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.
In other words, it would require every set of reals to be a Borel set. Since this is certainly not the case (at least under the standard assumptions of the axiom of choice), it follows that the Borel algebra is not a complete lattice.