I am reading Christopher Heil's Introduction to Real Analysis and there I found the following definition of the Borel $\sigma$-algebra. First denote $\mathcal{E}$ as a collection of subsets of $X$. Then the $\sigma$-algebra generated by $\mathcal{E}$, $\Sigma(\mathcal{E})$, is the smallest one containing $\mathcal{E}$. That is $$ \Sigma(\mathcal{E}) = \cap \left\{{\Sigma : \textrm{$\Sigma$ is a $\sigma$-algebra on $X$ and $\mathcal{E} \subseteq \Sigma$}}\right\} $$ Now let $\mathcal{U} = \left\{{U \subseteq X : \textrm{$U$ is open}}\right\}$ be the collection of all open subsets of $X$. Then the Borel $\sigma$-algebra on $X$ is $\Sigma(\mathcal{U})$. Moreover, the Borel sets are the members of $\Sigma(\mathcal{U})$.

On the other hand, the Wikipedia page of Borel sets says that Borel sets are the ones that can be written as countable combination of unions, intersections and complements of open sets, and that the Borel $\sigma$-algebra is the $\sigma$-algebra containing the Borel sets.

My question is the following: How can we conclude that the first definition implies that any member of $\Sigma(\mathcal{U})$ can be written as countable combination of unions, intersections and complements of open sets?

EDIT: From my understanding we can state something like this. Define the following four functions acting on $\mathcal{P}(X) \times \mathcal{P}(X) \rightarrow \mathcal{P}(X)$: $$ \begin{align} s(A,B) &= A \cup B \\ d(A,B) &= A \cap B \\ c(A,B) &= X - A \\ i(A,B) &= A \end{align} $$ Then for any Borel set $B$ there exists a sequence of open sets $\left\{{U_k}\right\}_{k \in \mathbb{N}}, U_k \in \mathcal{U}$ and a sequence of functions $\left\{{f_k}\right\}_{k \in \mathbb{N}}, f_k \in \left\{{s,d,c,i}\right\}$ such that (if that even makes sense) $$ B = \cdots f_3(f_2(f_1(U_1,U_2),U_3),U_4) \cdots $$ My question is how the intersection definition of the Borel $\sigma$-algebra implies the aforementioned representation of Borel sets.

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    $\begingroup$ note that the wiki page says that Borel sets "can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement". Thus, arbitrary combinations of these operations are allowed $\endgroup$
    – hgmath
    Mar 7, 2021 at 16:46
  • $\begingroup$ You can need infinitely many intersection and union steps to get to some given Borel set. The Wikipedia page just gives some examples of sets that are Borel. $\endgroup$ Mar 7, 2021 at 17:15
  • $\begingroup$ You also need countable operations which do not fit that framework, plus a countable union of sets from different “levels” to pass limit stages. Even that more liberal description after the edit is not enough. Read up on Borel hierarchy. It probably has a Wikipedia page too. There are all sorts of subclasses named $\Pi^0_\alpha$ and $\Sigma^0_\alpha$ for ordinal numbers less than $\omega_1$. $\endgroup$ Mar 7, 2021 at 22:08
  • $\begingroup$ The external definition vis $\sigma(U)$ is often easier to work with than the internal one, building up from open sets and sequences of operations with transfinite recursion. They yield the same sets at the end though. Topology prefers internal and measure theory external, in short. $\endgroup$ Mar 7, 2021 at 22:11
  • $\begingroup$ How can you assert that both constructions yield the same set? that's basically my question. $\endgroup$ Mar 8, 2021 at 0:25

2 Answers 2


No, there are many more Borel sets than those. The Borel sets contain (by definition ) all countable intersections of open sets, all complements of open sets ( ie all closed sets), so all countable unions of closed sets too. These are just the first steps in the Borel hierarchy which has $\omega_1$ levels for separable complete metric spaces like $\Bbb R$. It’s not an exhaustive list of the Borel sets and how they can be built up from open sets.

  • $\begingroup$ I have to wonder about the rationale for modern abstract treatments to basically only talk about open/closed sets, Borel sets, and measurable sets, thereby omitting all the various results in analysis which say that such and such set is $F_{\sigma}$ or $G_{\delta \sigma},$ etc. (only two Borel levels suffice for virtually everything that ever comes up) which are both more precise and tend to prevent this sort of confusion as to what Borel sets are. It's as if we're doing combinatorics and the answers are either none, one, or many! $\endgroup$ Mar 7, 2021 at 17:39
  • $\begingroup$ @DaveL.Renfro there are some recent results about absolute $F_{\sigma \delta}$ sets in $C_p(X)$ spaces etc but it’s getting rarer to have such results indeed. It was mostly an early 20th century thing to care about exact Borel classes of things. It does get rather technical quite quickly. $\endgroup$ Mar 7, 2021 at 18:30
  • $\begingroup$ Can you check my edit please? $\endgroup$ Mar 7, 2021 at 19:18

Consider $W = \{S \in \Sigma(U) : S $ can be formed from countable unions/intersections/complements of open sets$\}$. It can immediately be verified that since $\Sigma(U)$ is a $\sigma$-algebra, so too is $W$. And clearly $W$ contains all open sets. Then $\Sigma(U) \subseteq W$.

  • $\begingroup$ It is not so clear to me why $W$ is a $\sigma$-algebra, would mind elaborating a bit further, please? $\endgroup$ Mar 7, 2021 at 17:05
  • $\begingroup$ If $S$ can be formed from countable unions/intersections/complements of open sets, so too can its complement. If $S_1, S_2, ..., $ can be formed from countable unions, intersections, complements of open sets, then their countable union also can be. Finally, the empty set is open. $\endgroup$ Mar 7, 2021 at 17:32

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