Half-Open Real Interval is Closed in some Open Intervals

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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 half-open interval of $\R$.


Let $c < a$.

Then $\left[{a \,.\,.\, b}\right)$ is a closed set of $\left({c \,.\,.\, b}\right)$.


Similarly, let $d > b$.

Then the half-open interval $\left({a \,.\,.\, b}\right]$ is a closed set of $\left({a \,.\,.\, d}\right)$.


Proof

Consider:

$A := \left({c \,.\,.\, b}\right) \setminus \left[{a \,.\,.\, b}\right) = \left({c \,.\,.\, a}\right)$

Then $A$ is an open interval.

By Open Real Interval is Open Set, $A$ is open in $\R$.

By the definition of the subspace topology it follows that $A$ is open in $\left({c \,.\,.\, b}\right)$.

Thus by the definition of closed set, $\left({c \,.\,.\, b}\right) \setminus A = \left[{a \,.\,.\, b}\right)$ is closed in $\left({c \,.\,.\, b}\right)$.


Mutatis mutandis, the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is a closed set of $\left({a \,.\,.\, d}\right)$.

$\blacksquare$


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