Supremum of Bounded Above Set of Reals is in Closure
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Theorem
Let $\R$ be the real number line under the Euclidean metric.
Let $H \subseteq \R$ be a bounded above subset of $\R$ such that $H \ne \O$.
Let $u = \sup H$ be the supremum of $H$.
Then:
- $u \in \map \cl H$
where $\map \cl H$ denotes the closure of $H$ in $\R$.
Proof
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {B_\epsilon} u$ be the open $\epsilon$-ball of $u$ in $\R$.
From Distance from Subset of Real Numbers:
- $\map d {u, H} = 0$
Thus by definition of distance from subset:
- $\exists x \in H: \map d {u, x} < \epsilon$
Thus $x \in \map {B_\epsilon} u$.
As $x \in H$ and $x \in \map {B_\epsilon} u$, from the definition of intersection:
- $x \in H \cap \map {B_\epsilon} u$
The result follows from Condition for Point being in Closure.
$\blacksquare$
Also see
Sources
- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous): $2.28$
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.13$