# 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:

$I = \left({a \,.\,.\, b}\right)$
$I = \left[{a \,.\,.\, b}\right)$
$I = \left({a \,.\,.\, b}\right]$
$I = \left[{a \,.\,.\, b}\right]$

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 $I = \left({a \,.\,.\, b}\right)$.
$I^- = \left[{a \,.\,.\, b}\right]$

$\Box$

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

$\Box$

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

$\Box$

$(4): \quad$ Let $I = \left[{a \,.\,.\, b}\right]$.
$I$ is closed in $\R$.
$I^- = \left[{a \,.\,.\, b}\right]$

$\Box$

Thus all cases are covered.

The result follows by Proof by Cases.

$\blacksquare$