Non-Empty Bounded Above Subset of Banach Space with Archimedean Property has Supremum
Jump to navigation
Jump to search
Theorem
Let $\BB$ be a Banach space.
Let $\BB$ have the Archimedean property.
Let $S \subseteq \BB$ be a subset of $\BB$ which is bounded above.
Then $S$ admits a supremum.
Proof
![]() | This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.10$