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It might look quite stupid, but I had become little confused when understanding empty functions. Anyway, my question is,

If there is a statement $P(x)$ starting with "for $\forall x\in A$,..." and $A$ is an empty set, should I understand this as because the assumption is false, the conclusion is absolutely true? If not, how should I?

Well, the place where I got stuck was this: For every set X, there exists a unique empty function $f : \emptyset \rightarrow X$. To prove this I should set two empty functions $f_1, f_2$, and show that $\forall x\in \emptyset$, $f_1(x)=f_2(x)$. When thinking as I stated above, since the assumption is false, the conclusion is true. But instead if we think about a statement $\forall x\in \emptyset$, $f_1(x)\neq f_2(x)$, this may be also true....(?)


marked as duplicate by GNUSupporter 8964民主女神 地下教會, Claude Leibovici, mrp, Asaf Karagila elementary-set-theory Mar 21 '17 at 9:54

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  • $\begingroup$ there is no element: there is no x∈A. Nothing can still get mapped (to nothing) and it still fits the definition of a function. $\endgroup$ – user29418 Mar 21 '17 at 0:35
  • 2
    $\begingroup$ $\forall x\in\varnothing\ P(x)$ is automatically true, because there is no $x\in \varnothing$ for which $P(x)$ is false. $\endgroup$ – Aweygan Mar 21 '17 at 0:35
  • $\begingroup$ Why don't you tell us the entire statement as it will determine what the writer was trying to do. $\endgroup$ – Sentinel135 Mar 21 '17 at 0:38
  • $\begingroup$ Isn't this something like a vacuuous truth $\endgroup$ – theideasmith Mar 21 '17 at 0:38
  • $\begingroup$ Another way of seeing statement about "elements" of the empty set is that the statement's negation is not false. E.g. I assert that for any $x$ in the empty set, we have $x>0.$ Since you cannot find a counter example my statement is true. $\endgroup$ – Behnam Esmayli Mar 21 '17 at 0:38

Imagine this:

Everytime I have played the lottery I won the jackpot!

Why is this true? Am I the luckiest person on the planet? No. I just have never played the lottery.

That is: the set of all times T that I played the lottery is empty ... which is exactly why the claim $\forall t \in T: Jackpot!(t)$ is true.

And yes, unfortunately it is also true that $\forall t \in T: \neg Jackpot!(t)$

  • $\begingroup$ Yes the last truth is the thing that made me confused... $\endgroup$ – H. Jung Mar 21 '17 at 1:01
  • $\begingroup$ OK. Suppose I say that all flying pigs are blue. That is true, since there are no flying pigs. That is, all zero of them are blue. Put differently: there is no flying pig that is not blue. But note the statement all flying pigs are red is also true ... because all zero of them are red. Again, there is no flying pig that is not red ... exactly because there are no flying pigs. $\endgroup$ – Bram28 Mar 21 '17 at 1:06

A statement of the form

For all $x \in A$, it holds that $P(x)$
$\forall x \in A : P(x)$

can be re-formulated to the logically equivalent statement

There is no $x \in A$ such that $P(x)$ does not hold
$\neg \exists x \in A : \neg P(x)$

In order to show that a claim is true for all elements of a set, it is equally legitimate to show that there is no element for which it does not hold.

When applying this reasoning to, e.g., the formula

All green elephants talk
$\forall x \in E: Talk(x)$

where $E$ is the set of green elephants,

we can as equivalently word it as

There is no green elephant who does not talk
$\neg \exists x \in E: \neg Talk(x)$

Obviously, if there are no green elephants at all, there will be no elephant who does not talk. Therefore, since we found no counterexample to the claim, we can - a bit unintuitively, but logically correct in classical logic - draw the conclusion that all elephants do talk.

The pattern becomes a bit more obvious when formulating the sentence with material implication:

For all $x$ it holds that if it is a green elehant, it talks
$\forall x (GreenElph(x) \to Talk(x))$

This formula becomes true if and only if all entities in the model's domain that we map the variable $x$ to make the inner statement true, or conversely, that there is no entity for which the statement is false.
When recalling the truth table of material implication, we know that the only way for a formula of the form $\phi \to \psi$ become false is when $\phi$ is true and $\psi$ is false, or conversely, if $\phi$ is fase, the formula will become true no matter whether $\psi$ is true or false.
Now if there are no entities which are green elephants, there is no instance which makes the left side of the implication ($GreenElph(x)$) true. Thus, regardless of whether this thing talks or not, the implication becomes true. So there is no $x$ such that $GreenElph(x)$ is true and $Talk(x)$ is false, so the implication becomes true for all entities in the domain, and in consequence, the universally quantified formula is true.

This phenomena of a universally quantified formula being satisfied due to the antecedent being false for any entity (i.e. the set that satisfies the restriction is the empty set) is also called vacuous truth.


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