Understanding statement on algebras Statement: Let $\Omega = \mathbb R$ and $\mathscr F = \{(a,b):-\infty<a<b<\infty\}\cup\{\emptyset\}$. Then this is closed under finite intersections but this is not an algebra as it is not closed under countable unions.
I'm confused about this statement, I can see how this set is closed under intersections but I don't see how it is not closed under countable unions. 
Here is how I am trying to understand it, but not sure about my reasoning. For example, if we pick a,b to be natural numbers. Then $(0,1),(1,2),(1,3)...\in \mathscr F$ and to be closed under countable unions then we would need $(0,1)\cup(1,2)\cup(1,3)...= \mathbb R \setminus\mathbb N\in \mathscr F$ which clearly it is not as we cannot pick an interval (a,b) that skips over the natural numbers, for example (0,2) includes the natural number 1.
Also another question, the empty set is a subset of all sets, so why does this statement need to require the '$\cup\{\emptyset\}$' to satisfy being closed under finite unions?
 A: *

*The empty set is a subset of all sets, but it isn't an element of all sets. Let's write $$X = \{(a,b): -\infty < a< b < \infty\}$$ so that $\mathscr F = X\cup \{\emptyset\}$.  Then $\emptyset$ is not an element of $X$, because everything in $X$ is an interval of the form $(a,b)$ with $a<b$, and $\emptyset$ is not such an interval, so it is not in $X$.  But $X$ is not closed under intersections, because it includes $(1,2)$ and $(3,4)$, but not $(1,2)\cap (3,4) = \emptyset$.  So to make it closed we must at least include $\emptyset$.

*Your example for why $\mathscr F$ is not closed under countable unions is correct, but a simpler example is    $C = \{(0,n): n\in \Bbb N\}\subset \mathscr F$ and then $\bigcup C = (0,\infty)\notin \mathscr F$.
If you can use a finite countable union, then just observe that $\mathscr F$ includes $(0,1) $ and $(2,3)$ but does not include their union, since $(0,1)\cup (2,3)$ is neither an interval nor $\emptyset$.  If finite unions are not allowed (but I think they are) you can still rescue this example by taking $A= \left\{\left(0+\frac1n, 1-\frac1n\right) : n\in \Bbb N\right\}$ and
$B= \left\{\left(2+\frac1n, 3-\frac1n\right) : n\in \Bbb N\right\}$ and then $\bigcup A = (0,1)$ and $\bigcup B = (2,3)$ so $\bigcup (A\cup B) = (0,1)\cup (2,3)\notin \mathscr F$.
A: Take $I_n = (-\frac{1}{n},0), I'_n = (1,1+\frac{1}{n}) \to \cup I_n \cup \cup I'_n$ is not an interval.
