# 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 closed 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, an 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:

$\left ({a \,.\,.\, b} \right) \setminus \left ({c \,.\,.\, d} \right) = \left ({a \,.\,.\, c} \right] \cup \left [{d \,.\,.\, b} \right)$

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:

 $\displaystyle A$ $=$ $\displaystyle \left [{a \,.\,.\, c} \right] \cup \left ({c \,.\,.\, b} \right) \cup \left [{b \,.\,.\, b} \right)$ $\displaystyle$ $=$ $\displaystyle \left [{a \,.\,.\, c} \right] \cup \left ({c \,.\,.\, b} \right)$ as $\left [{b \,.\,.\, b} \right) = \varnothing$

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

• $a = c, d = b$:

Then:

 $\displaystyle A$ $=$ $\displaystyle \left [{a \,.\,.\, a} \right] \cup \left ({a \,.\,.\, b} \right) \cup \left [{b \,.\,.\, b} \right)$ $\displaystyle$ $=$ $\displaystyle \left\{ {a}\right\} \cup \left ({a \,.\,.\, b} \right)$ as $\left [{b \,.\,.\, b} \right) = \varnothing$

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.

$\blacksquare$