Half-Open Real Interval is not Closed in Real Number Line

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

Let $\left[{a \,.\,.\, b}\right) \subset \R$ be a half-open interval of $\R$.

Then $\left[{a \,.\,.\, b}\right)$ is not a closed set of $\R$.

Similarly, the half-open interval $\left({a \,.\,.\, b}\right] \subset \R$ is not a closed set of $\R$.



$A := \R \setminus \left[{a \,.\,.\, b}\right) = \left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$

Let $A:= \left({-\infty \,.\,.\, a}\right)$ and let $B := \left[{b \,.\,.\, \infty}\right)$.

Let $B_\epsilon \left({b}\right)$ be the open $\epsilon$-ball of $b$.

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

Now $b - \epsilon$ may itself lie in $A$.


$\left[{a \,.\,.\, b}\right) \cap B_\epsilon \left({b}\right) \ne \varnothing$

and so there are elements of $B_\epsilon \left({b}\right)$ which are not in $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$.

So $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$ is not an open set of $\R$.

Thus, by definition, $\left[{a \,.\,.\, b}\right)$ is not a closed set of $\R$.

Hence the result.

Mutatis mutandis, the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is neither an open set nor a closed set of $\R$.