Closure of Real Interval is Closed Real Interval/Proof 2/Lemma 2

Lemma for Closure of Real Interval is Closed Real Interval

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:

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

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

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

There are two possibilities:

$x < a$

or:

$x > b$

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 $I$.

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

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

$\blacksquare$