Question: Let $U$ denote a subset of $\mathbb{R}$. We say that $U$ is an open set when for every $x \in U$ there is an $\epsilon > 0$ for which $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is contained in $U$. Show:
(i) For $a,b \in \mathbb{R}$ the interval $(a,b)$ is open, but the interval $[a,b]$ is not.
(ii) If $U,W \subset \mathbb{R}$ are open, so are $U \cup W$ and $U \cap W$.
(iii) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cup_{i \in I}U_{i}$ is also open.
(iv) If $(U_{i})_{i \in I}$ is a set of open subsets from $\mathbb{R}$, $\cap_{i \in I}U_{i}$ is not necessarily open.
My attempt: I think that with (i) i can conceptually understand everything:
$\forall x \in (a,b): a< x - \frac{|a-x|}{2} < x < x + \frac{|b-x|}{2} < b$
If $x> \frac{|b-a|}{2}$ let $\epsilon = \frac{|b-x|}{2}$
If $x< \frac{|b-a|}{2}$ let $\epsilon = \frac{|a-x|}{2}$
$\Rightarrow \forall x \in (a,b) \exists \epsilon > 0 : U_{\epsilon}(x) = (x - \epsilon, x +\epsilon) \Rightarrow$ the interval (a,b) is open.
and
max$[a,b] = b \Rightarrow \forall x \in [a,b]: b \geq x$
If $\epsilon > 0, b \ngeq b+ \epsilon \Rightarrow$ there exists an $x$ for which $U_{\epsilon}(x) = (x - \epsilon, x +\epsilon)$ is not contained in $U \Rightarrow$ the interval is not open
With (ii) and (iii) i think i understand, but i am not sure how to start stating that they are true mathematically. to be sure, $\cup_{i \in I}U_{i}$ denotes several unions in a row, correct? i'm even reading a proof which goes:
Given $v \in U \cap W$, there exists an open ball $B$ of radius $r$ centered at $v$ contained in $U$, and there exists an open ball $B'$ of radius $r'$ centered at $v$ contained in $W$. Let $\delta =$ min$(r, r')$. Then the open ball of radius $\delta$ centered at $v$ is contained in $U \cap W$, which is therefore open.
Unfortunately i'm having trouble interpreting this and am wondering if there is another way to put it or approach it.
With (iv) i am not even sure intuitively why it should be true. i can't think of a situation where it wouldn't be true, i guess it has to do with whether the sets are finite or not, but i don't see how this would make a difference (especially given that openness is apparently preserved with infinite unions).
If anyone can help provide suggestions, corrections, hints, etc i appreciate it, thanks!