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?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ This question is unclear. Do you mean "does there exist an infimum or supremum"? $\endgroup$– Vishal GuptaFeb 27, 2017 at 17:24
-
$\begingroup$ $\inf = 1$ yes but $\sqrt{5} \notin \mathbb{Q}$, so the $\sup$ doesn't exist. $\endgroup$– Zain PatelFeb 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. $\endgroup$– Vishal GuptaFeb 27, 2017 at 17:28
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Feb 27, 2017 at 17:27
hmakholm left over Monicahmakholm left over Monica
282k23 gold badges418 silver badges678 bronze badges
-
$\begingroup$ but it doesn't matter '<' or ' ≤ ' , right ? $\endgroup$– KarryFeb 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