# Supremum and infimum definition

I’m currently learning it from scratch and I think that I pretty got the general idea: Supremum is largest limit of a set/sequence and infimum is the smallest.

I’ve encountered the following definition and found it difficult to fully understand the second statement:

$S$ is a supremum of a sequence/set $A$ if and only if: 1. $$x \le S , \forall x \in A$$ And 2. $$\forall \epsilon > 0 , \exists X_0 \in A$$ So that $$X_0 > S- \epsilon$$

I managed to understand the definition of sequence and functions limits which involves epsilons and deltas also bust this one isn’t as clear to me as them, I’d be glad to get a better explanation. Thanks for helping :)

• It's unclear what you're asking. Could you precisely state the definition to which you'd like to compare the above definition? – Dzoooks Aug 18 '18 at 13:55
• @Dzoooks what wasn’t clear about it? – Ozk Aug 18 '18 at 14:01
• @Dzoooks the second part “2.l of the definition isn’t completely understandable to me – Ozk Aug 18 '18 at 14:02
• Compared to what? If this is the definition, then it's the definition. Would you like an intuitive explanation for why it's a good definition? Do you know another definition to which you'd like to compare this one? – Dzoooks Aug 18 '18 at 14:03
• @Dzoooks Oh, umm no I don’t know another one and yes I’d like to understand intuitively and rigorously (if possible) why it is true. I know it’s true because it’s a definition but want to know why – Ozk Aug 18 '18 at 14:07

• The upper limit of the sequence (the $\limsup$).
Now, the definition of the supremum that you think is unclear is actually a rephrasing of the standard one. The usual intuition is that $\sup A$ is the least upper bound, in the sense that it is an upper bound, and no number less than it is an upper bound. Formally, given a a set $A \subseteq \mathbb{R}$, we say that $x = \sup A$ iff $a \leq x$ for all $a \in A$ ($x$ is an upper bound of $A$), and $y < x$ implies that $y$ is not an upper bound of $A$.
Specifically, if $y$ is not an upper bound for $A$, then there is some $a \in A$ such that $y < a$. This directly establishes your second point. For all $\epsilon > 0$, we have $\sup A - \epsilon < \sup A$, so there must be some $a \in A$ with $$\sup A - \epsilon < a \leq \sup A.$$ This inequality finds use throughout real analysis.
The upper limit of a sequence $\{s_k\}$ is denoted $\limsup_{n \to \infty} s_n$. The concept is more technical, but closely related.