Set of All Real Intervals is Semiring of Sets

Theorem
Let $\mathbb S$ be the set of all real intervals.

Then $\mathbb S$ is a semiring of sets, but is not a ring of sets.

Proof
Consider the types of real interval that exist.

In the following, $a, b \in \R$ are real numbers.

There are:
 * Closed intervals:
 * $\left [{a \, . \, . \, b \,} \right] = \left\{{x \in \R: a \le x \le b}\right\}$


 * Open intervals:
 * $\left ({a \, . \, . \, b} \right) = \left\{{x \in \R: a < x < b}\right\}$


 * Half-open intervals:
 * $\left [{a \, . \, . \, b} \right) = \left\{{x \in \R: a \le x < b}\right\}$
 * $\left ({a \, . \, . \, b \,} \right] = \left\{{x \in \R: a < x \le b}\right\}$


 * Unbounded half-open and unbounded open intervals:
 * $\left [{a \, . \, . \, \infty} \right) = \left\{{x \in \R: a \le x}\right\}$
 * $\left ({-\infty \, . \, . \, a\,} \right] = \left\{{x \in \R: x \le a}\right\}$
 * $\left ({a \, . \, . \, \infty} \right) = \left\{{x \in \R: a < x}\right\}$
 * $\left ({-\infty \, . \, . \, a} \right) = \left\{{x \in \R: x < a}\right\}$
 * $\left ({-\infty \, . \, . \, \infty} \right) = \left\{{x \in \R}\right\}$


 * The empty interval:
 * $\left ({a \, . \, . \, a} \right) = \left\{{x \in \R: a < x < a}\right\} = \varnothing$


 * Singleton intervals:
 * $\left [{a \, . \, . \, a \,} \right] = \left\{{x \in \R: a \le x \le a}\right\} = \left\{{a}\right\}$

Set of Intervals is Not a Ring
Consider, for example, a open interval:
 * $\left ({a \, . \, . \, b} \right) = \left\{{x \in \R: a < x < b}\right\}$

such that $a < b$.

Any subset $\left ({c \, . \, . \, d} \right) \subset \left ({a \, . \, . \, b} \right)$ such that $a < c, d < b$ is such that:
 * $\left ({a \, . \, . \, b} \right) - \left ({c \, . \, . \, d} \right) = \left ({a \, . \, . \, c \,} \right] \cup \left [{d \, . \, . \, b} \right)$

But $\left ({a \, . \, . \, c \,} \right] \cup \left [{d \, . \, . \, b} \right)$ is not in itself a real interval, and therefore is not an element of $\mathbb S$.

Hence $\mathbb S$ is not closed under the operation of set difference and is therefore not a ring of sets.

Set of Intervals is a Semiring
It is tedious but straightforward to examine each type of interval, and pass it through the same sort of exhaustive examination as follows.

We will take a general half-open interval:
 * $A = \left [{a \, . \, . \, b} \right)$

and note that the argument generalizes.

Let $c, d \in \R: a \le c < d \le b$.

Then $C = \left ({c \, . \, . \, d} \right)$ is a subset of $\left [{a \,. \, . \, b} \right)$.

There are four cases:


 * $a < c, d < b$:

Then:
 * $A = \left [{a \, . \, . \, c \,} \right] \cup \left ({c \, . \, . \, d} \right) \cup \left [{d \, . \, . \, b} \right)$

and it can be seen that this is a partition of $A$.


 * $a = c, d < b$:

Then:
 * $A = \left\{{a}\right\} \cup \left ({a \, . \, . \, d} \right) \cup \left [{d \, . \, . \, b} \right)$

and it can be seen that this is a partition of $A$.


 * $a < c, d = b$:

Then:

and it can be seen that this is a partition of $A$.


 * $a = c, d = b$:

Then:

and it can be seen that this is a partition of $A$.

The same technique can be used to generate a finite expansion of any interval of $\R$ from any subset of that interval.

Hence $\mathbb S$ is a semiring of sets.