Q:

How can an open set be equal to a union of half-open sets?

In discussing $\sigma$-algebras I have seen the following used in proofs:
$$(0,1)=\bigcup_{i=1}^\infty \left[ \frac{1}{n} , 1 \right)$$
In other words the open set is equal to a denumerable union of half-open/half-closed sets. The missing link is supposed to be the fact that $\sigma$-algebras are closed under countable unions. But what is the link? How can the equal sign be used since $\frac{1}{n}$ will always be greater than $0$?

A:

The fact that $\bigcup_{n=2}^\infty [\frac{1}{n},1) = (0,1)$ doesn't have anything to do with $\sigma$-algebras. 
The equality follows from the fact that if $0<x<1$ then there is some $n$ so that $0<\frac{1}{n}<x<1$, which gives that $(0,1)\subseteq \bigcup_{n=2}^\infty [\frac{1}{n},1)$. And, further that $[\frac{1}{n},1)\subseteq (0,1)$ for each $n$, and so $\bigcup_{n=2}^\infty [\frac{1}{n},1) \subseteq (0,1)$.

A:

Firstly, neither set is a closed set, so the question in your title is not the same as the question in the body.
To understand how $(0,1) = \cup_n [\frac 1n, 1)$, you might show that these two sets contain eachother. It is clear that the right is in the left. Let us show that the left is in the right.
Choose some number $\delta \in (0,1)$. Then since there is an $n$ such that $\frac{1}{n} < \delta$, we see that $\delta \in$ that $[\frac 1n, 1)$. This is true for every number on the left, and so the left set is contained in the right. Thus they are equal.
To answer the question in your title, both the empty set and the entire set are both open and closed. And depending on your topology, many other sets as well.

