in the LHS is also in the RHS.
Take some $x\in \text{LHS}$. Then for some $k\in\mathbb{N}, x \geq a+\frac{1}{k}$. But then $x>a+\frac{1}{k+1}$, so then $x\in(a+\frac{1}{k+1},\infty)$, so then it's inside some set in the RHS! Then it must be in the union of all the sets in the RHS. That means that $\text{LHS}\subseteq \text{RHS}$.
