# Closed Subset of Real Numbers with Lower Bound contains Infimum

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## Theorem

Consider the real number line as a metric space under the usual metric.

Let $A \subseteq \R$ such that $A$ is closed in $\R$ and $A \neq \varnothing$.

Let $A$ be bounded below.

Then $A$ contains its infimum.

## Proof

From Infimum of Bounded Below Set of Reals is in Closure:

- $\map \inf A \in \map \cl A$

From Set is Closed iff Equals Topological Closure:

- $A = \map \cl A$

Therefore $\map \inf A \in A$.

$\blacksquare$

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.6$: Open Sets and Closed Sets: Exercise $5$