'A set S is bounded' is defined by $\exists M >0, \forall x \in S : |x| \leq M$.

I know that proof that empty set is bounded.

$$\forall x \in \varnothing \Rightarrow \exists M >0, \forall x \in \varnothing : |x| \leq M$$

The assumption that $\forall x \in \varnothing$ is false, so it is true without reference to the conclusion.

I have a question that is it also true by the same way that empty set is not bounded?

Clearly, 'A set S is not bounded' is defined by $\forall M >0, \exists x \in S : |x| > M$.

Now by the same way

$$\forall x \in \varnothing \Rightarrow \forall M >0, \exists x \in \varnothing : |x| > M. $$

This assumption is also false so that it is true whatever the conclusion is.

However, I heard that it is not true. What is wrong in my proof?

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    $\begingroup$ $\forall x\in\emptyset\Rightarrow\ldots$" does not really make sense, syntactically $\endgroup$ – Hagen von Eitzen Aug 12 '15 at 20:39
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    $\begingroup$ Logical symbols are like $\forall$ are useful to those who are doing complicated proofs where the statements are hard to read in plain language. They still represent sentences in plain language and none of your proofs make any sense. Your first one directly translates to "For all $x$ in the empty set therefore there exists $M$ greater than zero for all $x$ in the empty set such that $|x|$ is less than $M$". This is nonsense. $\endgroup$ – PVAL-inactive Aug 12 '15 at 20:46
  • $\begingroup$ I think you mean that you believe that the empty set is bounded (when viewed as a subset of some ordered set), but you have heard that this is not true? If so, then you are right, but the details of your proof need to be tidied up a bit as other comments and answers are trying to explain. $\endgroup$ – Rob Arthan Aug 12 '15 at 20:48
  • $\begingroup$ To avoid unnecessary troubles with syntax one can apply an equivalent definition to bounded in a metric space: a set is bounded if it is a subset of some ball. Clearly, the empty set is a subset of any ball, thus, bounded. It is, in fact, even compact. $\endgroup$ – A.Γ. Aug 12 '15 at 20:53
  • $\begingroup$ @A.G.: (and nearly everyone else): so the OP is right! What he or she has "heard" is wrong. $\endgroup$ – Rob Arthan Aug 12 '15 at 20:56

"$\forall x\in\emptyset$" isn't actually a logical proposition, so it can't imply anything. A correct formulation of "bounded" would be: $$\exists M : \forall x \in \emptyset, \vert x\vert \leq M$$ which $\emptyset$ still passes, because it's vacuously true for all $x$. However, the correct formulation of "not bounded" is: $$\forall M>0, \exists x \in\emptyset : \vert x\vert > M$$ Which doesn't hold, because such an $x$ doesn't exist (as no $x$ exists in $\emptyset$).

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  • $\begingroup$ Why commas and not colons? $\endgroup$ – Carsten S Aug 12 '15 at 20:48
  • $\begingroup$ I'm not super confident I use commas and colons correctly, but my logic is to treat the colons as "such that", e.g., "There exists M such that...". But in the latter case, I want to say "For any M greater than 0, there exists ...". It doesn't make sense to say "such that", so I just put a comma instead. $\endgroup$ – Sam Jaques Aug 12 '15 at 20:56
  • $\begingroup$ I see, for linguistic reasons you treat existential and universal quantification differently. I do no think that that is a good idea, but it is consistent. $\endgroup$ – Carsten S Aug 12 '15 at 21:01
  • $\begingroup$ The commas and colons are completely optional; in most formal logic texts they are omitted completely. But there is no harm in including them. $\endgroup$ – Carl Mummert Aug 12 '15 at 21:50
  • $\begingroup$ I read again and again but I can't understand why the first is vacously true and the second is not. What exactly does it mean that comma and colon in predicate logic? Does it mean like $\wedge$? I want more predicate logical stage about this. $\endgroup$ – cokecokecoke Aug 13 '15 at 5:46

The issue is the existential quantifier (in the second statement): There does not exist $x\in\emptyset$ such that $|x|>1$ (using $1$ as a sample bound). Thus the empty set does have $-1$ and $1$ as lower and upper bounds respectively.

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