Closed Real Interval is Neighborhood Except at Endpoints

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Theorem

Let $\R$ be the real number line considered as an Euclidean space.

Let $\left[{a \,.\,.\, b}\right] \subset \R$ be a closed interval of $\R$.


Then $\left[{a \,.\,.\, b}\right]$ is a neighborhood of all of its points except $a$ and $b$.


Proof

Let $a < c < b$.

Let $\epsilon < \min \left({b - c, c - a}\right)$.

From the definition of positive it follows that $\epsilon \in \R_{>0}$.


Let $B_\epsilon \left({c}\right) = \left({c - \epsilon \,.\,.\, c + \epsilon}\right)$ be the open $\epsilon$-ball of $c$.

We have that $c + \epsilon < b$ and $a < c - \epsilon$.

Thus:

$B_\epsilon \left({c}\right) \subseteq \left[{a \,.\,.\, b}\right]$


It follows that, by definition, $\left[{a \,.\,.\, b}\right]$ is a neighborhood of $c$.


Now consider $a$.

Let $\epsilon \in \R_{>0}$.

We have that $a - \epsilon < a$ and so $B_\epsilon \left({a}\right) = \left({a - \epsilon \,.\,.\, a + \epsilon}\right)$ does not lie entirely in $\left[{a \,.\,.\, b}\right]$.


Similarly, consider $b$.

Let $\epsilon \in \R_{>0}$.

We have that $b < b + \epsilon$ and so $B_\epsilon \left({b}\right) = \left({b - \epsilon \,.\,.\, b + \epsilon}\right)$ does not lie entirely in $\left[{a \,.\,.\, b}\right]$.


It follows that, by definition, $\left[{a \,.\,.\, b}\right]$ is a neighborhood of neither $a$ nor $b$.

$\blacksquare$


Sources