1.3.1: Infimum Definitions

a

b

We prove the following lemma:

\begin{proof}
($\Rightarrow$). Suppose $g=\text{inf}(A)$, so then $g$ is both a lower bound and the greatest one. Let $\epsilon >0$ be arbitrary. Notice that then $g+\epsilon >g$, then sing $g$ is the greatest upper bound then $g+\epsilon$ must not be a lower bound for $A$, so then there is some $a\in A$ where $a<g+\epsilon$, per our requirements.

($\Leftarrow$). Suppose for all $\epsilon >0$ there is some $a\in A$ where $g+\epsilon >a$. We need to show that $g$ is the infimum for $A$. We already know that $g$ is a lower bound, satisfying (i). For (ii), let $g^{\prime }$ be any other lower bound for $A$. We need to show that $g\ge g^{\prime }$, so then for all $a\in A$ then $g^{\prime }\le a$. Choose $\epsilon =g^{\prime }-g>0$ since we've assumed that $g^{\prime }$ isn't the greatest lower bound. Thus, then $g+(g^{\prime }-g)=g^{\prime }>a$ is a contradiction as $g^{\prime }$ is a lower bound. Hence, $g^{\prime }$ must be the some other lower bound, so then $g^{\prime }-g<0$ and thus $g^{\prime }<g$ as required.
\end{proof}

1.3.3

a

Let $A$ be nonempty and bounded below, and: $$B = {b \in \mathbb{R} : b \text{ is a lower bound for $A$ }}$$Then $\text{sup}(B)=\text{inf}(A)$.

\begin{proof}
Clearly the suprema and infimum of $A,B$ exist since they are subsets of $\mathbb{R}$. Suppose that $b=\text{sup}(B)$. Then we have it that $b$ is an upper bound for $B$ as well as any other upper bounds $b^{\prime }$ for $B$ have it where $b\le b^{\prime }$. Namely:

• For all $b_o\in B$ then $b_o\le b$, so $b$ is a valid upper bound for $A$.
• For all $b^{\prime }\in B$ as other upper bounds, then $b_o\le b^{\prime }$ while still $b\le b^{\prime }$.

Likewise, for $a=\text{inf}(A)$ we get that:

• For all $a_o\in A$ then $a_o\ge a$, so $a$ is a valid lower bound for $A$.
• For all $a^{\prime }\in A$ as other lower bounds, then $a_o\ge a^{\prime }$ while still $a\ge a^{\prime }$.

We'll try to start on

\end{proof}