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Suppose we have a set $S$ that has an upper bound $a$. This means $\forall x\in S$ $\exists a$ s.t. $x \leq a$. The opposite of this is saying $a$ is not an upper bound of $S$. $\exists x \in S $ s.t. $a < x$

Is this correct? My understanding was if you want to negate a statement $\forall $ <--> $\exists$ are swapped. But in my negation I only modified one of them. What is the correct way to write "a is not an upper bound of S"?

$S$ is any set of real numbers.

user1068636
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1 Answers1

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Let $ A $ be a set and $ S $ be a subset of $ A $.($ S\subseteq A $)

Then $ a\in A $ is said to be an upper bound for $ S $ if $ \forall x\in S $, $ x\leq a $.

Hence if $ a\in A $ is not an upper bound for $ S $ then $ \exists x_{0}\in S $ such that $ x_{0}>a $. This is the contraposition of the above statement.

Also if $ S $ has an upper bound in $ A $ then $ \exists a\in A $ such that $ \forall x\in S $, $ x\leq a $.

The negation of "$ \exists a\in A $ such that $ \forall x\in S $, $ x\leq a $" is "$ \forall a\in A $, $ \exists x_{0}\in S $ such that $ x_{0}>a $."

Therefore if $ \forall a\in A $, $ \exists x_{0}\in S $ such that $ x_{0}>a $ then we say that $ S $ has no upper bound in $ A $.

ASB
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  • In my proof $A$ is a subset of $\mathbb{R}$ and $\leq$ is the partial ordering. – ASB Feb 04 '15 at 08:11