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Let $A\subset\mathbb{R}$ be an upper bounded set. Then

$$\forall\varepsilon>0~\exists x\in A\text{ such that }\sup{A}-\varepsilon< x \leq \sup A$$ I want to negate that statement. Would it be: $$ \exists \varepsilon>0~\forall x\in A\text{ such that } \sup A-\varepsilon\geq x\text{ or }x>\sup A~~?$$

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    $\begingroup$ In the most synthetic form I believe this is what you are looking for: $$x \in A: \sup A-\epsilon \ge x\Rightarrow \epsilon \gt 0$$ $\endgroup$
    – WindSoul
    Mar 24, 2023 at 18:34
  • $\begingroup$ @WindSoul I think this needs a quantifier to bind $x$ . $\endgroup$
    – mcd
    Mar 24, 2023 at 18:40

2 Answers 2

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The logic is essentially correct, but it is oddly phrased: I suggest $$ \exists \varepsilon>0\text{ such that }\forall x\in A \text{ either }x \leq \sup A-\varepsilon \text{ or }x>\sup A~.$$ However, you don't really need the 'or', as the elements of $A$ cannot be larger than $\sup A$, hence the negation can be written simply as $$ \exists \varepsilon>0\text{ such that }\forall x\in A \; x \leq \sup A-\varepsilon~.$$

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  • $\begingroup$ as an element of A, x can’t be greater than sup A. $\endgroup$
    – WindSoul
    Mar 24, 2023 at 18:24
  • $\begingroup$ Technically, the second statement is mathematically (though not logically) equivalent to the first statement, which is the actual negation. $\endgroup$
    – ryang
    Mar 24, 2023 at 18:30
  • $\begingroup$ It makes no difference mathematically, as ryang phrased it, but "either ... or" implies exclusive disjunction which is not how conjunction is negated. I prefer OP's phrasing. $\endgroup$
    – Ennar
    Mar 24, 2023 at 18:34
  • $\begingroup$ @Ennar Surely in mathematics (and logic) "or" is usually taken to be inclusive unless the conjunction is explicitly excluded? $\endgroup$
    – mcd
    Mar 24, 2023 at 18:42
  • $\begingroup$ Yes, that's correct, but you didn't write "or", you wrote "either ... or". $\endgroup$
    – Ennar
    Mar 24, 2023 at 18:43
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$$(\forall) x \in A: \sup A-\epsilon \ge x\Rightarrow \epsilon \gt 0$$

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