let $x\in\emptyset\subseteq\mathbb{R}$ why is $x\leq r,\forall r \in\mathbb{R}$ true

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 $$\infty$$?

• You have to consider the condition : "if $x \in \emptyset$, then $x \le r$", for an $r \in \mathbb R$ whatever. – Mauro ALLEGRANZA Jan 24 '19 at 16:33
• 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. – Mauro ALLEGRANZA Jan 24 '19 at 16:34

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.

• Why is it logical to say : a false premise implies all conclusions? – New2Math Jan 24 '19 at 19:59
• 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. – fleablood Jan 24 '19 at 20:08

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.