Limit Points of Open Real Interval

Theorem

Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.

Let $\openint a b$ be an open interval of $\R$.

Then the limit points of $\openint a b$ are:

all the points in $\openint a b$

as well as:

the points $a$ and $b$.

Proof

Let $x \in \openint a b$.

Then by definition $a < x < b$.

Let $N_x$ be an open neighborhood of $x$ in $\R$.

So by Open Superset is Open Neighborhood, $N_x$ is an open set of $\R$ containing $x$.

As $x \in N_x$, it follows that $\exists U \subseteq N_x$ such that $x \in U$ and:

$\exists p, q \in \R: U = \openint p q$

Let $r = \max \set {a, p}$ and $s = \min \set {b, q}$.

Then as $x > a, x > p$ and $x < b, x < q$ it follows that $x > \max \set {a, p}$ and $x < \min \set {b, q}$.

That is:

$x \in \openint r s$
$\exists y \in \R: r < y < x$

and so:

$y \in \openint r s$

By construction, it follows that:

$y \in \openint p q$

That is, $y$ is an element of $\R$ different from $x$ which is also in $N_x$.

As $N_x$ was arbitrary, it follows that every such open neighborhood of $x$ has the same property.

That is, $x$ is a limit point of $\openint a b$.

$\Box$

Let $N_a$ be an open neighborhood of $a$ in $\R$.

So by Open Superset is Open Neighborhood, $N_a$ is an open set of $\R$ containing $a$.

As $a \in N_a$, it follows that $\exists U \subseteq N_a$ such that $a \in U$ and:

$\exists p, q \in \R: U = \openint p q$

Let $s = \min \set {b, q}$.

As $a < b, a < q$ it follows that $a < \min \set {b, q}$.

That is:

$a \in \openint p s$
$\exists y \in \R: a < y < s$

and so:

$y \in \openint p s$

By construction, it follows that:

$y \in \openint p q$

That is, $y$ is an element of $\R$ different from $a$ which is also in $N_a$.

As $N_a$ was arbitrary, it follows that every such open neighborhood of $a$ has the same property.

That is, $a$ is a limit point of $\openint a b$.

By a similar argument it is shown that $a$ is a limit point of $\openint a b$.

$\Box$

Now let:

$x \notin \openint a b \cup \set {a, b}$

Then either:

$x < a$

or

$x > b$

Let $x < a$.

Then by Real Numbers are Densely Ordered:

$\exists q \in \R: x < q < a$

and so $\exists p \in \R$ such that:

$x \in \openint p q$

such that:

$\openint p q \cap \openint a b = \O$

That is, $\openint p q$ is an open set of $\R$ containing $x$ which contains no elements of $\openint a b$.

Hence by definition, $x$ is not a limit point of $\openint a b$.

By a similar argument it is shown that if $x > b$ then $x$ is not a limit point of $\openint a b$.

$\blacksquare$