Closure of Real Interval is Closed Real Interval

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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 1

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.

$\blacksquare$


Proof 2

Let $I$ be one of the intervals as specified in the exposition.


Note that:

$(1): \quad$ By Condition for Point being in Closure, $x \in I^-$ if and only if every open set in $\R$ containing $x$ contains a point in $I$.
$(2): \quad$ From Union of Open Sets of Metric Space is Open, every open set in $\R$ is a union of open intervals.

Thus we also have that $x \in I^-$ if and only if every open interval containing $x$ also contains a point in $I$.


This equivalence will be made use of throughout.


Lemma 1

$x \in \closedint a b \implies x \in I^-$

$\Box$


Lemma 2

$x \notin \closedint a b \implies x \notin I^-$

$\Box$


By the two lemmas proven above:

$\closedint a b = I^-$

$\blacksquare$


Sources

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