Result
R2971: Ordered vector space of real numbers is ordered vector space states that $\mathbb{R}$ forms an ordered vector space. Hence, we know that real summation preserves real order. That is, if $x, y, z \in \mathbb{R}$ such that $x \leq y$, then $x + z \leq y + z$. This property will be used in all of the following deductions.
If $I = (a, b)$, then
\begin{equation}
\begin{split}
I + c & = \{ x + c \mid a < x < b \} \\
& = \{ x \mid a < x - c < b \} \\
& = \{ x \mid a + c < x < b + c \} \\
& = (a + c, b + c)
\end{split}
\end{equation}
If $I = [a, b)$, then
\begin{equation}
\begin{split}
I + c & = \{ x + c \mid a \leq x < b \} \\
& = \{ x \mid a \leq x - c < b \} \\
& = \{ x \mid a + c \leq x < b + c \} \\
& = [a + c, b + c)
\end{split}
\end{equation}
If $I = (a, b]$, then
\begin{equation}
\begin{split}
I + c & = \{ x + c \mid a < x \leq b \} \\
& = \{ x \mid a < x - c \leq b \} \\
& = \{ x \mid a + c < x \leq b + c \} \\
& = (a + c, b + c]
\end{split}
\end{equation}
If $I = [a, b]$, then
\begin{equation}
\begin{split}
I + c & = \{ x + c \mid a \leq x \leq b \} \\
& = \{ x \mid a \leq x - c \leq b \} \\
& = \{ x \mid a + c \leq x \leq b + c \} \\
& = [a + c, b + c]
\end{split}
\end{equation}
$\square$
