Closed Real Interval is Closed in Real Number Line

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Theorem

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

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


Then $\left[{a \,.\,.\, b}\right]$ is closed (in the topological sense) in $\left({\R, \tau_d}\right)$.


Proof

From Open Sets in Real Number Line:

$U := \left({-\infty \,.\,.\, a}\right) \cup \left({b \,.\,.\, +\infty}\right)$

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

Consider the complement relative to $\R$:

$V := \complement_\R \left({U}\right) = \left[{a \,.\,.\, b}\right]$

By definition, $\complement_\R \left({U}\right)$ is closed (in the topological sense) in $\left({\R, \tau_d}\right)$.

But by construction:

$\complement_\R \left({U}\right) = \left[{a \,.\,.\, b}\right]$

Hence the result.

$\blacksquare$