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I have had trouble proving that the infimum of the set $A = \{ n - \frac{1}{n} : n \in \mathbb{N} \} $ is zero.

It is easy to prove that 0 is a lower bound. What I am struggling with is proving the following statement:

Statement: For every positive $\epsilon$, there exists $a \in A$ such that $a < \epsilon + 0$ .

How can I find such an element in the set?

What I have tried is using the Archimedean property to find such an element, but I have not been successful:

Since $\epsilon > 0$ , by the Archimedean property, there exists $ n \in \mathbb{N} $ such that $ \frac{1}{n} < \epsilon$. Then, $ -\epsilon < -\frac{1}{n} $, so $ n - \epsilon < n - \frac{1}{n} $, which is not what I want.

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    $\begingroup$ Pick $a = 1-\frac 11$? $\endgroup$
    – peterwhy
    Commented Sep 9 at 4:17
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    $\begingroup$ The sequence is strictly increasing, so $n=1$ yields the smallest element. $\endgroup$
    – AlvinL
    Commented Sep 9 at 4:26
  • $\begingroup$ Oh of course, but if I had only asked: "For every positive real number, is there an element of the form $ n - \frac{1}{n} < \epsilon $? How can I find such an element? Zero always works, but a more optimal element, for example, for $\epsilon = 2$ , $ 2 - \frac{1}{2} $ works, and for $ \epsilon = 5 $ , $ 5 - \frac{1}{5} $ works. I need an element not as trivial as zero. Just out of curiosity (and without using convergence) $\endgroup$
    – C25
    Commented Sep 9 at 4:27
  • $\begingroup$ the only element that works for small enough $\varepsilon$ is zero, just plug in a few values of $n$ to see how the sequence behaves. The definition of infimum/supremum doesn't say anything about "optimality". We are required for every $\varepsilon$ to pick a suitable element, that's it. $\endgroup$
    – AlvinL
    Commented Sep 9 at 4:28
  • $\begingroup$ For a given positive $\epsilon$, this is an inequality to be solved: $$\begin{align*}n-\frac1n &< \epsilon\\ n^2 - \epsilon n - 1 &< 0 \end{align*}$$together with $n \in \mathbb N$. $\endgroup$
    – peterwhy
    Commented Sep 9 at 4:33

2 Answers 2

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$0$ is an element of the set, and is also the lower bound of the set, so it is the minimum of the set. If a set has a minimum, it is also always the infimum.

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As others said, you can always pick $1-\frac11=0$. In fact, if $n\neq1$, then $$n-\frac1n\ge2-\frac12=\frac32$$ So for any $\varepsilon<3/2$, you have to pick $n=1$.

Hope this helps. :)

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