Limit Points of Open Real Interval

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Theorem

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

Let $\left({a \,.\,.\, b}\right)$ be an open interval of $\R$.


Then the limit points of $\left({a \,.\,.\, b}\right)$ are:

the points $\left({a \,.\,.\, b}\right)$ itself

and

the points $a$ and $b$.


Proof

Let $x \in \left({a \,.\,.\, b}\right)$.

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 $x$.

By Open Sets in Real Number Line, $N_x$ is a countable union of pairwise disjoint open intervals.

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

$\exists p, q \in \R: U = \left({p \,.\,.\, q}\right)$

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

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

That is:

$x \in \left({r \,.\,.\, s}\right)$

From Real Numbers are Close Packed:

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

and so:

$y \in \left({r \,.\,.\, s}\right)$

By construction, it follows that:

$y \in \left({p \,.\,.\, q}\right)$

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 $\left({a \,.\,.\, b}\right)$.

$\Box$


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 $a$.

By Open Sets in Real Number Line, $N_a$ is a countable union of pairwise disjoint open intervals.

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

$\exists p, q \in \R: U = \left({p \,.\,.\, q}\right)$

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

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


That is:

$a \in \left({p \,.\,.\, s}\right)$

From Real Numbers are Close Packed:

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

and so:

$y \in \left({p \,.\,.\, s}\right)$

By construction, it follows that:

$y \in \left({p \,.\,.\, q}\right)$

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 $\left({a \,.\,.\, b}\right)$.


By a similar argument it is shown that $a$ is a limit point of $\left({a \,.\,.\, b}\right)$.

$\blacksquare$


Now let:

$x \notin \left({a \,.\,.\, b}\right) \cup \left\{{a, b}\right\}$

Then either:

$x < a$

or

$x > b$

Let $x < a$.

Then by Real Numbers are Close Packed:

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

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

$x \in \left({p \,.\,.\, q}\right)$

such that:

$\left({p \,.\,.\, q}\right) \cap \left({a \,.\,.\, b}\right) = \varnothing$

That is, $\left({p \,.\,.\, q}\right)$ is an open set of $\R$ containing $x$ which contains no elements of $\left({a \,.\,.\, b}\right)$.

Hence by definition, $x$ is not a limit point of $\left({a \,.\,.\, b}\right)$.


By a similar argument it is shown that if $x > b$ then $x$ is not a limit point of $\left({a \,.\,.\, b}\right)$.

$\blacksquare$