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.