Closure of Real Interval is Closed Real Interval/Proof 2

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

Let $x \in \closedint a b$.

Let $\openint c d$ be an open interval in $\R$ such that $x \in \openint c d$.

We must show that $\openint c d$ contains a point in $I$.

One of the following three possibilities holds:

$a < x < b$
$x = a$
$x = b$

Case: $a < x < b$

In this case, $x \in I$ and $x \in \openint c d$.

Therefore $\openint c d$ contains a point in $I$.

$\Box$

Case: $x = a$

If $I$ contains $a$, then this means $x \in I$, and the proof is complete.

So, assume that $a \notin I$.

Since $I$ is nonempty but does not contain $a$, we must have $a < b$.

Let $r$ be the minimum of $d$ and $b$, so that $r \le d$ and $r \le b$.

Since $a = x < d$ by choice of $d$ and since $a < b$ by assumption, we must have $a < r$.

Thus, by Real Numbers are Densely Ordered, there exists some $s \in \R$ such that $a < s < r$.

To summarize, we have $c < x = a < s < r$, where $r \le d$ and $r \le b$.

This means that $s$ satisfies both $c < s < d$ and $a < s < b$.

Hence, $s$ is a point in $\openint c d$ which is also in $I$.

The existence of such a point is what we wanted to show.

$\Box$

Case: $x = b$

This case is analogous to case when $x = a$.

Here we instead let $l$ be the maximum of $c$ and $a$, and select an $s$ such that $l < s < x = b < d$, where $c \le l$ and $a \le l$.

$\Box$

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

Suppose $x \notin \notin \closedint a b$.

We must find an open interval containing $x$ which does not contain a point in $I$.

There are two possibilities:

$x < a$

or:

$x > b$

Case: $x < a$

By Real Numbers are Densely Ordered, there exists $r \in \R$ such that $x < r < a$.

Thus $\openint {x - 1} r$ is an open interval, all of whose elements are less than $a$, and hence not in $I$.

Case: $x > b$

This is similarly to the case when $x < a$.

Here instead we pick $r$ such that $b < r < x$, and consider the interval $\openint r {x + 1}$.

$\Box$

By the two lemmas proven above:

$\closedint a b = I^-$

$\blacksquare$