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So I am trying to prove a proposition. It goes like this

Let there be $\emptyset\neq X\subset\mathbb{R}$ which is bounded from above. The next two statements are equivalent about $s\in\mathbb{R} $

1) i)$ x\leq s$ $\forall x\in X$

ii) If $x\leq u$ $ \forall x\in X$ then $s\leq u$

2) i)$x\leq s$ $\forall x\in X$

ii*)$\forall\epsilon>0$ $ \exists x\in X$ s.t $s-\epsilon<x\leq s$

So I am trying to prove $2\Rightarrow1$, in other words to prove $ii* \Rightarrow ii$. I wanted to prove it by contradiction but I couldn't figure out how to negate ii. As I know it, if we have two statements A and B with $A \Rightarrow B$ then the negation would be $\neg B \Rightarrow \neg A$.Here, for example A=If $u \leq x$ $\forall x \in X$ and B is $s \leq u$. But when I applied it, it really didn't make any sense. Maybe I have the wrong idea of negation. Can someone explain me where I am wrong with the negation? Thank you.

Charles Carmichael
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2 Answers2

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The negation of $A \rightarrow B$ is :

$\lnot (A \rightarrow B)$ --- (*).

To "move inside" the negation sign, we have to use the equivalence between $p \rightarrow q$ and $\lnot p \lor q$ and De Morgan : $\lnot(p \lor q)$ is equivalent to $\lnot p \land \lnot q$.

Thus (*) is :

$\lnot (\lnot A \lor B)$ i.e. $A \land \lnot B$.

Mauro ALLEGRANZA
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If ii) is not true then some $u$ must exist with $x\leq u$ for each $x\in X$ and secondly $u<s$.

Taking $\varepsilon=s-u>0$ we then find that ii*) cannot be true.

Proved is now that ii*) implies ii).

drhab
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