Test 1 Review Guide

This is a place to post helpful things related to the exam like problems handed out in class and other questions that arise while studying for the test on Wednesday and Friday.

This is one of the problems he indicated should be discussed:

Assume that $x_n$ is Cauchy. Prove that $\lambda x_n$ is Cauchy where $\lambda \in \mathbb{R}$.

One strategy is to use the definition of Cauchy and the other is to use convergence of a sequence.

I am out of time at the moment but hope to add some ideas later......anyone want to jump in?

Definitions and Theorems

1. A real number $s$ is the supremum of a set $A\subseteq \mathbb{R}$ if
   i) $s$ is an upper bound of $A$ and
   ii) if $b\in\mathbb{R}$ is an upper bound of $A$, then $s\le b$.
   
   
2. Axiom of Completeness: Every nonempty set of real numbers that is bounded above has a least upper bound.
3. A sequence $(a_n)$ converges to $a\in\mathbb{R}$ if $$(\forall \varepsilon >0) (\exists N\in\mathbb{N}) ( n\ge N \implies \left| a_n-a \right|<\varepsilon).$$
4. A sequence $(a_n)$ is increasing if $a_n\le a_{n+1}$ $\forall n\in\mathbb{N}$. A sequence $(a_n)$ is decreasing if $a_n\ge a_{n+1}$ $\forall n\in\mathbb{N}$. A sequence is monotone if it is either increasing or decreasing.
5. Let $(a_n)$ be a sequence of real numbers and let $n_1 < n_2 < n_3 < \dots$ be an increasing sequence of natural numbers. Then the sequence $$a_{n_1},a_{n_2},a_{n_3},\dots$$ is called a subsequence of $(a_n)$.
6. Belzano-Weierstrass Theorem: Every bounded sequence contains a convergent subsequence.
7. A sequence $(a_n)$ is called Cauchy if $$(\forall \varepsilon >0)(\exists N\in\mathbb{N} )(m,n\ge N \implies \left| a_n - a_m\right| < \varepsilon).$$
8. Cauchy Criterion: A sequence converges iff it is a Cauchy sequence.

A sequence converges if and only if it is a Cauchy sequence. Since $x_n$ is Cauchy it converges to a limit say $L$. Let $\lambda x_n$ be the sequence with $\lambda \in \mathbb{R}$. Then the $\lim{ \lambda x_n}= \lambda \lim{x_n}= \lambda L $. Therefore $\lambda x_n$ converges and is by definition Cauchy. Is this wrong or impolite?

To use the definition of Cauchy, would I do the following?

Assume that $x_n$ is Cauchy. Prove that $\lambda x_n$ is Cauchy where $\lambda \in \mathbb{R}$.

Proof: Let $\epsilon > 0$ and choose $N > \dfrac{\epsilon}{\lvert\lambda\rvert}$. Then whenver $n,m \geq N$, $$\lvert \lambda x_n - \lambda x_m \rvert = \lvert \lambda \rvert \lvert x_n - x_m \rvert < \lvert \lambda \rvert \frac{\epsilon}{\lvert \lambda \rvert} = \epsilon$$.

End Proof

Defining $N$ in relation to $\epsilon$ is only really useful when we have a sequence in our hands that has $n$'s in the definition. In this case $N > \dfrac{\epsilon}{|\lambda|}$ doesn't constrain $x_n$ in a way that we can use.

I think what you would want to do is choose $N$ such that whenever $n,m \geq N$, $|x_n - x_m| < \dfrac{\epsilon}{|\lambda|}$.

Am I off-base here?

Why would you not have $| x_n - x_m | < \epsilon $ ?

In the context of the problem @cseagrav was doing that choice of $N$ yields

$|x_n - x_m| < \dfrac{\epsilon}{|\lambda|} \implies |\lambda||x_n - x_m| = |\lambda x_n - \lambda x_m | < \epsilon$

Therefore $\lambda x_n$ is Cauchy.

@violincounter I think you're right, choosing $N > \frac{\varepsilon}{|\lambda |}$ does not necessarily mean $\left| x_n - x_m \right| < \frac{\varepsilon}{|\lambda |}$ because we have no idea when our sequence terms actually get that close because we don't have the sequence. Instead we would choose $N\in\mathbb{N}$ such that $n,m\ge N \implies \left| x_n - x_m \right| <\frac{\varepsilon}{|\lambda |}$ like you mentioned. We know we can do this by the definition of a Cauchy sequence and this allows us to make assertions we can use.

I'm not completely sure if this is something that we should be focusing on for Wednesday's part of the test, but I was working on it anyway and since it's on one of our review sheets, I figured I might as well post it here now. I'm not completely sure that this is right or not, so hopefully if I have made some mistakes someone else will be able to point them out for us.

This is for 2c on the first review sheet, where we're supposed to use the definition of the limit to prove that

$\lim(2a_n - 3b_n) = 2\lim a_n - 3\lim b_n$

Proof:

Assume that $a_n$ and $b_n$ have limits, such that $\lim a_n = a/2$ and $\lim b_n = b/2$. Let $\epsilon > 0$. Choose $N \in \mathbb{N}$ such that $n \geq N$ implies $\left|2a_n - a\right| < \frac{\epsilon}{2}$ and $\left|3b_n - b\right| < -\frac{\epsilon}{2}$. Then $n \leq N$ implies

$$\left|2a_n - 3b_n - (a+b) \right| = \left|2a_n - a - 3b_n - b \right| \leq \left|2a_n - a\right| - \left|3b_n - b\right|$$
$$= \left|2(a_n - \frac{a}{2}) \right| - \left|3(b_n - \frac{b}{3})\right| = \left|2\right|\left|a_n - \frac{a}{2} \right| - \left|3\right|\left|b_n - \frac{b}{3}\right| $$
$$= 2\left|a_n - \frac{a}{2} \right| - 3\left|b_n - \frac{b}{3}\right| < \frac{\epsilon}{2} - \frac{-\epsilon}{2} = \epsilon$$.

Therefore, $\lim(2a_n - 3b_n) = 2\lim a_n - 3\lim b_n$.

Thanks @violincounter and @jmincey for clarifying this. I think I understand that I was wrong to

but instead woud want to do as @violincounter said,

Then I would be able to start with what I want to control, that is $\lvert \lambda x_n - \lambda x_m \rvert$, and manipulate this into something I can control, $\lvert \lambda \rvert \lvert x_n - x_m \rvert < \lvert \lambda \rvert \dfrac{\epsilon}{\lvert \lambda \rvert} = \epsilon$.

So when proving limits with specified or defined sequences is when I would choose $N$ specifically?

You imply that $| a_n - a| < \frac{\epsilon}{2}$ and $| b_n - b| < - \frac{\epsilon}{2}$, but in your proof, you imply that $|a_n - \frac{a_n}{2}| < \frac{\epsilon}{2}$ and $| b_n - \frac{b}{3}| < - \frac{\epsilon}{2}$ which isn't what you implied earlier.

I was thinking that if we let $\epsilon > 0$ then choose $N \in \mathbb{N} \ni n\geq$ where $N > \displaystyle\frac{\epsilon}{6}$ for all n.

\begin{align*}
\left|2a_n - 3b_n - (2a-3b) \right| &= \left|2a_n - 2a + 3b - 3b_n \right|\\
&\leq \left| 2a_n - 2a \right| + \left|3b_n - 3b\right|\\
&\leq \left| 2 \right| \left| a_n -a \right| + \left| 3 \right| \left| b_n - b \right|\\
\end{align*}

Here we can state that since $N > \displaystyle\frac{\epsilon}{6}$ for all n, we can also say that $N > \displaystyle\frac{\epsilon}{4}$ for all n allowing us to do the following....

\begin{align*} &\leq \left| 2 \right| \displaystyle\frac{\epsilon}{4} + \left| 3 \right| \displaystyle\frac{\epsilon}{6}\\
&\leq \displaystyle\frac{\left| 2\right|\epsilon}{4} + \displaystyle\frac{\left| 3\right|\epsilon}{6}\\
&\leq \displaystyle\frac{\epsilon}{2} + \displaystyle\frac{\epsilon}{2}\\
&= \epsilon
\end{align*}.

What do you think?

That's because I messed up on what I was typing, oddly enough. Before the proof started, I had MEANT to have it so that $\left|2a_n -a\right| < \frac{\epsilon}{2}$ and $\left|3b_n -b\right| < \frac{\epsilon}{2}$, so that when you eventually factored out you'd get what I had in my proof.

From what I see from what you've done though, yours also works and is probably the way that we should do it.

@notjeremy we can't say $N> \frac{\varepsilon}{6} \implies N>\frac{\varepsilon}{4}$. Example $\varepsilon = 20$. For $N> \frac{\varepsilon}{6}$ we could choose $N=4$ but $4\ngeq \frac{\varepsilon}{4}=5$.

But further than that, I'm still skeptical at the choosing of the $N$. The idea of $\varepsilon$ is that it is a very small number and if you divide it you only make it smaller. That means you would choose $N=1$ in an application of your proof which I can't imagine can really work for all sequences. You have the same problem that was encountered in the above proof where you are trying to relate $N$ to $\varepsilon$ when we don't have a sequence. Instead you should say something closer to choose $N$ such that $N\ge n$ implies $\left| a_n - \lim{(a_n)} \right| < \frac{\varepsilon}{2}$ and $\left| a_n - \lim{(b_n)} \right| <\frac{\varepsilon}{2} $. In order to assert we can do this you would have to mention the special case when one or both of the sequences diverge and handle them separtely. Then do some algebra and it should work.

Does this sound right?

2.6.1 Give an example of each of the following, or argue that such a request is impossible.
(a) A Cauchy sequence that is not monotone.
(b) A monotone sequence that is not Cauchy.
(c) A Cauchy sequence with a divergent subsequence.
(d) An unbounded sequence containing a subsequence that is Cauchy.

(b) (1,2,3,4,5…)
(c) Impossible, it converges by the Cauchy criterion then by the. 2.5.2 (on page 55) then subsequences of a convergent sequences have to also converge.

(a) $(-1,\frac{1}{2},-\frac{1}{3},\dots , (-1)^n\frac{1}{n},\dots )\rightarrow 0$ so we know it is Cauchy by Theorem 2.6.2.

(d) Take the sequence $(-1,1,-1,\dots,(-1)^n,\dots)$ which has subsequence $(1,1,1,\dots)$. It converges so we know it is Cauchy.

I like this!! Is it right though? It seems more straightforward than the other method.

I feel as though I haven't done anything illegal, its just not going through the inequality route.

Problem 1.2.5:

$||a|-|b|| \leq |a - b|$

By the triangle in equality we know,

$||a|-|b|| = ||a - b + b| - |b||$

Thus,
$\leq || a - b| + |b| - |b||$
$ = ||a - b||$
$ = |a - b|$

Therefore, $|| a| - |b|| \leq |a - b|$.

Objections??