Lecture A2

January 9, 2025

Lemma

If $B\subset A$ and $B$ has a Supremum, then the $\sup$ is unique. Similar statement for Infimum.

Proof:

• Supremum:

Let $a,a^{\prime }$ satisfy the definition of the Supremum. We show uniqueness by proving $a=a^{\prime }$. For contradiction, suppose $a<a^{\prime }$ or $a^{\prime }<a$.

Case 1: $a<a^{\prime }$ Since $a^{\prime }=\sup B$ and $a<a^{\prime }$, but then by property (2), $\exists b\in B$ such that $a<b$. OTOH, $a$ also satisfies $\forall b\in B,b\le a$. But then $a<b\le a$. This is a contradiction.

Case 2: $a^{\prime }<a$ The proof is similar, or you can do WLOG.

Therefore $a=a^{\prime }$.

• Infimum

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Infimum

Fields