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:

  • Half-open intervals:
    • $\hointr a b = \set {x \in \R: a \le x < b}$
    • $\hointl a b = \set {x \in \R: a < x \le b}$
  • Unbounded closed and unbounded open intervals:
    • $\hointr a {+\infty} = \set {x \in \R: a \le x}$
    • $\hointl {-\infty} a = \set {x \in \R: x \le a}$
    • $\openint a {+\infty} = \set {x \in \R: a < x}$
    • $\openint {-\infty} a = \set {x \in \R: x < a}$
    • $\openint {-\infty} {+\infty} = \R$


Set of Intervals is Not a Ring

Consider, for example, an open interval:

$\openint a b = \set {x \in \R: a < x < b}$

such that $a < b$.

Any subset $\openint c d \subset \openint a b$ such that $a < c, d < b$ is such that:

$\openint a b \setminus \openint c d = \hointl a c \cup \hointr d b$

But $\hointl a c \cup \hointr d b$ 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:

$A = \hointr a b$

and note that the argument generalizes.

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

Then $C = \openint c d$ is a subset of $\hointr a b$.


There are four cases:


  • $a < c, d < b$:

Then:

$A = \closedint a c \cup \openint c d \cup \hointr d b$

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


  • $a = c, d < b$:

Then:

$A = \set a \cup \openint a d \cup \hointr d b$

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


  • $a < c, d = b$:

Then:

\(\ds A\) \(=\) \(\ds \closedint a c \cup \openint c b \cup \hointr b b\)
\(\ds \) \(=\) \(\ds \closedint a c \cup \openint c b\) as $\hointr b b = \O$

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


  • $a = c, d = b$:

Then:

\(\ds A\) \(=\) \(\ds \closedint a a \cup \openint a b \cup \hointr b b\)
\(\ds \) \(=\) \(\ds \set a \cup \openint a b\) as $\hointr b b = \O$

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.

$\blacksquare$

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