Question: Is there any supremum and the infimum of the set

$$ \{x ∈ \mathbb{Q} \mid 1<x<\sqrt{5} \}$$

my answer is $\sup= \sqrt 5$, $\inf=1$.

Am I right ?

So $\mathbb{Q}$ in this question doesn't matter?

  • $\begingroup$ This question is unclear. Do you mean "does there exist an infimum or supremum"? $\endgroup$ Feb 27, 2017 at 17:24
  • $\begingroup$ $\inf = 1$ yes but $\sqrt{5} \notin \mathbb{Q}$, so the $\sup$ doesn't exist. $\endgroup$
    – Zain Patel
    Feb 27, 2017 at 17:27
  • 1
    $\begingroup$ I and Henning made some edits to your post. Please try to follow them so you know how to format questions properly here. $\endgroup$ Feb 27, 2017 at 17:28

1 Answer 1


It matters whether you consider your set as a subset of the ordered set $(\mathbb Q,{\le})$ or the ordered set $(\mathbb R,{\le})$.

In the former case there is no supremum (because the supremum of a subset has to be an element of the ordered set you're considering); in the latter case $\sqrt 5$ is correct.

Lesson to learn: the terms "supremum" and "infimum" depends not only on what the set is, but also on what you consider it as a subset of.

  • $\begingroup$ but it doesn't matter '<' or ' ≤ ' , right ? $\endgroup$
    – Karry
    Feb 27, 2017 at 17:37
  • $\begingroup$ @Karry: Yeah, that is just a question of which convention you're using for specifying an ordered set. $\endgroup$ Feb 27, 2017 at 17:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .