Set of All Real Intervals is Semiring of Sets

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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:

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

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


  • $a = c, d = b$:

Then:

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

$\blacksquare$