Closure of Real Interval is Closed Real Interval/Proof 1

Theorem
Let $I$ be a non-empty real interval such that one of these holds:
 * $I = \left({a \,.\,.\, b}\right)$
 * $I = \left[{a \,.\,.\, b}\right)$
 * $I = \left({a \,.\,.\, b}\right]$

or
 * $I = \left[{a \,.\,.\, b}\right]$

Let $I^-$ denote the closure of $I$.

Then $I^-$ is the closed real interval $\left[{a \,.\,.\, b}\right]$.

Proof
There are four cases to cover:


 * $(1): \quad$ Let $I = \left({a \,.\,.\, b}\right)$.

From Closure of Open Real Interval is Closed Real Interval:
 * $I^- = \left[{a \,.\,.\, b}\right]$


 * $(2): \quad$ Let $I = \left[{a \,.\,.\, b}\right)$.

From Closure of Half-Open Real Interval is Closed Real Interval:
 * $I^- = \left[{a \,.\,.\, b}\right]$


 * $(3): \quad$ Let $I = \left({a \,.\,.\, b}\right]$.

From Closure of Half-Open Real Interval is Closed Real Interval:
 * $I^- = \left[{a \,.\,.\, b}\right]$


 * $(4): \quad$ Let $I = \left[{a \,.\,.\, b}\right]$.

From Closed Real Interval is Closed in Real Number Line:
 * $I$ is closed in $\R$.

From Closed Set equals its Topological Closure:
 * $I^- = \left[{a \,.\,.\, b}\right]$

Thus all cases are covered.

The result follows by Proof by Cases.