# Thread: intro analysis help

1. ## intro analysis help

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

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

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

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

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

thanks

2. For Q1, $\displaystyle A \subset \mathbb{R}$. But to show $\displaystyle \sup A$ exists, you have to also show that $\displaystyle A$ 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 $\displaystyle A = \{a_{n} : n\in\mathbb{ N}\}$. Show that every right end $\displaystyle b_{n}$ is an upper bound of $\displaystyle A$.
Let $\displaystyle I_j = [a_j,b_j]$ be your intervals for $\displaystyle j\geq 1$. They are nested, so, $\displaystyle I_i\subseteq I_j$ if $\displaystyle j>i$. Assume, by contradiction, not every right end was an upper bound of $\displaystyle \{ a_n\}$. Thus, it means there exists, a left end $\displaystyle a_m$ so that $\displaystyle a_m$ which was not an upper bound for $\displaystyle A$. That means, there exists $\displaystyle b_n$ so that $\displaystyle a_m>b_n$. Now $\displaystyle a_1\leq a_2\leq ... \leq a_n\leq b_n$, therefore if $\displaystyle a_m > b_n$ then it must mean that $\displaystyle m>n$. But we know that $\displaystyle I_m \subseteq I_n$. Now $\displaystyle a_m\in I_m$ but $\displaystyle a_m\not \in I_n$ because $\displaystyle a_m > b_n$. This is a contradiction by the nested interval property.

4. Can I simply break Q1 into cases?

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

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

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

Do I need to add any thing more to Q2?

Thanks for your help!