# Negation of a statement with an inequality.

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: $$\exists \varepsilon>0~\forall x\in A\text{ such that } \sup A-\varepsilon\geq x\text{ or }x>\sup A~~?$$

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

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 $$\exists \varepsilon>0\text{ such that }\forall x\in A \; x \leq \sup A-\varepsilon~.$$

• as an element of A, x can’t be greater than sup A. Mar 24, 2023 at 18:24
• Technically, the second statement is mathematically (though not logically) equivalent to the first statement, which is the actual negation. Mar 24, 2023 at 18:30
• 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. Mar 24, 2023 at 18:34
• @Ennar Surely in mathematics (and logic) "or" is usually taken to be inclusive unless the conjunction is explicitly excluded?
– mcd
Mar 24, 2023 at 18:42
• Yes, that's correct, but you didn't write "or", you wrote "either ... or". Mar 24, 2023 at 18:43

$$(\forall) x \in A: \sup A-\epsilon \ge x\Rightarrow \epsilon \gt 0$$