Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 8707

What property causes the supremum to be adherent point for $\mathbb{R}$? [closed]

$
0
0

In general, supremum is defined as the least upper bound. In the case of $\mathbb{R}$, we have an alternative characterization that $\text{sup} S $ is the supremum of a set $S$ if and only if for all epsilon, there exist an $s \in S$ so that $ \text{sup} S -\epsilon< s$. Theorem-1

I am trying to figure out which property of $\mathbb{R}$ allows us to show the alternative characterization of supremum. I know that it can't directly be out of the supremums axiom since $\mathbb{N}$ and $\mathbb{Z}$ also satisfy this but do not have this property *.


Viewing all articles
Browse latest Browse all 8707

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>