By the definition, Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement. Do the Borel sets must satisfy all the conditions of "countable union","countable intersection" and "relative complement" , or just one of them?
Generating conditions of Borel sets
borel-sets
Related Solutions
Re: your first question, often - and annoyingly - the topology is omitted so we speak of $\mathcal{B}(X)$ when we should be talking about $\mathcal{B}(X,\mathcal{T})$. In general though this is only done when there is a "natural" choice of topology involved - e.g. if $X=\mathbb{R}^n$, then we're using the Euclidean topology - and so this abuse of notation has survived.
As to your second question, as Henno says you can form discs via countable unions. Here's a construction which is reasonably simple and will help build intuition:
This isn't exactly what you ask, but I think it's a bit easier, and understanding it first will make the instance you're looking at much more intuitive.
The open unit disc centered at the origin, $U$, is the union of countably many closed rectangles.
(Here we're working in $\mathbb{R}^2$ with the usual topology.)
Proof: the initial idea is to put a small closed rectangle $R_p$ around each point $p$ in $U$. Of course that would lead to uncountably many closed rectangles, so we combine this with the fact that $\mathbb{R}^2$ has a countable dense set - namely, the set $\mathbb{Q}^2$ of points both of whose coordinates are rational. Of course, this means that it is nontrivial to show that $U$ is in fact covered ...
Specifically, we do the following:
For $(a,b)\in\mathbb{Q}^2\cap U$, let $\epsilon={\sqrt{a^2+b^2}\over 3}$. Note that the open ball of radius $\epsilon$ centered at $(a,b)$ is wholly contained in $U$ and contains $(a,b)$ as an element.
We now draw a square: let $R_{(a,b)}$ be the closed square centered at $(a,b)$ of side length $\epsilon\sqrt{2}$. This is contained wholly inside the open ball of radius $\epsilon$ centered at $(a,b)$.
Finally, we claim that $$U=\bigcup_{(a,b)\in\mathbb{Q}^2}R_{(a,b)}.$$ One inclusion is trivial. In the other direction, we want to show that an arbitrary point $(x,y)\in U$ is in $\bigcup_{(a,b)\in\mathbb{Q}^2}R_{(a,b)}$. If $(x,y)\in\mathbb{Q}^2$ this is of course obvious, but what if it's not?
The solution to this problem is to find a rational point $(a,b)\in U\cap\mathbb{Q}^2$ which is "sufficiently close" to $(x,y)$. This winds up being a bit of annoying "epsilon-juggling" ... so I'm going to leave it as an exercise because I'm lazy and suffering builds character.
The real takeaway from the construction in the previous section is that the collection of sets we can build from closed sets via countable intersections (these are variously called $F_\sigma$ sets or $\Sigma^0_2$ sets - I'll use the former notation since it's more common in topology, even though there's a very good reason to prefer the latter) is richer than just the collection of closed sets themselves. In particular, a quick extension of the above argument is that every open set is $F_\sigma$. Even more simply, the set $\mathbb{Q}^2$ itself is $F_\sigma$ since it's countable, and $\mathbb{Q}^2$ is neither open nor closed.
And of course, this is only (half of) the second "layer" of the Borel hierarchy. Practically any set you can describe is Borel.
On that note, the Borel hierarchy as a more "constructive" description of the collection of Borel sets is extremely useful for building intuition as well as proving theorems.
If you actually fully expand out, you'll realize you actually have an uncountable intersection, because the index set of your intersection is actually the set of all sequences $\mathbb{N} \to \mathbb{N}$. For an explicit example, $\mathbb{Q}$ is a countable union of countable intersections of open sets, but it is not a countable intersection of open sets itself. See the example section of https://en.wikipedia.org/wiki/G%CE%B4_set
Best Answer
Borel sigma is the smallest sigma algebra containing open sets and closed under all three operations.