1. ## intro analysis help

Q1: Consider the statement of the Nested Intervals Property. As in the proof, consider the
set of all left ends $A =
\{a_{n} : n\in\mathbb{ N}\}$
. Show that every right end $b_{n}$ is an upper bound of $A$.

Q2: Consider the set $A =
\{ \frac{1}{n} : n\in\mathbb{ N} \}$
. Prove that $inf(A) = 0$.

A1: By the axiom of completeness, $A$ has a greatest upper bound $x=sup(A)$. Thus, for $\forall\\n\in\mathbb{N}$, $a_{n}\leq\\x$ and for any $\epsilon>0$ there exists a $b_{0}\in\\B$ such that $x. Therefore, for $\forall\\n\in\mathbb{N}$, $b_{n}$ is an upper buond for $A$.

A2: Suppose there exists a lower bound $l$ such that $inf(A)=0. Then, $l-0>0$, so we can consider $\epsilon=l-0>0$. By lemma 1.3.7, there exists an $a\in\\A$ such that, for any $\epsilon>0$, $a. But, if we consider $\epsilon=l-0$, then $a<2l$. Therefore, $\frac{a}{2}. This contradicts $l$ being a lower bound for $A$. Hence, $inf(A)=0$.

How do this proofs look? I feel I am not showing enough in the first one.

thanks

2. For Q1, $A \subset \mathbb{R}$. But to show $\sup A$ exists, you have to also show that $A$ is bounded above to invoke the least upper bound property. So I think you are using what you are trying to prove.

3. Originally Posted by Danneedshelp
Q1: Consider the statement of the Nested Intervals Property. As in the proof, consider the
set of all left ends $A =
\{a_{n} : n\in\mathbb{ N}\}$
. Show that every right end $b_{n}$ is an upper bound of $A$.
Let $I_j = [a_j,b_j]$ be your intervals for $j\geq 1$. They are nested, so, $I_i\subseteq I_j$ if $j>i$. Assume, by contradiction, not every right end was an upper bound of $\{ a_n\}$. Thus, it means there exists, a left end $a_m$ so that $a_m$ which was not an upper bound for $A$. That means, there exists $b_n$ so that $a_m>b_n$. Now $a_1\leq a_2\leq ... \leq a_n\leq b_n$, therefore if $a_m > b_n$ then it must mean that $m>n$. But we know that $I_m \subseteq I_n$. Now $a_m\in I_m$ but $a_m\not \in I_n$ because $a_m > b_n$. This is a contradiction by the nested interval property.

4. Can I simply break Q1 into cases?

Let $I_{n}=[a_{n},b_{n}]$. Since $I_{n+1}\subseteq\\I_{n}$, we can see $a_{1}\leq\\b_{1}, a_{2}\leq\\b_{2},...$. Thus, there are three cases:

Case 1: If $i=j$, then it is a given that $a_{i}\leq\\b_{j}$.
Case 2: If $i, note that $a_{i}\leq\\a_{j}\leq\\b_{j}$.
Case 3: If $i>j$, then $a_{i}\leq\\b_{i}\leq\\b_{j}$.

Therefore, $b_{n}$ is an upper bound for $A$, for every $n\in\mathbb{N}$.

Do I need to add any thing more to Q2?