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The emtyset has no element one can compare it with so why the statement is true can somebody explain the logic of this. I ask because I have seen a post where one claims that $sup(\emptyset)=-\infty$, the argument was that the above statement is always true and if there would exist a $r\in\mathbb{R}$ then one can always find a smaller upperbound. Id on't see why this argument is always true and hope somebody can explain it to me.

For reference

Why is the supremum of the empty set $-\infty$ and the infimum $\infty$?

In particular the answer of Clive Newstead

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    $\begingroup$ You have to consider the condition : "if $x \in \emptyset$, then $x \le r$", for an $r \in \mathbb R$ whatever. $\endgroup$ – Mauro ALLEGRANZA Jan 24 '19 at 16:33
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    $\begingroup$ Due to the fact that $x \in \emptyset$ is always false the above condition is always true, for every $r \in \mathbb R$, whatever small $r$ is. $\endgroup$ – Mauro ALLEGRANZA Jan 24 '19 at 16:34
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Three explanations: (But they all involve that as there are no $x \in \emptyset$ we can not negate $x \le r$ for any $r \in \mathbb R$)

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Either $x\in \emptyset \implies x \le r :\forall r \in \mathbb R$ is true or it is false.

If it is false then that implies that there is an $x \in \emptyset$ and an $r \in \mathbb R$ so that $x > r$. That is impossible as there are no $x \in \emptyset$.

If it is true then for every $x \in \emptyset$ then ... something. We can never test this because we can never find an $x \in \emptyset$.

However we can do logic the statement $P \implies Q$ will be true if $P$ and $Q$ are both true, or if $P$ is false. It will only be false if $P$ is true and $Q$ is false. If $P= : x \in \emptyset$ and $Q = : x \le r :\forall r \in \mathbb R$ then $P$ is always false. And $Q$ can never be true for any actuall $x \in \mathbb R$. So $P \implies Q$ is true.

Also a statement is equivalent to the contrapostive. The contrapositive of $x\in \emptyset \implies x \le r:\forall r \in \mathbb R$ is:

$\exists r\in \mathbb R: r < x \implies x\not \in \emptyset$. Well that is certainly true! If $r < x$ then $x$ must exist. And if $x$ exists then $x \not \in \emptyset$!.

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If $x\in \emptyset$ then $x$ is a green cheese eating alien who is the reincarnation of Elvis Presley is true because logically a false premise implies all conclusions.

So if $x \in \emptyset$ then $x \le r$ for all $r \in \mathbb R$. (it's also true that $x >r$ fora all $r \in \mathbb R$ and that $x = r$ for all $r \in \mathbb R$ and so on.... as $x$ does not exist we can say anything we want to be true about it.)

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Or looking at it another way: If $A_r = (-\infty, r)$ then $\emptyset \subset A_r$ because the empty set is a subset of all sets. (The emptyset has no elements, so no elements aren't in any set.) So if $x \in \emptyset \implies x \in A_r$.

That is true for all $A_r: r\in \mathbb R$. So $x \in (-\infty, r)$ for all $r\in \mathbb R$ so $x \le r$ for all $r \in \mathbb R$.

You might say "there's got to be a trick in there somewhere; nothing can be in all of those intervals" and you'd be right. The trick is that no such $x\in \emptyset$ does exist. But if such an $x$ DID exist, it would have to be in every set including every interval including those intervals.

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  • $\begingroup$ Why is it logical to say : a false premise implies all conclusions? $\endgroup$ – New2Math Jan 24 '19 at 19:59
  • $\begingroup$ because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true. $\endgroup$ – fleablood Jan 24 '19 at 20:08
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Given $r$, every element of the empty set is at most $r$. This statement is true because its negation is false. Namely, its negation is that there exists $r$ and an element of the empty set that is at greater than $r$. But the empty set has no elements.

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