inf or sup of a sequence including $(-1)^n$

Question

find $infA$ where $A=\{(-1)^n(1-\frac{1}{n}), n=1,2,3,...\}$
My work is as below:
if n is even : claim : $infA=\frac{1}{2}$
if n is odd : claim : $infA=-1$
Now should I choose the lowest number for the final claim?( and the greatest if it was asking for Sup(A)?).
Now let claim $infA=-1$
1- I prove -1 is a lower bound for A:
$\frac{1}{n}<1$ so $-\frac{1}{n}>-1>-2$ from here $1-\frac{1}{n}>1-2=-1$
hence all elements of A are greater than -1, hence -1 in a lower bound for A.
2- Now I want to prove there can not be any number greater than -1 which is a lower bound. or equivalently I want to show $-1+\epsilon$ can not be a lower bound.
I do not know I think I am confused. I guess I know hoe to prove sup lim questions. I do not know how $A=\{(-1)^n$ affect the problem. I would appreciate your help.

Answer 1

The infimum of a set is unique.

I get the point of your idea, but if you want to write it down like that, you should split $A$ into the two sets consisting of even and odd positiv integers. Then $A$ is the disjoint union of these sets.

However, this is not really necessary. But you are absolutly right. Indeed we have $\operatorname{inf}(A)=-1$.

For that we show that $-1$ is a lower bound of $A$ and the greatest lower bound.

-1 is lower bound of $A$

We have to show that $-1\leq a$ for every $a\in A$.

We seperate the cases. Either $a=1-\frac1m$ for some even $m\in\mathbb{N}$, or $a=-1+\frac1m$ for some odd $m\in\mathbb{N}$

In the first case it is clear, since $-1\leq 1-\frac{1}m\Leftrightarrow \frac1m\leq 2\Leftrightarrow 1\leq 2m$, which is true, as $m\geq 1$

In the second case we have $-1\leq -1+\frac1m\Leftrightarrow 0\leq \frac1m\Leftrightarrow 0\leq 1$.

So $-1$ is indeed a lower bound.

Why is $-1$ the greatest lower bound of $A$?

Suppose there is a lower bound $b$ of $A$ with $b>-1$. Then we can write $b=-1+\varepsilon$ for some fixed $\varepsilon >0$.

$b$ is supposed to be a lower bound, so

$-1+\varepsilon\leq a$ for every $a\in A$. For arbitrary $n=2m+1$ we have:

$-1+\varepsilon\leq -1+\frac{1}{2m+1}\Leftrightarrow \varepsilon\leq \frac{1}{2m+1}$ for every $m\in\mathbb{N}$!

But then $2m+1\leq \frac1\varepsilon$ for every $m\in\mathbb{N}$, which is false, since the LHS is unbounded. (For every real number there is a natural number greater then that.)

Answer 2

By following your steps,

• you made a small mistake in step 1: you proved that for all $n$ even elements of $A$ are greater than $-1$. For $n$ odd you have to prove that $-1+\frac{1}{2n+1}>-1$, but that is pretty simple, so you might have omitted it because it's trivial.

• for step 2: given any $\epsilon>0$ you can find elements of $A$ which are smaller than $-1+\epsilon$: take $n$ odd, so we look only at elements of the type $-1+\frac{1}{2n+1}$.Now you can always choose $n$ large enough so that $\frac{1}{2n+1} < \epsilon$, for instance take any $n>\frac{1}{2\epsilon}-\frac{1}{2}$. Thus proving that, for any $\epsilon>0$, $-1+\epsilon$ cannot be the $\inf$.

The trick here is to separate $A$ into two subsets: $A_{even}=\{1-\frac{1}{2n}, n=1,2,...\}$ and $A_{odd}=\{-1+\frac{1}{2n+1}, n=1,2,...\}$. You clearly have that $A$ is the union of these two, therefore the inf of $A$ is the smallest of the infs of $A_{odd}$ and $A_{even}$.

I hope it clears it up :)