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 :)
 A: We should clear some things up. You talk about sequences and sets, which are different things. "The supremum" of a sequence could be interpreted in two different ways:


*

*The supremum of the image of the sequence (the set of values); or

*The upper limit of the sequence (the $\limsup$).


You mean the first, so let's talk about that.
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.
