Closed Real Interval is Closed in Real Number Line

Theorem

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.

Let $\closedint a b$ be a closed interval of $\R$.

Then $\closedint a b$ is closed (in the topological sense) in $\struct {\R, \tau_d}$.

Proof

From Open Sets in Real Number Line:

$U := \openint \gets a \cup \openint b \to$

is an open set in $\struct {\R, \tau_d}$.

Consider the complement relative to $\R$:

$V := \relcomp \R U = \closedint a b$

By definition, $\relcomp \R U$ is closed (in the topological sense) in $\struct {\R, \tau_d}$.

But by construction:

$\relcomp \R U = \closedint a b$

Hence the result.

$\blacksquare$

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): closed set (of points)
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): closed set (of points)