# prove the set of all finite unions of all intervals in [0,1] is an algebra but not a sigma-algebra

let’s say

\$mathbf{A}=\${all intervals contained in [0,1]} and

\$mathbf{B}\$={all finite unions of elements of \$mathbf{A}\$}

proving that \$mathbf{B}\$ is an algebra is mostly straightforward, except i’m not sure how to prove that \$emptyset subset mathbf{B}\$. And for that matter, how do I even prove that \$emptyset subset mathbf{A}\$? Can I just say the ‘interval’ for example \$[frac{1}{2},frac{1}{2}]\$ it not actually an interval is therefore \$emptyset\$ ?

To prove that \$mathbf{B}\$ is not a sigma-algebra, do I need to construct a countable union or intersection of subsets of \$mathbf{B}\$ and show that it’s not contained in \$mathbf{B}\$? Can I say that \$bigcup_{j in mathbf{N}} [frac{1}{2j+1},frac{1}{2j}] notin mathbf{B}\$ (where \$mathbf{N}\$ is the set of natural numbers) and thus \$mathbf{B}\$ is not closed under countable union and therefore cannot be a sigma-algebra?

All topic