Closure of Real Interval is Closed Real Interval/Proof 1

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

From Closure of Open Real Interval is Closed Real Interval:

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

$\Box$


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

From Closure of Half-Open Real Interval is Closed Real Interval:

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

$\Box$


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

From Closure of Half-Open Real Interval is Closed Real Interval:

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

$\Box$


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

From Closed Real Interval is Closed in Real Number Line:

$I$ is closed in $\R$.

From Set is Closed iff Equals Topological Closure:

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

$\Box$


Thus all cases are covered.

The result follows by Proof by Cases.

$\blacksquare$