I'm asked to tell what the supremum and maximum is of the following set: $$\begin{Bmatrix}n:n\in\mathbb{N}\end{Bmatrix}$$

I don't know that I should say $\text{sup}\begin{Bmatrix}n:n\in\mathbb{N}\end{Bmatrix}\overset{?}{=}\infty$ because that seems an abuse, but I'm not sure like I said.

  • $\begingroup$ Is it "is" or "are?" $\endgroup$ Aug 26, 2013 at 4:17
  • $\begingroup$ Look at the numbers, start listing them. The biggest is $1/2$. And next you may be asked about min and infimum. No min, infimum $0$. $\endgroup$ Aug 26, 2013 at 4:19
  • $\begingroup$ Does the definition of the natural numbers you need to work with include 0 or not? If it does, you may be correct about $\infty$, otherwise, you have a strictly sequence, so you have a well-defined supremum which should be the maximal element too. $\endgroup$
    – Avraham
    Aug 26, 2013 at 4:20
  • $\begingroup$ No, the author doesn't acknowledge $0$. $\endgroup$ Aug 26, 2013 at 4:25
  • 2
    $\begingroup$ @Avraham: You are wrong. Even if $0$ was taken as part of the natural numbers in this context, the supremum isn't going to be $\infty$. The reason is that $\frac10$ is not infinity; it is simply undefined. $\endgroup$
    – Asaf Karagila
    Aug 26, 2013 at 4:42

3 Answers 3


Try writing out the first few numbers:

$$\frac{1}{2}, \frac{1}{4}, \frac 1 6, ...$$

So the supremum and the maximum agree with each other, since there's a largest element, namely $1/2$.

Edit: For clarification, the question originally asked what the supremum and maximum of $\{\frac{1}{2n} : n \in \Bbb{N}\}$ are.

  • $\begingroup$ The OPs seems to be confused about $0\in\Bbb N$. $\endgroup$
    – Pedro
    Aug 26, 2013 at 4:25
  • 1
    $\begingroup$ What a blunder. Yes, I know. I copied the wrong set in. I meant $\{n:n\in\mathbb{N}\}$. $\endgroup$ Aug 26, 2013 at 4:25
  • $\begingroup$ @LoieBenedicte I see. In that case, again try writing out elements: Note that there are arbitrarily large elements, so the supremum must be infinity. $\endgroup$
    – user61527
    Aug 26, 2013 at 4:29
  • $\begingroup$ It's just that the definition of supremum doesn't seem to allow for $\infty$. I mean what does $n<\infty$ mean? $\endgroup$ Aug 26, 2013 at 4:30
  • 1
    $\begingroup$ @Avraham That's not entirely correct - for example, the interval $(0, 1)$ certainly has no largest element, but its supremum is $1$. $\endgroup$
    – user61527
    Aug 26, 2013 at 4:33

As I and others mentioned in the comments, if the author does not have 0 in his or her definition of the natural numbers, you have a strictly decreasing sequence of rational numbers with a well-defined maximal element and supremum at 1/2. You have no well defined minimal element, but you have an infimum at 0.


The idea of supremum and maximum come only for a bounded set. You are considering the set $\{n : n \in \mathbb{N}\} = \{1, 2, 3, \dots\}$. This is an unbounded set in $\mathbb{R}$, as for any positive real number $G$, you shall get an element $m \in \mathbb{N}$ s.t. $m > G$.

So Supremum of the set does not exists, and same for maximum.


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