# Why are those the Borel sets?

I have read several definitions and examples, but I still don't understand the Borel-sets clearly. I found the following: Let $$(X,\mathcal{T})$$ a topological space. We say $$\sigma(\mathcal{T})$$ generated $$\sigma$$-algebra is the Borel-sets of $$X$$, and we denote it as: $$\mathcal{B}(X)$$.

First, I don't understand how should I know what is the $$\mathcal{T}$$ , when I read $$\mathcal{B}(X)$$ somewhere?

Second, I don't understand for example in $$\mathbb{R}^{n}$$ how can the $$\mathcal{B}(\mathbb{R}^{n})=\sigma([a_{i},b_{i}):a_{i}\leq b_{i} , i=1,2,...n)$$ be the Borel sets? If I have a disk somewhere in $$\mathbb{R}^{2}$$, how can it be generated by rectangels? Or a disk is not a Borel set? If its not a Borel set, then why?

Another definition is the following (from wikipedia): In mathematics, a 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.

In this case, why can't I form a disk through the operations of countable union, countable intersection, and relative complement? If I use other disks it wouldn't be a problem at all.

I know I'm looking forward the smallest $$\sigma$$-algebra which contains $$\mathcal{T}$$, but how do I know what is $$\mathcal{T}$$ for example in the case of $$\mathcal{B}(\mathbb{R}^{n})$$? If they say $$\mathcal{T}$$ is the set of rectangels in $$\mathbb{R}^{2}$$ then it would be all clear. However they not mention it. How should I know it if they don't mention such things!?

I hope I was clear. I don't want to write nonsenses, so sorry for that if I did. I just want to clarify this question once and for all. Thank you in advance.

• Yes, the topology $\tau$ on $\mathbb{R}^n$ is usually the Euclidean topology (unless stated otherwise), The rectangles are not open sets. However, the half-open intervals generate the same $\sigma$-algebra as the open sets and are usually an easier generator to wok with. Also, you can form a disk out of half-open intervals using these operations, though an explicit description of how to do so would be rather tedious. Dec 29, 2019 at 14:58
• For instance, you have that $$\prod_{i=1}^n (a_i,b_i)=\bigcup\left\{\prod_{i=1}^n [q_i,b_i)\,:\, \forall i, q_i\in \Bbb Q\cap (a_i,b_i)\right\}$$ The "cubic" open balls are therefore in the generated topology. They form a basis for the euclidean topology, which is second countable: thus all open sets are countable unions of such "balls".
– user239203
Dec 29, 2019 at 15:01
• Because you have countable unions you can make disks from rectangles. It's a limiting process.. Dec 29, 2019 at 15:08

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.