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:

$I = \openint a b$

$I = \hointr a b$

$I = \hointl a b$

$I = \closedint a b$

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

Then $I^-$ is the closed real interval $\closedint a b$.

Proof

There are four cases to cover:

$(1): \quad$ Let $I = \openint a b$.

From Closure of Open Real Interval is Closed Real Interval:

$I^- = \closedint a b$

$\Box$

$(2): \quad$ Let $I = \hointr a b$.

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

$I^- = \closedint a b$

$\Box$

$(3): \quad$ Let $I = \hointl a b$.

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

$I^- = \closedint a b$

$\Box$

$(4): \quad$ Let $I = \closedint a b$.

From Closed Real Interval is Closed in Real Number Line:

$I$ is closed in $\R$.

From Set is Closed iff Equals Topological Closure:

$I^- = \closedint a b$

$\Box$

Thus all cases are covered.

The result follows by Proof by Cases.

$\blacksquare$