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So I need to find the infimum of$\{\frac{1}{n^2} | n \in \Bbb N\}$

I know that this means that I need to find some $x$ where $x < \frac{1}{n^2} \forall n \in \Bbb N$.

By intuition I know that $\lim_{n \to \infty}$, is $0$, which means that the infimum should be $0$, but is there a more formal way of proving that without using limits?

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  • $\begingroup$ Clearly $\frac{1}{n^2}> 0$ for all $n > 0$. For $a > 0$ you can find $n > 0$ such that $a > \frac{1}{n^2}$ and $a$ isn't a lower bound for the set. Now by definition $0$ is the infimum (largest lower bound) of the set. $\endgroup$ Feb 22, 2017 at 15:52
  • $\begingroup$ @MatiasHeikkilä I reached the same conclusion, and the part that I wasn't so sure about was saying that "you can find an $n > 0$ such that $a > \frac{1}{n^2}$" Do I need to somehow prove that statement? I know it's true intuitively, but is there some formal property of natural numbers or anything else I can base that on? Do I even need to be that rigorous? $\endgroup$
    – m0meni
    Feb 22, 2017 at 15:59
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    $\begingroup$ Well you can solve the inequality: $a > \frac{1}{n^2}$ if and only if $n > \frac{1}{\sqrt{a}}$. Any $n$ satisfying this condition suffices (we do believe there are arbitrarily large natural numbers, right? ;) ) $\endgroup$ Feb 22, 2017 at 16:01
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    $\begingroup$ @AR7 That comes via archimedean property. $\endgroup$
    – Error 404
    Feb 22, 2017 at 16:01

4 Answers 4

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Maybe this can help. Let us write $$A=\bigg\{\frac{1}{n^2}:n\in\Bbb N\bigg\}.$$ Clearly, $0$ is a lower bound of $A$. So, the set $A$ is a non-empty set bounded below. Thus, we can find a real number $w$ such that $$\inf A=w.$$ Then, $$w\geq 0.$$ Let $\epsilon >0$. Then, using the Archimedean Property, we can find $n\in\Bbb N$ such that $$\frac{1}{n}<\sqrt{\epsilon}.$$ Since $w$ is a lower bound of $A$ and $\frac{1}{n^2}\in A$, we get $$w\leq\frac{1}{n^2}<\epsilon.$$ Hence, $$w< \epsilon\quad \forall \epsilon>0.$$ Thus, $$w\leq 0$$ and hence $$w=0.$$

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We already know that $0<\frac{1}{n^2}$ for any $n\in\mathbb{N}$. By the definition of infimum you only need to show that $0$ is the biggest number with this property.

Assume it is not true, so there exists $x>0$ such that $x < \frac{1}{n^2}$ for any $n\in\mathbb{N}$. Then $xn^2<1$ for any $n\in\mathbb{N}$.

Now without a loss of generality we may assume that $x<1$. Otherwise $n=1$ would contradict the inequality. Put $n := \lceil\frac{1}{x}\rceil$ (here $\lceil\cdot\rceil$ denotes the ceiling function). Note that $n\in\mathbb{N}$ and furthermore

$$xn^2=x\bigg\lceil\frac{1}{x}\bigg\rceil^2 \geq x\bigg(\frac{1}{x}\bigg)^2=\frac{1}{x}>1$$

Last inequality since $x<1$. Contradiction. $\Box$.

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    $\begingroup$ It works, but I would say it's easier to just say that for all $n \geq 1/\sqrt{x}$ the inequality is not satisfied. No need to use floor functions. $\endgroup$
    – TMM
    Feb 22, 2017 at 16:05
  • $\begingroup$ @TMM Right, so the floor function is simply to give explicite construction of $n$, without saying "there exists a natural bigger then $\frac{1}{\sqrt{x}}$". $\endgroup$
    – freakish
    Feb 22, 2017 at 16:09
  • $\begingroup$ "The first integer bigger than $1/\sqrt{x}$" is also an explicit construction of $n$, again avoiding the floor function. $\endgroup$
    – TMM
    Feb 22, 2017 at 20:42
  • $\begingroup$ @TMM But that's exactly what I did. You just have to recall how the floor function is defined. Well, almost, I could use the ceiling function instead, but for nonintegers $\lceil r \rceil=\lfloor r\rfloor + 1$. Yeah, ceiling will be better, more readable. Fixed. $\endgroup$
    – freakish
    Feb 23, 2017 at 8:37
  • $\begingroup$ I know what the floor and ceiling functions are; I'm just saying it is unnecessary to introduce new notation. And I still don't get why you're setting $n = 1/x$ instead of $n^2 = 1/x$ - sure, you could also set $n = 712/x^2$, but that's also not the most natural, tightest choice. $\endgroup$
    – TMM
    Feb 23, 2017 at 11:17
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Hint:

prove it indirectly, assume there is a bigger lower bound: $\epsilon >0$.

Now what can you say about $\frac{1}{n^2}$ where $n=\lfloor\sqrt{1/\epsilon}\rfloor+1$?

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  • $\begingroup$ Why down vote tho? $\endgroup$
    – Shashi
    Sep 3, 2021 at 11:06
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There is no inf that is in the set, because all values are positive but for any value, there is a smaller one.

However, there is a lim inf, and it is zero, because, for any $c> 0$, all values beyond a $n$ that depends on $c$ are within $c$ of zero.

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    $\begingroup$ Infimum of a set can be outside of the set. $\endgroup$
    – Error 404
    Feb 22, 2017 at 15:54
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    $\begingroup$ There is an inf since the set is bounded from below. $\endgroup$ Feb 22, 2017 at 15:54
  • $\begingroup$ I edited it to reflect what I meant. $\endgroup$ Feb 22, 2017 at 15:57
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    $\begingroup$ Inf has a very widely accepted definition you aren't using. $\endgroup$
    – user223391
    Feb 22, 2017 at 15:59

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