If we define
$$\mathcal{J} = \{\text{ all intervals contained in [0,1]}\} $$ Then $$ B_0 = \{ \text{ all finite unions of elements of }\mathcal{J}\} $$ Is an algebra (so it is closed under the formation of complements (but it is not a $\sigma$ algebra) . Now consider $$ B_1 = \{\text{ all finite or countable unions of elements of }\mathcal{J}\} $$ and let $K$ be the Cantor set (everything that is left over after removing all the middle thirds over the interval $[0,1]$. Then the complement of the Cantor set is in $B_1$. That is, $$ K^c \in B_1 $$ which I believe is because the cantor set $K$ can be written as the countable intersection of closed intervals, and therefore its complement $K^c$ can be written as a countable union of open intervals. By definition, then, $$ K^c \in B_1 $$ However, $$ K \not\in B_1 $$ Since the complement of something in $B_1$ is not in $B_1$, then $B_1$ is not an algebra, even though $B_0$ is? I don't see how allowing the countable union of intervals would ruin closure under complements? Can someone explain this to me or point me in the right direction? Or let me know if there is something else going on that I am missing?
Thanks.
You're absolutely right (assuming that, in the definition of $B_0$, "interval" means "open or closed interval") - $B_0$ is an algebra, but $B_1$ is not, even though $B_1$ is gotten from $B_0$ by adding a closure property. What's going on, basically, is that the complement of an infinite union is more complicated to describe than you might suspect based on just looking at finite unions. In particular, the complement of a finite union of intervals is again a finite union of intervals, so closure under complementation is in $B_0$ "for free," but this fails badly for countable unions.
For a simpler example of how adding closure in one direction can weaken closure in another, consider the following: let $C_0$ be the set of all subsets of $[0, 1]$ which are either finite or cofinite (=have finite complement). Then clearly $C_0$ is an algebra.
Now let $C_1$ be $C_0$, closed off under countable unions. Now $C_1$ is not an algebra, since e.g. the set of irrationals is not in $C_1$ but the set of rationals is.