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