# Is the borel $\sigma$-algebra over reals a complete lattice?

I've read in some forum answer that "the Borel sigma-algebra on the real numbers $$\mathscr{B}(\mathbb{R})$$ is not a complete lattice" and I was wondering why and hope you can help.

Def. a complete lattice is a partially ordered set in which all subsets have both a supremum (join) and an infimum (meet).

Def. The Borel $$\sigma$$-algebra on the reals is the smallest $$\sigma$$-algebra that contains all the open sets

One way for the above statement to hold would be if $$[-\infty,\infty] \notin \mathscr{B}(\mathbb{R})$$ -is this the case and why?
I would have guessed that $$[-\infty,\infty] \in \mathscr{B}(\mathbb{R})$$; because $$\mathscr{B}(\mathbb{R})$$ is closed under countable union and all open and closed sets exist in it, thus, $$(0,1),[1,2]\in \mathscr{B}(\mathbb{R})\Rightarrow [-\infty,1),[1,\infty]\in \mathscr{B}(\mathbb{R}) \Rightarrow [-\infty,\infty]\in \mathscr{B}(\mathbb{R})$$.

PS: I hope the tags are correct.

Note that every singleton is a Borel set in the case of $\Bbb R$. So for the Borel sets to form a complete lattice, any collection of singletons must have a join.
• There are $\mathfrak{c}$ many Borel sets (does this depend on choice?) and $2^\mathfrak{c}$ many subsets of $\mathbb{R}$ so counting gives us non-Borel sets already. Nov 26 '17 at 12:32
• @Henno: If $\Bbb R$ is a countable union of countable sets, then every set is Borel. Nov 26 '17 at 12:54
• So this could be formulated as: Every singleton set of $\mathbb{R}$ exists in $\mathscr{B}(\mathbb{R})$. In order for $\mathscr{B}(\mathbb{R})$ to form a complete lattice, any collection of singletons must have a join. This is not true for a set containing all but one of the singletons, e.g. $\{\{r\} \mid r \in \mathbb{R}\setminus \{0\}\} \notin \mathscr{B}(\mathbb{R})$. Nov 30 '17 at 13:56