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I have the sequence:

$A = \{ 1 - \frac{1}{n} | n \in \mathbb{N}\}$

I need to find the surpremum with correct mathmatical notation. I got some help from my math instructor, i thought i understood it, but some of the steps i don't understand. I'll make the steps and write when the logic fails for me.

First let's find an upper bound:

$1 - \frac{1}{n} \leq 1$

Check of it is the lowest upper bound:

pick and epsilon $\forall \epsilon > 0 $ or: $1 - \epsilon$

Now comes the step i don't understand. He said: $\exists n \in \mathbb{N} \space | \space 1 - \epsilon < 1 - \frac{1}{n} \leq 1$

This part i don't understand. Why should $1- \epsilon < 1 - \frac{1}{n} $? then it is not in the sequence.. Shouldn't we only check it is smaller than 1, and still be inside the sequence?

the last steps he did was:

$\Rightarrow n > \frac{1}{\epsilon}$

For $n > \frac{1}{\epsilon}$:

$1 - \epsilon < 1 - \frac{1}{n} \leq 1$

and then this should be the conclusion.. But i'm not sure what we just concluded.

I would very much appreciate your perspective..

nammerkage
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  • The inequality $1 - \epsilon < 1 - \frac{1}{n}$ is exactly the part that proves that $1-\epsilon$ cannot be an upper bound, since there are elements if $A$ that are larger. Whether $1-\epsilon$ is an actual element of the sequence is irrelevant, as bounds do not have to be part of the set that they bound. – Arthur Nov 13 '17 at 15:55
  • This actually makes a lot of sense Thank you. @Arthur. But why did i have to isolate n? – nammerkage Nov 13 '17 at 16:14
  • And will you please explain as well, why we set $1 - \epsilon $ to be smaller than the sequence itself? I don't understand that part.. Thank you for the help :D – nammerkage Nov 13 '17 at 16:18

2 Answers2

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For every $n,$ $n \in \mathbb{N+}:$

$a_n = 1 - 1/n <1$, I.e.

$S:= 1$ is an upper bound.

To show that it is the least upper bound , the supremum, we assume that there is a smaller upper bound $\gamma $.

Let $\epsilon:= S - \gamma \gt 0$.

We need find an $a_n$ with:

$\gamma \lt a_n \lt S $, I.e.

$ 0 < a_n - \gamma \lt S -\gamma=\epsilon.$

Choose $n$ such that $1/n \lt \epsilon.$

(Possible since for $1/\epsilon$ there is a $n$, $n \in \mathbb{N}$, such that $n > 1/\epsilon$.) (Archimedes)

For this $n$ we have: $1/n \lt \epsilon =S - \gamma$, or

$-1 +1/n \lt S -\gamma -1,$ or

$a_n = 1-1/n \gt \gamma -S +1 =\gamma$

Recall: $S =1$.

Hence $\gamma$ is not an upper bound.

Implies: $\sup(A) =1.$

Peter Szilas
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So we know that $1-\frac1n ≤1$, hence $1$ is an upper bound of A. To show that $1$ is actually the lowest upper bound, we assume that there is a lower one and lead that to a contradiction. If there is a lower upper bound than $1$, it must be lower than $1$ and hence we can write it as $1-\epsilon$ for a (possibly very small) $\epsilon >0$. But then we can choose $n \in \Bbb N$ with $n>\frac1\epsilon$ and we see $$1-\frac1n>1-\frac{1}{\frac1\epsilon}=1-\epsilon.$$ So $1-\epsilon$ can't be an upper bound, because we just found a term of the sequence which is greater. Hence we conclude that there can't be an upper bound of the form $1-\epsilon$ for $\epsilon >0$ and thus 1 is the lowest upper bound of $A$.

blat
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