Closure of Real Interval is Closed Real Interval/Proof 2
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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
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:
- $\closedint a b = I^-$
$\blacksquare$