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to find infimum of set $ \{\frac{1}{2^n} : n \in \mathbb{N}\}$

Clearly $0$ is lower bound of set. Let $l$ be another lower bound such that $ l > 0 $ .Now by Archimedian property we have $\frac{1}{n} < l $ for some $n$.

Also we have for all $n \in \mathbb{N}$, $2^n> n$. so $\frac{1}{2^n} < \frac{1}{n}$. In particular we have ,$\frac{1}{2^n} < l$. hence a contradiction

I am not entirely sure about my proof and wondering what other ways can be used to prove this ?

Olivia
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    Looks good to me. – terran Mar 25 '23 at 06:54
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    Maybe I am picky, but saying that you are contradicting the assumption that $\ell$ was a a lower bound could be good. Also, I would add, as a last sentence : "Hence 0 is the largest lower bound for the set." – C614 Mar 25 '23 at 14:04

2 Answers2

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Your proof works. You can notice that, given an arbitrary $\epsilon>0$, it is $0+\epsilon>1/n_{\epsilon} \iff n_{\epsilon}> 1/\epsilon$ (you should actually take the ceiling function, since $n_{\epsilon}$ must be a positive integer; this $n_{\epsilon}$ exists because $\mathbb{N}$ is unbounded above). So, $0$ is the maximum of the lower bounds of your set. Hence, $0$ is the infimum of your set (greatest lower bound).

Bernkastel
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$A=\{\frac{1}{2^n} : n \in \mathbb{N}\}$

$i=\inf(A)=0$

$\max(A)=\frac{1}{2}$

$i=\inf(A) \iff \forall\varepsilon>0\,\, \exists x \in A:i+\varepsilon>x $

Proof by contradiction:

Suppose there exists a bigger lower bound $i'$:

$i'>i=0 \,\,\land\,\, i'\leq \frac{1}{2}$

Now let $\varepsilon:=i'-i=i'>0$

Therefore $a:=i+\frac{\varepsilon}{2} \in A$, because:

$a=\frac{\varepsilon}{2}>0$ and $a=\frac{\varepsilon}{2}=\frac{i'}{2}\leq\frac{1}{4}$.

But: $a=\frac{\varepsilon}{2}<\varepsilon=i' \implies a<i'$, which is a contradiction to $i'$ is the biggest lower bound of $A$.

Therefore $i$ must be the infimum of $A$.

Think
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