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: {{begin-eqn}} {{eqn | l=A       | r=\left [{a \,.\,.\, c} \right] \cup \left ({c \,.\,.\, b} \right) \cup \left [{b \,.\,.\, b} \right)       | c= }} {{eqn | r=\left [{a \,.\,.\, c} \right] \cup \left ({c \,.\,.\, b} \right)       | c=as $\left [{b \,.\,.\, b} \right) = \varnothing$ }} {{end-eqn}} and it can be seen that this is a partition of $A$. * $a = c, d = b$: Then: {{begin-eqn}} {{eqn | l=A       | r=\left [{a \,.\,.\, a} \right] \cup \left ({a \,.\,.\, b} \right) \cup \left [{b \,.\,.\, b} \right)       | c= }} {{eqn | r=\left\{ {a}\right\} \cup \left ({a \,.\,.\, b} \right)       | c=as $\left [{b \,.\,.\, b} \right) = \varnothing$ }} {{end-eqn}} 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. {{qed}} Category:Semirings of Sets Category:Analysis qy3dzdf6fd5htzazwphb11jv8i5eb9j