Closed Real Interval is Closed in Real Number Line
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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)