Definition for supremum Is the below definition for supremum correct? If no, then how to define it in similar way?
$$\sup E = s \Longleftrightarrow(\forall t, \; \forall x, \; x \le t \Longrightarrow x \le s \le t) $$
We suppose that $E \subseteq \mathbb R$; $x\in E$; $s,t \in \mathbb R$.
 A: No. $s$ is $\le$ any other upper bound of $E$. It is not $\le$ any number that happens to be above an element of $E$. For instance if $E = (0,1)$ and $x = 1/2$ then $s = 1$, and $1/2 < 3/4$, but $s \not \le 3/4$.
A: If by "similar way" you mean the definition should be written as a single logical formula, then we can write $$\textrm{sup}\:⁡E=s⟺\bigg(\big(∀x,\:x≤s\big)  ∧\big(\left(∃t\:∀x,\:x≤t\right)⟹s≤t\big)  \bigg),$$ where $E⊆\mathbb{R};\:x∈E;\: s,t∈\mathbb{R}$.
A: I believe you forgot an important part of the definition of the supremum, which is that it is the lowest upper-bound. For that, you need to have that there exists an element in E bigger or equal to sup E - k, for k>0.
A: Paul Sinclair already noted that your definition is faulty. I'll explain how the supremum is defined.
If there exists a number $s$ such that


*

*$s$ is an upper bound for $E$, that is, $\forall x \in E \implies s \geq x$;

*$s$ is the lowest upper bound, that is, $\forall \epsilon > 0, \exists x \in E\quad\text{s.t.}\quad s - \epsilon \lt x \leq s$;


we call it supremum of $E$ and write $\sup E = s$.
To understand the second point, think about it in these terms: what does it mean to be the lowest upper bound? It means that if we subtract a positive amount, however small, we instantly find ourselves inside $E$. This does not happen if we are considering $E = (0, 1)$ and the upper bound $2$, for example.
A: To be a supremum of $E$ (i.e. $s = \sup E$), 


*

*s must be a upper bound (i.e. $s \ge e, \forall e \in E$).

*s must be the smallest upper bound of $E$ (i.e. $s \le b, \forall b$ where $b$ is an upper bound of $E$).


Similarly, the infimum can be defined.
