Closure of Real Interval is Closed Real Interval/Proof 2

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

Note that:
 * $(1): \quad$ By Condition for Point being in Closure, $x \in I^-$ 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 $x \in I^-$ 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 $I$.

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 Close Packed, 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.

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 $a \le l$.

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 Close Packed, 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 $\openint r {x + 1}$.

By the two lemmas proven above:
 * $\closedint a b = I^-$