\begin{definition}[Definition:Real Interval/Empty]
Let $a, b \in \R$.
Let $\left [{a \,.\,.\, b} \right]$, $\left [{a \,.\,.\, b} \right)$, $\left ({a \,.\,.\, b} \right)$ and $\left ({a \,.\,.\, b} \right)$ be real intervals: closed, half-open and open as defined.
{{ExtractTheorem|this needs to be proved}}
When $a > b$:
{{begin-eqn}}
{{eqn | l = \closedint a b
      | m = \set {x \in \R: a \le x \le b}
      | mo = =
      | r = \O
}}
{{eqn | l = \hointr a b
      | m = \set {x \in \R: a \le x < b}
      | mo = =
      | r = \O
}}
{{eqn | l = \hointl a b
      | m = \set {x \in \R: a < x \le b}
      | mo = =
      | r = \O
}}
{{eqn | l = \openint a b
      | m = \set {x \in \R: a < x < b}
      | mo = =
      | r = \O
}}
{{end-eqn}}
When $a = b$:
{{begin-eqn}}
{{eqn | ll= \hointr a b
      | lo= =
      | l = \hointr a a
      | m = \set {x \in \R: a \le x < a}
      | mo= =
      | r = \O
}}
{{eqn | ll= \hointl a b
      | lo= =
      | l = \hointl a a
      | m = \set {x \in \R: a < x \le a}
      | mo= =
      | r = \O
}}
{{eqn | ll= \openint a b
      | lo= =
      | l = \openint a a
      | m = \set {x \in \R: a < x < a}
      | mo= =
      | r = \O
}}
{{end-eqn}}
Such empty sets are referred to as '''empty intervals'''.
\end{definition}