Quick question about the notation $\bigcup_{n \in\mathbb{N}} \cdots$

Question

My textbook is using the notation $$ \bigcup_{n \in \mathbb{N}} \left [ 2 - \frac{1}{n}, 4+\frac{1}{n} \right] $$ Does the comma here mean that the stuff inside the brackets represents a set? The set below is equivalent to the one above, right? $$ \left\{ 2-\frac{1}{1}, 4+\frac{1}{1} \right\} \cup \left\{ 2-\frac{1}{2}, 4+\frac{1}{2}, \right\} \cup \dots \cup\left\{ 2-\frac{1}{n}, 4+\frac{1}{n} \right\} \cup \dots $$

Answer 1

$$\bigcup_{n \in \mathbb{N}} \left [ 2 - \frac{1}{n}, 4+\frac{1}{n} \right] = \left\{x: \exists n \in \mathbb{N}, x \in \left [ 2 - \frac{1}{n}, 4+\frac{1}{n} \right]\right\}=\\= \left[ 2-\frac{1}{1}, 4+\frac{1}{1} \right] \cup \left[ 2-\frac{1}{2}, 4+\frac{1}{2}, \right] \cup \dots \cup\left[ 2-\frac{1}{n}, 4+\frac{1}{n} \right] \cup \dots=[1,5]$$

Addition (may be you find this useful): $$\bigcap_{n \in \mathbb{N}} \left [ 2 - \frac{1}{n}, 4+\frac{1}{n} \right] = \left\{x: \forall n \in \mathbb{N}, x \in \left [ 2 - \frac{1}{n}, 4+\frac{1}{n} \right]\right\}=\\= \left[ 2-\frac{1}{1}, 4+\frac{1}{1} \right] \cap \left[ 2-\frac{1}{2}, 4+\frac{1}{2}, \right] \cap \dots \cap\left[ 2-\frac{1}{n}, 4+\frac{1}{n} \right] \cap \dots=[2,4]$$

Answer 2

$[a,b] \subset \mathbb R$ is the set of all real numbers $x$ such that $a \le x \le b$. Can you use that to calculate the object your answer writes about? By the way, $(a,b)$ is the same but with $<$ instead of $\le$.