Closure of Real Interval is Closed Real Interval

Theorem
Let $I$ be a real interval such that:
 * $I = \left({a \, . \, . \, b}\right)$;
 * $I = \left[{a \, . \, . \, b}\right)$;
 * $I = \left({a \, . \, . \, b}\right]$, or
 * $I = \left[{a \, . \, . \, b}\right]$.

Then $\operatorname{cl} \left({I}\right)$, the closure of $I$, is the closed interval $\left[{a \,. \, . \, b}\right]$.

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

We are to show that $x \in \left[{a \,. \, . \, b}\right]$ iff every open set in $\R$ containing $x$ contains a point in $I$.

Let $x \in \left[{a \,. \, . \, b}\right]$.

Then one of the following three possibilities holds:


 * $a < x < b$;
 * $x = a$;
 * $x = b$.

Let $\left({c \, . \, . \, d}\right)$ be an open interval in $\R$ such that $x \in \left({c \, . \, . \, d}\right)$.

Then $\left({c \, . \, . \, d}\right)$ is an open set in $\R$.

By definition, there exists an $\epsilon$-neighborhood $N_\epsilon \left({x}\right)$ of $x$ such that $N_\epsilon \left({x}\right) \subseteq \left({c \, . \, . \, d}\right)$.


 * If $a < x < b$ then $x \in I$ and so $\left({c \, . \, . \, d}\right)$ contains a point in $I$.


 * If $x = a$, then $\exists \epsilon > 0: a + \epsilon < b$.

Hence $\exists y \in N_\epsilon \left({x}\right)$ such that $y > a$ and $y < b$.

Hence $y \in I$.


 * If $x = b$, then $\exists \epsilon > 0: b - \epsilon > a$.

Hence $\exists y \in N_\epsilon \left({x}\right)$ such that $y > a$ and $y < b$.

Hence $y \in I$.

Now suppose $x \notin \left[{a \,. \, . \, b}\right]$.

Then $x < a$ or $x > b$.


 * Let $x < a$.

Then $\exists \epsilon > 0: x + \epsilon < a$.

Hence $\exists N_\epsilon \left({x}\right): N_\epsilon \left({x}\right) \cap \left[{a \,. \, . \, b}\right] = \varnothing$.


 * Similarly for $x > b$.

The result follows from Condition for Point being in Closure.