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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 $E \subseteq \mathbb R$; $x\in E$; $s,t \in \mathbb R$.

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5 Answers 5

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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 $s \not \le 3/4$.

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  • $\begingroup$ Oh, thanks! I've made a stupid mistake. $\endgroup$ Oct 24, 2015 at 10:23
  • $\begingroup$ Stupid mistakes are the best mistakes - they are the easiest to correct! $\endgroup$ Oct 24, 2015 at 16:24
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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 $E⊆\mathbb{R};\:x∈E;\: s,t∈\mathbb{R}$.

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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.

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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.

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  • $\begingroup$ I don't think it's necessary to assume the existence of a minus operator to define supremum. $\endgroup$
    – Jan Stout
    Oct 24, 2015 at 6:38
  • $\begingroup$ @JanStout You are right, but since OP seemed to be working over $\mathbb R$ I adapted the definition. This one is also very easy to understand. $\endgroup$
    – rubik
    Oct 24, 2015 at 6:46
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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.

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