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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Nov 26 '17 at 12:32
  • $\begingroup$ @Henno: If $\Bbb R$ is a countable union of countable sets, then every set is Borel. $\endgroup$ – Asaf Karagila Nov 26 '17 at 12:54
  • $\begingroup$ You write "In other words, it would require every set of reals to be a Borel set". That seems an oversimplification. For a sublattice to be a complete lattice, it is not necessary that it is a complete sublattice. (in the sense that the infinite joins must all be the same) E.g. the closed sets do form a complete lattice. $\endgroup$ – Niels J. Diepeveen Nov 26 '17 at 17:18
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    $\begingroup$ @Niels: Yeah, okay, I see what you mean there. But if a set has a smallest Borel superset, then it has to be equal to that set. Proof: simply remove one point. $\endgroup$ – Asaf Karagila Nov 26 '17 at 17:23
  • $\begingroup$ 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. $\{\{r\} \mid r \in \mathbb{R}\setminus \{0\}\} \notin \mathscr{B}(\mathbb{R})$. $\endgroup$ – Kirkeby Nov 30 '17 at 13:56

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