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Definition: We say that $X\subset \mathbb{R}$ is bounded above if $\exists C\in \mathbb{R}$ s.t. $\forall x\in X$ we have $x\leq C$.

$(X \ \text{is bounded above}):=\exists C\in \mathbb{R}((\forall x\in X)\Rightarrow (x\leq C))$

So I want to take the negation of it: $$\neg(X \ \text{is bounded above})\equiv \neg(\exists C\in \mathbb{R}((\forall x\in X)\Rightarrow (x\leq C)))\equiv$$ $$\equiv\forall C\in \mathbb{R} \neg((\forall x\in X)\Rightarrow (x\leq C))$$ But we know that $\neg (P\Rightarrow Q)\equiv P\land \neg Q$ which implies that $$\neg(X \ \text{is bounded above})\equiv \forall C\in \mathbb{R} ((\forall x\in X)\land (x> C)) $$

But I know that the correct negation should be the following: $X\subset \mathbb{R}$ is not bounded above if $\forall C\in \mathbb{R}$ $\exists x\in X$ s.t. $x>C$.

What am I doing wrong in the above?

RFZ
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2 Answers2

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Right in the begining on the second line. The big quantifier $\forall$ is not part of the implication. It is:

$\exists C\in R (\forall x\in X (x\leq C))$

Edit: Expression $(\forall x\in X) \rightarrow (x\leq C) $ isn't meaningful expression because $(\forall x\in X)$ has no logical value (it is not true-false statement) it is just quantification.

dmk
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  • Why it is not a part of the implication? What it wrong with this statement? I mean this one: $\exists C\in \mathbb{R}((\forall x\in X)\Rightarrow (x\leq C))$ – RFZ Jan 05 '21 at 00:34
  • Thanks! But I have not studied math logic. My knowledge is quite very poor on this subject. Could you give more details why my statement is false. – RFZ Jan 05 '21 at 00:38
  • You are building formulas (long expressions) from smaller atomic formulas. Those are just some statements that are either true or false ($2<3$, seven is even, whatev...) and you build bigger expressions. You can build using connectives. You have two atomic and you say they are both true or one implies the other...or you can build bigger blog like saying it is true for every number or there is number for which it is true. But you used $\forall x \in X$ as a first part of implication but this is void. You say for all x in X so: but you should say ´for all x in X this happens so:. – dmk Jan 05 '21 at 01:17
  • Sorry but I am a bit confused with your last sentence. – RFZ Jan 05 '21 at 01:25
  • I meant that I do not see any difference between "for all $x$ in $X$ so" and "for all $x$ in $X$ this happens so" – RFZ Jan 05 '21 at 01:28
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If we write definition as $$(\exists C\in \mathbb{R})(\forall x\in X)(x\leq C)\quad (1)$$ Then negation gives $$(\forall C\in \mathbb{R})(\exists x\in X)(x> C)\quad (2)$$ Now step by step: using, that $(\forall x\in X)Q(x)$ is same as $(\forall x)(x\in X \Rightarrow Q(x))$ and $(\exists x\in X)Q(x)$ is same as $(\exists x)(x\in X \land Q(x))$, definition $(1)$ can be decoded as $$(\exists C)\big(C\in \mathbb{R} \land (\forall x)(x\in X \Rightarrow x\leq C)\big)\quad (1)$$ Taking negation gives $$(\forall C)\big(C\notin \mathbb{R} \lor (\exists x)(x\in X \land x> C)\big)$$ which is $$(\forall C)\big(C\notin \mathbb{R} \lor (\exists x\in X)( x> C)\big)$$ $$(\forall C)\big(C\in \mathbb{R} \Rightarrow (\exists x\in X)( x> C)\big)$$ $$(\forall C\in \mathbb{R})(\exists x\in X)(x> C)\quad (2)$$

zkutch
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  • Thanks for your reply! I am reading it right now. But what is wrong with my reasoning? – RFZ Jan 05 '21 at 00:53
  • Your record for definition "$:=$" should be $(\exists C\in \mathbb{R})(\forall x) (x \in X \Rightarrow x\leq C)$ – zkutch Jan 05 '21 at 01:00
  • But it does not explain what is wrong with my record, right? Could you explain it, please? – RFZ Jan 05 '21 at 01:01
  • In my answer I brought definition of $(\forall x\in X)Q(x)$. Accordingly it your record $(\forall x\in X)\Rightarrow (x\leq C)$ formally should be written as $(\forall x)(x\in X \Rightarrow (\Rightarrow (x\leq C)))$, which, as you can see, have no sense. – zkutch Jan 05 '21 at 01:11